Differential Equations Tutorial

Chris L T521 · #1 $Old$ May 13th, 2008, 01:12 AM

Once in a while, differential equation questions pop up, so I'm going to point out the various techniques on how to solve them:

1. Direct Integration:

If you have a differential equation in the form $\frac{dy}{dx}=f(x)$ , we can use direct integration to solve the DE.

Example 1:

Solve $\frac{dy}{dx}=3x^2$ .

To solve, simply integrate both sides of the equation:

$\int\frac{dy}{dx}\,dx=\int 3x^2\,dx \implies y+C_1=x^3+C_2 \implies \color{red}\boxed{y=x^3+C}$ . Note that we can combine the two constants into a new constant C.

Also, we may encounter differential equations with given conditions. These types of differential equations are called initial value problems (IVP). When solving a DE without conditions, we always find the General Solution to the DE. When an initial condition is applied, then we are finding a Particular Solution. Let's go through a quick example.

Example 2:

Solve $\frac{dy}{dx}=xe^{-x}$ ; $y(0)=0$ .

Directly integrate the DE:

$\int\frac{dy}{dx}\,dx=\int xe^{-x}\,dx$ .

We need to apply integration by parts to the integral on the right side.

$\int xe^{-x}\,dx$
let $u=x$ and $dv=e^{-x}$ . $\therefore \,du=\,dx$ and $v=-e^{-x}$ .

$\therefore \int xe^{-x}\,dx=-xe^{-x}+\int e^{-x}\,dx=-xe^{-x}-e^{-x}+C=-e^{-x}(x+1)+C$ .

Now apply the initial condition $y(0)=1$ .

$1=-e^{0}(0+1)+C \implies C=2$ .

Thus, our particular solution is : $\color{red}\boxed{y=-e^{-x}(x+1)+2}$ .

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2. Separation of Variables

Another technique in solving differential equations is separation of variables. As the name suggests, we "separate" one variable from another in order to find a solution. Some of these are very straight forward, whereas some of the DE's require some thought. I will go through an easier example, and then a harder one:

Example 3:

Solve $\frac{dy}{dx}=4x^3y-y$ ; $y(1)=-3$ .

First, factor out a y and then separate the variables.

$\frac{dy}{dx}={\left(4x^3-1\right)}y \implies \frac{dy}{y}={\left(4x^3-1\right)}\,dx$ .

Integrate both sides and solve for y.

$\int\frac{dy}{y}=\int{\left(4x^3-1\right)}\,dx \implies ln{\left|y\right|}=x^4-x+C \implies {\left|y\right|}=e^{C}e^{x^4-x} \implies y=Ce^{x^4-x}$ .

Apply the initial condition $y(1)=-3$ .

$-3=Ce^{0} \implies C=-3$ .

$\therefore \color{red}\boxed{y=-3e^{x^4-x}}$ .

Example 4:

Solve $x^2\frac{dy}{dx}=1-x^2+y^2-x^2y^2$ .

Factor the right hand side of the equation.

$x^2\frac{dy}{dx}=1-x^2+(1-x^2)y^2 \implies x^2\frac{dy}{dx}=(1-x^2)(1+y^2)$ .

Separate the variables and integrate.

$\frac{dy}{1+y^2}=\frac{1-x^2}{x^2}\,dx$

$\int \frac{dy}{1+y^2}= \int\frac{1-x^2}{x^2}\,dx$
$tan^{-1}y=-\frac{1}{x}-x+C$ .

Solve for y.

$\color{red}\boxed{y=tan{\left(-\frac{1}{x}-x+C\right)}}$ .

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I will post more later on today...after I sleep

Chris L T521 · #2 $Old$ May 13th, 2008, 12:30 PM

3. The Integrating Factor

The method of the integrating factor is used when we have differential equations in the form $\frac{dy}{dx}+P(x)y=Q(x)$ . Multiplying the equation through by the integrating factor $e^{\int P(x)\,dx}$ , we would have the equation ${\left[e^{\int P(x)\,dx}y\right]}^{/}=Q(x)e^{\int P(x)\,dx}$ . Integrating both sides and solving for y, we get:

$y=e^{-\int P(x)\,dx}{\left[\int Q(x)e^{\int P(x)\,dx}\,dx\right]}$ .

Let us go through an easy example, and then a challenging one.

Example 5:

Solve $x\frac{dy}{dx}+y=3xy$ ; $y(1)=0$ .

$x\frac{dy}{dx}+y=3xy \implies x\frac{dy}{dx}+(1-3x)y=0$

In order to apply the integrating factor, the coefficient of $\frac{dy}{dx}$ must be equal to 1.

$x\frac{dy}{dx}+(1-3x)y=0 \implies \frac{dy}{dx}+{\left(\frac{1}{x}-3\right)}y=0$ .

Now find the integrating factor:

$\rho(x)=e^{\int P(x)\,dx}=e^{\int {\left(\frac{1}{x}-3\right)}\,dx}=e^{lnx-3x}=xe^{-3x}$ .

Multiplying through, we should get:

${\left[xe^{-3x}y\right]}^{/}=0$

Integrating, we find that:

$xe^{-3x}y=C$

Imposing the initial condition $y\left(1\right)=0$ , we see that $e^{-3}\cdot0=C\implies C=0$

Therefore, the solution to the differential equation is $y=0\cdot x^{-1}e^{3x}\implies \color{red}\boxed{y \equiv 0}$ .

Example 6:

Solve $(x^2+1)\frac{dy}{dx}+3x^3y=6xe^{-\frac{3}{2}x^2}$ ; $y(0)=1$ .

Divide through by $x^2+1$ :

$\frac{dy}{dx}+\frac{3x^3}{x^2+1}y=\frac{6xe^{-\frac{3}{2}x^2}}{x^2+1}$ .

Now find the integrating factor:

$\rho(x)=e^{\int P(x)\,dx}=e^{\int \frac{3x^3}{x^2+1}\,dx}$

Apply long division to simplify the integrand:

(Verify): $\rho(x)=e^{\int{\left(3x-\frac{3x}{x^2+1}\right)}\,dx}=e^{\frac{3}{2}x^2-\frac{3}{2}ln(x^2+1)}=(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}$ .

Multiplying through by the integrating factor, we should get:

(Verify): ${\left[(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}y\right]}^{/}=\frac{6x}{(x^2+1)^{\frac{5}{2}}}$

Integrating both sides and then solving for y, we get:

$(x^2+1)^{-\frac{3}{2}}e^{\frac{3}{2}x^2}y=-\frac{2}{(x^2+1)^{\frac{5}{2}}}+C \implies y=e^{-\frac{3}{2}x^2}{\left(C(x^2+1)^{\frac{3}{2}}-2\right)}$ .

Now apply the initial condition: y(0)=1:

$1=C-2 \implies C=3$ .

Therefore, our particular solution will be:

$\color{red}\boxed{y=e^{-\frac{3}{2}x^2}{\left(3(x^2+1)^{\frac{3}{2}}-2\right)}}$ .

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4. Exact Equations

In order to use the technique to solve exact equations, the equations must be in the form:

$M(x,y)dx+N(x,y)dy=0$ ,

And they must satisfy this one condition:

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

If this relationship is true, we'll continue on with this technique. If its not true, we will resort to 2 other possible techniques which will be discussed later.

When we go about solving this, we should make known that $\frac{\partial f}{\partial x}=M(x,y)$ and that $\frac{\partial f}{\partial y}=N(x,y)$ .

Step one: find f(x,y). You can do it two ways, but I will do it this way because its the most common way.

$\frac{\partial f}{\partial x}=M(x,y) \implies \int\frac{\partial f}{\partial x}\,dx=\int M(x,y)\,dx \implies f(x,y)=\int M(x,y)\,dx + g(y)$

Step 2: Find g(y). To do this, partially differentiate f(x,y) with respect to y.

$\frac{\partial f}{\partial y}=\frac{\partial}{\partial y}\int M(x,y)\,dx+g^{/}(y)$

Since $N(x,y)=\frac{\partial f}{\partial y}$ ,

$N(x,y)=\frac{\partial}{\partial y}\int M(x,y)\,dx+g^{/}(y)$

Solving for $g^{/}(y)$ , we get:

$N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx=g^{/}(y)$ .

Integrate to find g(y):

$g(y)=\int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy$ .

Step 3: write solution in general form.

The general solution of an exact equation will have the form:

$f(x,y)=C$ .

Since $f(x,y)=\int M(x,y)\,dx + \int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy$ , the general solution will be:

$\int M(x,y)\,dx + \int{\left(N(x,y)-\frac{\partial}{\partial y}\int M(x,y)\,dx\right)}\,dy=C$ .

Example 7:

Solve $(cosx+lny)dx+{\left(\frac{x}{y}+e^y\right)}dy=0$ .

$M(x,y)=cosx+lny$
$N(x,y)=\frac{x}{y}+e^y$

Test for exactness:

$\frac{\partial M}{\partial y}=\color{red}\frac{1}{y}$
$\frac{\partial N}{\partial x}=\color{red}\frac{1}{y}$

They are equal, so they are exact equations.

Find f(x,y):

$f(x,y)=\int (cosx+lny)\,dx \implies f(x,y)=sinx+xlny+g(y)$ .

Now find g(y):

$\frac{\partial f}{\partial y}=\frac{x}{y}+g^{/}(y)$ .

Since $N(x,y)=\frac{\partial f}{\partial y}$ ,

$\frac{x}{y}+e^y=\frac{x}{y}+g^{/}(y) \implies g^{/}(y)=e^y \implies g(y)=e^y$

Therefore,

$f(x,y)=sinx+xlny+e^y$ .

Therefore, the general solution is:

$\color{red}\boxed{sinx+xlny+e^y=C}$ .

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I will be back later with more...

Chris L T521 · #3 $Old$ May 13th, 2008, 07:08 PM

5. Bernoulli Equations

A differential equation that has the form $\frac{dy}{dx}+P(x)y=Q(x)y^n$ is known as a Bernoulli's Equation. When n=0 or n=1, the equation is linear. However, when n>1, we make a substitution $v=y^{n-1}$ , which then transforms it into a linear DE of the form : $\frac{dv}{dx}+(1-n)P(x)v=(1-n)Q(x)$ .

Example 8:

Solve $x\frac{dy}{dx}+6y=3xy^{\frac{4}{3}}$ .

This can easily be recognized as a Bernoulli's Equation where $n=\frac{4}{3}$ . $\therefore 1-n=-\frac{1}{3}$ .

Make the substitution $v=y^{-\frac{1}{3}} \implies y=v^{-3}$ .

Find $\frac{dy}{dx}$ .

$\frac{dy}{dx}=\frac{dy}{dv}\frac{dv}{dx}=-3v^{-4}\frac{dv}{dx}$

Substituting these values into the differential equation, we get:

$-3xv^{-4}\frac{dv}{dx}+6v^{-3}=3xv^{-4} \implies \frac{dv}{dx}-\frac{2}{x}v=-1$ . We now have a linear DE, which we already know how to solve.

Find the integrating factor:

$\rho(x)=e^{\int P(x)\,dx}=e^{-2\int \frac{\,dx}{x}}=x^{-2}$ .

Multiply throughout by the integrating factor, and then simplify:

${\left[x^{-2}v\right]}^{/}=-\frac{1}{x^2} \implies x^{-2}v=\frac{1}{x}+C \implies v=x+Cx^2$ .

We don't want to know what v is. We want to know what y is. Since $v=y^{-\frac{1}{3}}$ , we see that:

$y^{-\frac{1}{3}}=x+Cx^2 \implies \color{red}\boxed{y=\frac{1}{(x+Cx^2)^3}}$

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Second Order Differential Equations

A homogeneous second order differential equation has the form :

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=0$ .

If we assume that a solution has the form $y=e^{rx}$ , then the differential equation becomes:

$ar^2e^{rx}+bre^{rx}+ce^{rx}=0$ . Knowing that $e^{rx}>0$ , we can divide both sides by $e^{rx}$ , which gives us:

$ar^2+br+c=0$ .

The equation above is know as the characteristic or auxillary equation. Solving for r, we use the quadratic formula:

$r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Depending on the value of $b^2-4ac$ , we have three different ways of finding the particular solution to a DE.

$\bold{b^2-4ac>0}$ :

If this is the case, then we have 2 real distinct roots: $r_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $r_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$ .

Due to the principle of superposition, if f and g are a solution to a DE, then any linear combination of f and g is also a solution (I will not prove this). Since $y_1=e^{\frac{-b+\sqrt{b^2-4ac}}{2a}x}$ and $y_2=e^{\frac{-b-\sqrt{b^2-4ac}}{2a}x}$ are solutions to the DE, then any linear combination of the two is a solution. Thus the general solution in this case will be:

$\color{red}y(x)=c_1y_1+c_2y_2=c_1e^{\frac{-b+\sqrt{b^2-4ac}}{2a}x}+c_2e^{\frac{-b-\sqrt{b^2-4ac}}{2a}x}$

$\bold{b^2-4ac=0}$ :

If this is the case, we have real, repeated roots $r_1=r_2=r=-\frac{b}{2a}$ . However, it will have a different solution, due to the fact that each of the solutions must be linearly independent of each other (this will be discussed later). As a result, the solutions will be $y_1=e^{-\frac{b}{2a}x}$ and $y_2={\color{red}x}e^{-\frac{b}{2a}}$ (This is known as reduction of order, which will be proved when we discuss linear independence). Thus, the general solution in this case will be:

$\color{red}y=c_1e^{-\frac{b}{2a}}+c_2xe^{-\frac{b}{2a}}$

$\bold{b^2-4ac<0}$ :

If this is the case, then we have a pair of complex conjugate roots:

$r_1=\alpha+\beta i$ and $r_2=\alpha - \beta i$ .

Thus, the solutions to the DE will be $y_1=e^{(\alpha+\beta i)x}=e^{\alpha x}e^{\beta i x}$ and $y_2=e^{(\alpha-\beta i)x}=e^{\alpha x}e^{-\beta i x}$ . Thus, the general solution will have the form:

$y=e^{\alpha x}{\left(c_1e^{\beta i x}+c_2e^{-\beta i x}\right)}$ .

We really don't want the complex numbers in here, so what we will do is use Euler's Formula to clean it up.

Euler's Formula states that $e^{i\theta}=\cos(\theta)-i\sin(\theta)$ .

Thus,

$e^{\beta i x}=\cos(\beta x)+i\sin(\beta x)$
$e^{-\beta i x}=\cos(\beta x)-i\sin(\beta x)$ .

Substituting this back into the general solution, we have:

$y=e^{\alpha x}{\left(c_1(\cos(\beta x)+i\sin(\beta x))+c_2(\cos(\beta x)-i\sin(\beta x))\right)}$
$=e^{\alpha x}{\left((c_1+c_2)\cos(\beta x)+(c_1-c_2)i\sin(\beta x))\right)}$ .

By defining new constants $C_1=c_1+c_2$ and $C_2=(c_1-c_2)i$ , the general solution is:

$\color{red}y(x)=e^{\alpha x}{\left(C_1\cos(\beta x)+C_2\sin(\beta x)\right)}$ .

Example 9:

Solve $\frac{d^2y}{dx^2}-3\frac{dy}{dx}+2y=0$ ; $y(0)=1$ ; $y^{/}(0)=0$ .

Assuming a solution of $y=e^{rx}$ , we have:

$r^2-3r+2=0 \implies (r-1)(r-2)=0 \implies r=1$ or $r=2$ .

Thus, $y_1=e^x$ and $y_2=e^{2x}$ . Therefore, the general solution will be:

$y(x)=c_1e^{x}+c_2e^{2x}$ .

However, we are given two initial conditions, one for $y(x)$ and one for $y^{/}(x)$ . Let us first find $y^{/}(x)$ :

$y^{/}(x)=c_1e^x+2c_2e^{2x}$ .

Now apply the initial conditions:

$1=c_1+c_2$
$0=c_1+2c_2$ .

We see that $-2c_2=c_1 \implies 1=-c_2\implies c_2=-1$ . Thus, $c_1=2$ .

Therefore, the particular solution is:

$\color{red}\boxed{y(x)=2e^x-e^{2x}}$

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I will post more examples later...

Chris L T521 · #4 $Old$ May 15th, 2008, 11:15 PM

Linear Dependence and Linear Independence

Linear Dependence

The $n$ -number of functions $f_1,f_2,f_3,...,f_n$ are said to be linearly dependent on the interval $I$ , provided that there exists constants $c_1,c_2,c_3,...,c_n$ not all zero such that
$c_1f_1+c_2f_2+...+c_nf_n=0$ on $I$ ; that is, when
$c_1f_1(x)+c_2f_2(x)+...+c_nf_n(x)=0\ \forall \ x\in I$ .

Another way that a set of functions can be determined to be linear dependent is when the n x n Wronksian of n-number of functions is exactly equal to zero:

$W= \left|\begin{array}{cccc}f_1 & f_2 & \cdots & f_n \\ f_{1}^{/} & f_{2}^{/} & \cdots & f_{n}^{/}\\ \vdots & \vdots & \ddots & \vdots\\ f_{1}^{(n-1)} & f_{2}^{(n-1)} & \cdots & f_{n}^{(n-1)} \end{array}\right|\equiv 0$

Example 10:

Show that the functions $y_1=\sin(2x), \ y_2=\sin(x)\cos(x) \ and \ y_3=e^x$ are linearly dependent.

We will go about it two ways: finding the constants $c_1, \ c_2, \ c_3$ such that $c_1y_1+c_2y_2+c_3y_3=0$ . The second way will be by using the Wronskian.

It is alright for one of the constants to be zero, as long as all of the constants aren't zero. If we choose $c_3=0$ , we are left with $c_1\sin(2x)+c_2\sin(x)\cos(x)$ . Noting that $sin(2x)=2sin(x)cos(x)$ , we can pick the remaining constants such that $c_1\sin(2x)+c_2\sin(x)\cos(x)=0$ . If we chose $c_1=1 \ and \ c_2=-2$ , we have:

$sin(2x)-2\sin(x)\cos(x)=sin(2x)-sin(2x)=\color{red}\boxed{0}$ .

The alternative way is by using the Wronskian:

$W= \left|\begin{array}{ccc} sin(2x) & sin(x)cos(x) & e^{x} \\ 2cos(2x) & cos(2x) & e^{x} \\ -4sin(2x) & -2sin(2x) & e^{x} \end{array}\right|=sin(2x)\left| \begin{array}{cc} cos(2x) & e^x \\ -2sin(2x) & e^{x}\end{array} \right|$
$-\frac{1}{2}sin(2x) \left| \begin{array}{cc} 2cos(2x) & e^{x} \\ -4sin(2x) & e^{x} \end{array} \right|+e^x \left| \begin{array}{cc} 2cos(2x) & cos(2x) \\ -4sin(2x) & -2sin(2x) \end{array} \right|$
$=e^{x}sin(2x)(cos(2x)+2sin(2x))-e^{x}sin(2x)(cos(2x)+2sin(2x))+0=\color{red}\boxed{0}$

Since $W\equiv 0$ , this set of functions are linearly dependent.

Linear Independence

A function is linearly independent when the n x n Wronskian of n-number of functions is not equal to zero:

$W= \left|\begin{array}{cccc}f_1 & f_2 & \cdots & f_n \\ f_{1}^{/} & f_{2}^{/} & \cdots & f_{n}^{/}\\ \vdots & \vdots & \ddots & \vdots\\ f_{1}^{(n-1)} & f_{2}^{(n-1)} & \cdots & f_{n}^{(n-1)} \end{array}\right|\neq 0$

Example 11:

Show that the functions $y_1=e^{-3x}, \ y_2=cos(2x) \ and \ y_3=sin(2x)$ are linearly independent.

Find the Wronskian:

$W= \left|\begin{array}{ccc} e^{-3x} & cos(2x) & sin(2x) \\ -3e^{-3x} & -2sin(2x) & 2cos(2x) \\ 9e^{-3x} & -4cos(2x) & -4sin(2x)\end{array}\right|=e^{-3x}\left| \begin{array}{cc} -2sin(2x) & 2cos(2x) \\ -4cos(2x) & -4sin(2x)\end{array} \right|$
$+3e^{-3x} \left| \begin{array}{cc} cos(2x) & sin(2x) \\ -4cos(2x) & -4sin(2x) \end{array} \right|+9e^{-3x} \left| \begin{array}{cc} cos(2x) & sin(2x) \\ -2sin(2x) & 2cos(2x) \end{array} \right|$
$=8e^{-3x}+0+18e^{-3x}=\color{red}\boxed{26e^{-3x} \neq 0 \ \forall \ x \in \mathbb{R}}$

Since $W\neq 0$ , this set of functions are linearly independent.

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Reduction of Order

When solving a homogeneous DE, we noted there were three cases:

1) Real and Distinct roots
2) Real repeated roots
3) Complex conjugate roots

Let us re-examine the second case.

We said that if we had real repeated roots, then the solutions would have the form $y_1=e^{rx}$ and $y_2={\color{red}x}e^{rx}$ . We will now see where the x comes from.

We want the solutions to be linearly independent. In other words,

$y_2\neq cy_1$ or $y_1\neq cy_2$ . But $y_2=u(x)y_1$ where $u(x)$ is not a constant.

In solving $y^{//}+P(x)y^{/}+Q(x)y=0$ , we will assume that $y_2=u(x)y_1$ is a solution to this DE.

$y_2=u(x)y_1$
$y_{2}^{/}=u^{/}(x)y_1+u(x)y_{1}^{/}$
$y_{2}^{//}=u^{//}(x)y_1+2u^{/}(x)y_{1}^{/}+u(x)y_{1}^{//}$

Substitute these values into the differential equation:

$u^{//}(x)y_1+2u^{/}(x)y_{1}^{/}+u(x)y_{1}^{//}+P(x)(u^{/}(x)y_1+u(x)y_{1}^{/})+Q(x)u(x)y_1=0$

Rewrite the DE so it's in terms of u (i.e. au"+bu'+cu=0)

$u^{//}y_1+u^{/}(2y_{1}^{/}+P(x)y_1)+u({\color{blue}y_{1}^{//}+P(x)y_{1}^{/}+Q(x)y_1})=0$

The part highlighted in blue is equal to zero, since $y_1$ is a solution to the DE.

Thus, we are left with:

$u^{//}y_1+u^{/}(2y_{1}^{/}+P(x)y_1)=0$

If we let $w=u^{/} \implies w^{/}=u^{//}$ , we get a first order DE:

$w^{/}y_1+w(2y_{1}^{/}+P(x)y_1)=0 \implies w^{/}y_1=-w(2y_{1}^{/}+P(x)y_1)$

Separate the variables:

$\frac{dw}{w}=-\left(\frac{2y_{1}^{/}+P(x)y_1}{y_1}\right)\,dx$
$\frac{dw}{w}=-\left(\frac{2y_{1}^{/}}{y_1}+\frac{P(x)y_1}{y_1}\right)\,dx$
$\frac{dw}{w}=-\left(\frac{2}{y_1}\frac{dy_1}{dx}+P(x)\right)\,dx$
$\int\frac{dw}{w}=-\int\left(\frac{2}{y_1}\frac{dy_1}{dx}+P(x)\right)\,dx$
$\ln\left|w\right|=-2\ln\left|y_1\right|-\int P(x)\,dx$
$w=e^{-2\ln\left|y_1\right|-\int P(x)\,dx}$
$w=y_{1}^{-2}e^{-\int P(x)\,dx}$

Since $w=u^{/}(x)$ ,

$u^{/}=\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}$

$\therefore \color{red}\boxed{u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx}$

$\therefore \color{red}\boxed{y_2=\left(\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx\right)y_1}$ .

Example 12:

Solve $y^{//}-2y^{/}+y=0$

Assuming a solution of the form $y=e^{rx}$ , we have:

$r^2-2r+1=0 \implies r=1$ with a multiplicity of 2 (Thus, we have repeated roots).

The solutions will thus have the form $y_1=e^x$ and $y_2=u(x)e^x$ where $u=\int \left[\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\right]\,dx$ .

Find u(x):

$u(x)=\int \left[\frac{e^{\int 2\,dx}}{(e^x)^{2}}\right]\,dx=\int\frac{e^{2x}}{e^{2x}}\,dx=\int\,dx=\color{red}\boxed{x}$ .

Therefore, $\color{red}\boxed{y_2=xe^x}$ .

Therefore, $\color{red}\boxed{y(x)=c_1e^x+c_2xe^x}$ .

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Will post more when I'm done with finals...

Chris L T521 · #5 $Old$ May 22nd, 2008, 10:25 PM

Cauchy-Euler Equations:

A second order homogeneous Cauchy-Euler Equation takes the form:

$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0$ .

This time, we assume that a solution to the equation has the form $y=x^r$ . Substituting this in for y, we get the new DE:

$ax^2\left[r(r-1)x^{r-2}\right]+bx\left[rx^{r-1}\right]+cx^r=0$

Simplifying, we get:

$a(r^2-r)x^{r}+brx^{r}+cx^{r}=0$

Pulling out a common factor of $x^r$ , we get:

$\left(a(r^2-r)+br+c\right)x^{r}=0$

Assuming that $x^r\neq 0$ , we divide both sides by $x^r$ and get:

$ar^2-ar+br+c=0 \implies ar^2+(b-a)r+c=0$ .

This is the characteristic equation for the Cauchy-Euler Equation.

Again, just with the second order homogeneous DE with constant coefficients, there are three general cases:

1) Real distinct roots.
2) Real repeated roots.
3) Complex conjugate roots.

Case 1:

If we have two real and distinct roots $r=r_1 \ and \ r=r_2$ , then the solutions to the DE are $y_1=x^{r_1} \ and \ y_2=x^{r_2}$ . Thus the general solution would be:

$\color{red}\boxed{y=c_1x^{r_1}+c_2x^{r_2}}$

Example 13:

Solve $x^2\frac{d^2y}{dx^2}+7x\frac{dy}{dx}+5y=0$ .

Assuming a solution of $y=x^r$ , we get the characteristic equation:

$r^2+(7-1)r+5=0 \implies r^2+6r+5=0 \implies (r+1)(r+5)=0 \implies r=-1 \ or \ r=-5$ .

Therefore, $y_1=x^{-1} \ and \ y_2=x^{-5}$

$\color{red}\boxed{\therefore y=c_1x^{-1}+c_2x^{-5}}$

Case 2:

If we have real repeated roots, then $r=r_1=r_2$ . Thus, the solutions take the form $y_1=x^{r} \ and \ y_2=u(x)x^{r}$ where $u(x)=\int\frac{e^{-\int P(x)\,dx}}{y_{1}^{2}}\,dx$ . This is the case because $y_1 \ \text{and} \ y_2$ must be linearly independent solutions. To use the reduction of order formula, we need to manipulate the original DE:

$ax^2\frac{d^2y}{dx^2}+bx\frac{dy}{dx}+cy=0 \implies \frac{d^2y}{dx^2}+\frac{b}{ax}\frac{dy}{dx}+\frac{c}{ax^2}=0$ .

Now apply Reduction of order:

$u(x)=\int\frac{e^{\int \frac{b}{ax} \,dx}}{x^{2r}}\,dx$ where $r=\frac{-(b-a)}{2a}$ . Thus, we get:

$u(x)=\int\frac{e^{-\int \frac{b}{ax} \,dx}}{x^{2\frac{a-b}{2a}}}\,dx=\int\frac{x^{-\frac{b}{a}}}{x^{1}x^{-\frac{b}{a}}}\,dx=\int\frac{1}{x}\,dx=\color{red}\boxed{\ln x}$

Thus, the solutions are $y_1=x^r \ \text{and} \ y_2=x^r \ln x$ .

Therefore, the general solution is:

$\color{red}\boxed{y=c_1x^r+c_2x^r \ln x}$ .

Example 14:

Solve $x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+4y=0$ .

Assuming a solution of $y=x^r$ , we get:

$r^2+(-3-1)r+4=0 \implies r^2-4r+4=0 \implies (r-2)^2=0$
$\implies r=2\ w. \ \text{multiplicity 2}$

Thus, the solutions are $y_1=x^2 \ \text{and} \ y_2=x^2 \ln x$ .

Therefore, the general solution is:

$\color{red}\boxed{y=c_1x^2+c_2x^2 \ln x}$ .

Case 3:

Whenever we have complex conjugate roots $r = \alpha \pm \beta i$ , we see that our two solutions to the DE take on the form: $y_1 = x^{\alpha + \beta i} {\text{ and }}y_2 = x^{\alpha - \beta i}$ . However, it doesn't look pleasant with complex numbers in the solution. To take care of this, we will use Euler's Formula, which states that:

$e^{i\theta } = \cos \theta + i\sin \theta$

Since $x^r=e^{r\ln x}$ , we can say that

$\begin{gathered} y_1 = e^{r_1 \ln x} = e^{\left( {\alpha + \beta i} \right)\ln x} = e^{\alpha \ln x} e^{\left( {\beta \ln x} \right)i} \hfill \\ {\text{and }} \hfill \\ y_2 = e^{r_2 \ln x} = e^{\left( {\alpha - \beta i} \right)\ln x} = e^{\alpha \ln x} e^{ - \left( {\beta \ln x} \right)i}. \hfill \\ \end{gathered}$

Now applying Euler's Formula, we get:

$\begin{gathered} e^{\left( {\beta \ln x} \right)i} = \cos \left( {\beta \ln x} \right) + i\sin \left( {\beta \ln x} \right) \hfill \\ e^{ - \left( {\beta \ln x} \right)i} = \cos \left( { - \beta \ln x} \right) + i\sin \left( { - \beta \ln x} \right). \hfill \\ \end{gathered}$

Due to the even and odd properties of Cosine and Sine, we get that

$e^{ - \left( {\beta \ln x} \right)i} = \cos \left( {\beta \ln x} \right) - i\sin \left( {\beta \ln x} \right)$

Thus, the general solution in this case would be:

$y = c_1 y_1 + c_2 y_2 = x^{\alpha} \left[ {c_1 \left( {\cos \left( {\beta \ln x} \right) + i\sin \left( {\beta \ln x} \right)} \right) + c_2 \left( {\cos \left( {\beta \ln x} \right) - i\sin \left( {\beta \ln x} \right)} \right)} \right]$

$\Rightarrow y = x^{\alpha} \left[ {\left( {c_1 + c_2 } \right)\cos \left( {\beta \ln x} \right) + \left( {c_1 - c_2 } \right)i\sin \left( {\beta \ln x} \right)} \right]$

Letting $\left( {c_1 + c_2 } \right) = C_1$ and $\left( {c_1 - c_2 } \right)i = C_2$ , we have the general solution:

$\color{red}\boxed{y = x^{\alpha} \left[ {C_1 \cos \left( {\beta \ln x} \right) + C_2 \sin \left( {\beta \ln x} \right)} \right]}$

Example 15:

Solve $x^2 \frac{{d^2 y}}{{dx^2 }} + 2x\frac{{dy}}{{dx}} + y= 0.$

Assuming a solution of the form $y=x^r$ , we get the characteristic equation:

$r^2 + r + 1 = 0 \Rightarrow r = \frac{{ - 1 \pm \sqrt {1 - 4} }}{2} \Rightarrow r = - \frac{1}{2} \pm \frac{{\sqrt 3 }}{2}i$

$\color{red}\boxed{\therefore y = x^{ - \frac{1}{2}} \left[ {c_1 \cos \left( {\frac{{\sqrt 3 }}{2}\ln x} \right) + c_2 \sin \left( {\frac{{\sqrt 3 }}{2}\ln x} \right)} \right]}$

Chris L T521 · #6 $Old$ May 24th, 2008, 09:24 PM

Second Order NON-Homogeneous Differential Equations

Ah...the dreadful second order non-homogeneous differential equation has the form:

$a\frac{{d^2 y}}{{dx^2 }} + b\frac{{dy}}{{dx}} + cy = f\left( x\right)$

I will go through three techniques on how to solve these nasties:

Method of Undetermined Coefficients
The "Annihilator" Method (Similar to #1)
Variation of Parameters

Technique 1 : Method of Undetermined Coefficients:

In the case we have a differential equation like $a\frac{{d^2 y}}{{dx^2 }} + b\frac{{dy}}{{dx}} + cy = f\left( x\right)$ , we can guess what the particular solution to a DE may be, depending on what $f(x)$ is. For example, let us say that $f(x)=3x+7$ . We would assume that a particular solution to the DE would be $y_p=Ax+B$ . To find the Undetermined Coefficients, plug $y_p$ back into the original DE.

If $f(x)=5e^{-3x}$ , we assume that the particular solution to the DE would have the form of $y_p=Ae^{-3x}$ . We too would substitute $Ae^{-3x}$ into the DE to find the unknown coefficient value.

If $f(x)=3\cos(x)$ , we assume that the particular solution to the DE would have the form of $y_p=A\cos(x)+B\sin(x)$ . Again, to find the unknown coefficients, substitute $y_p$ into the original DE.

Sometimes, the guess of $y_p$ isn't that obvious. Try to think outside the box when solving these problems!

Note that the solution to the non-homogeneous DE is a linear combination of the complimentary solution (solution to the homogeneous equation) and the particular solution (solution to the non-homogeneous equation)

Example 16:

Solve $y'' - 3y' + 2y = 3e^{ - x} - 10\cos \left( {3x} \right)$ ; $y(0)=1$ ; $y'(0)=2$ .

Solve the homogeneous equation $y'' - 3y' + 2y = 0$ first.

$y''-3y'+2y=0 \implies r^2-3r+2=0 \implies r=2 \ \text{or} \ r=1$ .

Thus, $y_c=c_1e^{x}+c_2e^{2x}$ .

Now solve the non-homogeneous equation.

$y''-3y'+2y=3e^{ - x} - 10\cos \left( {3x} \right)$ .

Using the method of Undetermined Coefficients, we guess and assume that the particular solution will have the form:

$y_p=Ae^{-x}+B\cos(3x)+C\sin(3x)$

$\therefore y_{p}^{/}=-Ae^{-x}-3B\sin(3x)+3C\cos(3x)$

$\therefore y_{p}^{//}=Ae^{-x}-9B\cos(3x)-9C\sin(3x)$ .

Substituting these values into the original DE, we get:

$(Ae^{-x}-9B\cos(3x)-9C\sin(3x))-3(-Ae^{-x}-3B\sin(3x)+3C\cos(3x))+2(Ae^{-x}+$
$B\cos(3x)+C\sin(3x))=3e^{ - x} - 10\cos \left( {3x} \right)$

$\implies 6Ae^{-x}+(-7B-9C)\cos(3x)+(9B-7C)\sin(3x)=3e^{-x}-10\cos(3x)$

Now compare the coefficients (like in Partial Fractions)

$\begin{aligned}6A&=3 \\7B-9C&=-10 \\9B-7C&=0\end{aligned}$

Solving for the constants, we get $A=\frac{1}{2}$ , $B=\frac{7}{13}$ and $C=\frac{9}{13}$ (verify).

Thus, $y_p=\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)$ .

Therefore, the solution is

$y=c_1e^{x}+c_2e^{2x}+\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)$ .

Find $y^{/}$ so we can apply the initial conditions:

$y^{/}=c_1e^{x}+2c_2e^{2x}-\frac{1}{2}e^{-x}+\frac{1}{13}\left(21\cos(3x)+27\sin(3x)\right)$ .

Apply the initial conditions.

$y(0)=1=c_1+c_2+\frac{1}{2}+\frac{7}{13}$
$y^{/}(0)=2=c_1+2c_2-\frac{1}{2}+\frac{21}{13}$ .

Solving for $c_1 \ \text{and} \ c_2$ , get $c_1=-\frac{1}{2} \ \text{and} \ c_2=\frac{6}{13}$ (verify)

Therefore, the solution to the initial value problem is:

$\color{red}\boxed{y=-\frac{1}{2}e^{x}+\frac{6}{13}e^{2x}+\frac{1}{2}e^{-x}+\frac{1}{13}\left(7\cos(3x)+9\sin(3x)\right)}$ .

I will discuss the next two sometime later...

Chris L T521 · #7 $Old$ May 25th, 2008, 02:23 AM

Technique #2 : The "Annihilator" Method:

An alternative to the first technique would be the Annihilator method. As it's name foretells, we annihilate the non-homogeneous term and make the equation homogeneous.

To use the Annihilator technique, you must rewrite the DE using Differential Operator Notation:

$a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+cy=f(x) \implies aD^2y+bDy+cy=f(x) \implies \left(aD^2+bD+c\right)(y)=f(x)$

Note that we "factor" out a y (I use this term very loosely; you really can't factor out the y, but as you will see, it will work out to our advantage

)

You can apply the annihilator to any of the following families of functions that $f(x)$ can be:

1. $1,x,x^2,x^3,\cdots ,x^{n-1}$

2. $e^{\alpha x}, xe^{\alpha x}, x^2e^{\alpha x},\cdots ,x^{n-1}e^{\alpha x}$

3. $e^{\alpha x}\cos(\beta x), xe^{\alpha x}\cos(\beta x), x^2e^{\alpha x}\cos(\beta x),\cdots,x^{n-1}e^{\alpha x}\cos(\beta x)$

4. $e^{\alpha x}\sin(\beta x), xe^{\alpha x}\sin(\beta x), x^2e^{\alpha x}\sin(\beta x),\cdots,x^{n-1}e^{\alpha x}\sin(\beta x)$

The Annihilators:

$\color{red}\boxed{D^{n}}$ will annihilate $1,x,x^2,x^3,\cdots ,x^{n-1}$ .

I will leave it for you to prove the other two:

$\color{red}\boxed{\left(D-\alpha\right)^n}$ will annihilate $e^{\alpha x}, xe^{\alpha x}, x^2e^{\alpha x},\cdots ,x^{n-1}e^{\alpha x}$ .

$\color{red}\boxed{\left(D^2-2\alpha D+(\alpha^2+\beta^2)\right)^n}$ will annihilate $e^{\alpha x}\cos(\beta x), xe^{\alpha x}\cos(\beta x), x^2e^{\alpha x}\cos(\beta x),\cdots,x^{n-1}e^{\alpha x}\cos(\beta x)$ and
$e^{\alpha x}\sin(\beta x), xe^{\alpha x}\sin(\beta x), x^2e^{\alpha x}\sin(\beta x),\cdots,x^{n-1}e^{\alpha x}\sin(\beta x)$ .

Example 17:

Solve $y''-4y=\cosh (2x)$ .

Solving the homogeneous equation $y'' -4y=0$ , we see that $r^2-4=0 \implies r=\pm 2$ . Thus the complimentary solution is $y_c=c_1e^{-2x}+c_2e^{2x}$ .

Now, to apply the annihilator to the DE, we need to rewrite it in differential form:

$y''-4y=\cosh (2x) \implies \left(D^2-4\right)(y)=\cosh (2x)$

noting that $\cosh u =\frac{1}{2}\left( e^{u}+e^{-u}\right)$ , we have the DE:

$\left(D^2-4\right)(y)=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

Let us determine the proper annihilator:

$D-2 \ \text{annihilates} \ e^{2x} \ \text{and} \ D+2 \ \text{annihilates} \ e^{-2x}$ . This may pose a problem: we have two different annihilators! so which one do we apply to the DE? the answer is both. There is a theorem that states something like the following:

If there are two functions $f(x) \ \text{and} \ g(x)$ and their annihilators are $L_1 \ \text{and} \ L_2$ respectively,
then the product of the two annihilators $\left(L_1L_2\right)$ will annihilate $f(x)+g(x)$ .

Thus, $\left(D-2\right)\left(D+2\right)$ will annihilate $\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$ .

Applying the newly found annihilator to both sides we get:

$\left(D+2\right)\left(D-2\right)\left(D^2-4\right)(y)=0$

Now rewrite the equation so we get the characteristic equation:

$\left(r+2\right)\left(r-2\right)\left(r^2-4\right)=0$

Solving for r, we get $r=2$ with multiplicity 2 and $r=-2$ with multiplicity 2.

Note that 2 of the r values were values used to determine the complimentary solution! Thus, the general solution will be:

$y=\underbrace{c_1e^{-2x}+c_2e^{2x}}_{y_c}+\underbrace{c_3xe^{-2x}+c_4xe^{2x}}_{y_p}$

Now find the coefficients $c_3 \ \text{and} \ c_4$ (which I will denote by A and B, respectively).

$y_p=Axe^{-2x}+Bxe^{2x}$
$y_{p}^{/}=\left(A-2Ax\right)e^{-2x}+\left(2Bx+B\right)e^{2x}$
$y_{p}^{//}=\left(4Ax-4A\right)e^{-2x}+\left(4Bx+4B\right)e^{2x}$

Substituting these values into the DE, we get:

$\left(4Ax-4A\right)e^{-2x}+\left(4Bx+4B\right)e^{2x}-4\left[Axe^{-2x}+Bxe^{2x}\right]=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

$\implies -4Ae^{-2x}+4Be^{2x}=\frac{1}{2}\left(e^{2x}+e^{-2x}\right)$

Comparing the coefficients, we get:

$\begin{aligned}-4A &= \frac{1}{2}\\4B &= \frac{1}{2}\end{aligned}$

This gives us $A=-\frac{1}{8}$ and $B=\frac{1}{8}$

Therefore the general solution is:

$\color{red}\boxed{y=c_1e^{-2x}+c_2e^{2x}-\frac{1}{8}\left(xe^{-2x}-xe^{2x}\right)}$

Example 18:

Solve $y''-6y'+13y=xe^{3x}\sin(2x)$ (WARNING!! THIS IS A VERY TEDIOUS PROBLEM TO SOLVE!!!

)

Here's the easy part (solve the homogeneous equation):

$y''-6y+13=0 \implies r^2-6r+13=0\implies r=\frac{6\pm \sqrt{36-52}}{2}$
$\implies r=\frac{6\pm \sqrt{-16}}{2}\implies r=3\pm 2i$

Thus the complimentary solution will be $y_c=e^{3x}\left(c_1\cos(2x)+c_2\sin(2x)\right)$

The next part isn't that bad (solving the non-homogeneous equation):

$y''-6y+13=xe^{3x}\sin(2x) \implies \left(D^2-6D+13\right)(y)=xe^{3x}\sin(2x)$

Now we can find the annihilator:

$\left(D^2-2(3)D+(3^2+2^2)\right)^2 \implies \left(D^2-6D+13\right)^2$ will annihilate $xe^{3x}\sin(2x)$ .

Applying the annihilator to both sides, we get:

$\left(D^2-6D+13\right)^3(y)=0$

Converting it to the characteristic equation, we have:

$\left(r^2-6r+13\right)^3=0 \implies r=3\pm 2i$ with multiplicity three (note that one of these roots form the complementary solution)

Now here comes the nasty part: find $y_p$

$y_p=xe^{3x}\left(A\cos(2x)+B\sin(2x)\right)+x^2e^{3x}\left(C\cos(2x)+D\sin(2x)\right)$
$\implies y_p=\left(Ax+Cx^2\right)e^{3x}\cos(2x)+\left(Bx+Dx^2\right)e^{3x}\sin(2x)$

I'll leave it for you to show that

$\begin{aligned}y_{p}^{/}=&\left[\left(3A+2B+2C\right)x+\left(3C+2D\right)x^2+A\right]e^{3x}cos(2x)\\&+\left[\left(3B-2A+2D\right)x+\left(3D-2C\right)x^2+B\right]e^{3x}\sin(2x)\end{aligned}$

$\begin{aligned}y_{p}^{//}=&\left[\left(5A+12B+12C+8D\right)x+\left(12D+5C\right)x^2+3A+2B\right]e^{3x}\cos(2x)\\&+\left[\left(5B-12A-8C+12D\right)x+\left(5D-12C\right)x^2+3B-2A\right]e^{3x}\sin(2x)\end{aligned}$

Substituting this into the DE (

), we have (...get ready...)

$\begin{aligned}&\left(\left[\left(5A+12B+12C+8D\right)x+\left(12D+5C\right)x^2+3A+2B\right]e^{3x}\cos(2x)\right.\\&+\left.\left[\left(5B-12A-8C+12D\right)x+\left(5D-12C\right)x^2+3B-2A\right]e^{3x}\sin(2x)\right)\\&-6\left[\left[\left(3A+2B+2C\right)x+\left(3C+2D\right)x^2+A\right]e^{3x}cos(2x)\right.\end{aligned}$
$\begin{aligned}&+\left.\left[\left(3B-2A+2D\right)x+\left(3D-2C\right)x^2+B\right]e^{3x}\sin(2x)\right]\\&+13\left[\left(Ax+Cx^2\right)e^{3x}\cos(2x)+\left(Bx+Dx^2\right)e^{3x}\sin(2x)\right]=xe^{3x}\sin(2x)\end{aligned}$

...well, after a decent amount of cancellations, we get:

$\left[2D-3A+8Dx\right]e^{3x}\cos(2x)+\left[-3B-2A-8Cx\right]e^{3x}\sin(2x)=xe^{3x}\sin(2x)$

Comparing the coefficients, we get:

$\begin{aligned}2D-3A&=0 \\8D&=0 \\-8C&=1 \\-3B-2A&=0\end{aligned}$

Solving this system, we get $A=0 \text{,} \ B=0 \text{,} \ C=-\frac{1}{8} \text{,} \ \text{and} \ D=0$

Therefore, $y_p=-\frac{1}{8}x^2e^{3x}\cos(2x)$

Therefore, the general solution is:

$\color{red}\boxed{y=e^{3x}\left(c_1\cos(2x)+c_2\sin(2x)\right)-\frac{1}{8}x^2e^{3x}\cos(2x)}$

I will discuss Variation of Parameters tomorrow (hopefully)

Chris L T521 · #8 $Old$ May 27th, 2008, 01:59 AM

Technique #3: Variation of Parameters:

We have dealt with non-homogeneous equations where $f(x)$ had the form of $x^n \text{,} \ x^ne^{\alpha x} \text{,} \ x^ne^{\alpha x}\sin(\beta x) \text{,} \ etc$ . However, what if $f(x)$ was $\sec x$ ? $\sin^{-1}x$ ? $\ln x$ ? We would now need a new technique to conquer these non-homogeneous equations. That technique is known as variation of parameters.

If we have a differential equation in the form $\frac{d^2y}{dx^2}+P(x)\frac{dy}{dx}+Q(x)y=f(x)$ , we want $y_p=u_1(x)y_1+u_2(x)y_2$ . But how do we find $u_1(x)$ and $u_2(x)$ ?? We will substitute $y_p$ into the differential equation. This will give us the following:

$u_1^{//} y_1 + u_1^/ y_1^/ + u_1^/ y_1^/ + u_1 y_1^{//} + u_2^{//} y_2 + u_2^/ y_2^/$ $+ u_2^/ y_2^/ + u_2 y_2^{//} + P(x)\left[ {u_1^/ y_1 + u_1 y_1^/ + u_2^/ y_2 + u_2 y_2^/ } \right]$ $+ Q(x)\left[ {u_1 y_1 + u_2 y_2 } \right]$
$= u_1 \left[ {y_1^{//} + P(x)y_1^/ + Q(x)y_1 } \right] + u_2 \left[ {y_2^{//} + P(x)y_2^/ + Q(x)y_2 } \right]$ $+ u_1^{//} y_1 + u_1^/ y_1^/ + u_2^{//} y_2 + u_2^/ y_2^/ + P(x)\left[ {u_1^/ y_1 + + u_2^/ y_2 } \right]$ $+ u_1^/ y_1^/ + u_2^/y_2^/$
$= \frac{d}{{dx}}\left[ {u_1^/ y_1 + u_2^/ y_2 } \right] + P(x)\left[ {u_1^/ y_1 + + u_2^/ y_2 } \right] + u_1^/ y_1^/ + u_2^/ y_2^/$
$=f(x)$

Since we seek $u_1$ and $u_2$ , we need two equations. To get these equations, we need to impose 2 conditions onto $u_1$ and $u_2$ The first condition is that $L\left[y_p\right]=f(x)$ . The second one we can apply is not given, thus, we need to come up with one.

Recall that $y_p^/ = u_1^/ y_1 + u_1 y_1^/ + u_2^/ y_2 + u_2 y_2^/$ . Rearranging, we get $y_p^/ = (u_1 y_1^/ + u_2 y_2^/ ) + (u_1^/ y_1 + u_2^/ y_2 )$ .

To avoid the appearance of the second derivatives $u_1^{//}$ and $u_2^{//}$ , we impose the second condition that the second sum must equal zero:

$u_1^/ y_1 + u_2^/ y_2 =0$

With these assumptions, the DE becomes $u_1^/ y_1^/ + u_2^/ y_2^/ = f(x)$ .

Now, we have a system of equations:

$\left\{\begin{array}{l}u_1^/ y_1 + u_2^/ y_2 =0 \\u_1^/ y_1^/ + u_2^/ y_2^/ = f(x)\\\end{array}\right.$

To solve this, we will use Cramer's Rule:

If Ax = b is a system of n linear equations in n unknowns such that $\det (A) \ne 0$ , then the system has a unique solution. This solution is
$x_1 = \frac{{\det (A_1 )}}{{\det (A)}}$ ,..., $x_2 = \frac{{\det (A_2 )}}{{\det (A)}}$ ,..., $x_n= \frac{{\det (A_n )}}{{\det (A)}}$

where $A_j$ is the matrix obtained by replacing the terms in the $j^{th}$ column of A by the entries in matrix b.

To apply Cramer's Rule, we write the system in matrix form:

$\left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {u_1^/ } \\ {u_2^/ } \\ \end{array} } \right] = \left[ {\begin{array}{*{20}c} 0 \\ {f(x)} \\ \end{array} } \right]$

Thus, according to Cramer's rule,

$u_1^/ = \frac{{\det \left[ {\begin{array}{*{20}c} 0 & {y_2 } \\ {f(x)} & {y_2^/ } \\ \end{array} } \right]}}{{\det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]}}$ and $u_2^/ = \frac{{\det \left[ {\begin{array}{*{20}c} {y_1 } & 0 \\ {y_1^/ } & {f(x)} \\ \end{array} } \right]}}{{\det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]}}$

We represent the determinant matrices as $W_1$ , $W_2$ , and $W$ . Particularly,

$W_1^{} = \det \left[ {\begin{array}{*{20}c} 0 & {y_2 } \\ {f(x)} & {y_2^/ } \\ \end{array} } \right]$ , $W_2^{} = \det \left[ {\begin{array}{*{20}c} {y_1 } & 0 \\ {y_1^/ } & {f(x)} \\ \end{array} } \right]$ , and $W = \det \left[ {\begin{array}{*{20}c} {y_1 } & {y_2 } \\ {y_1^/ } & {y_2^/ } \\ \end{array} } \right]$

We recongnize $W$ as the Wronskian of $y_1$ and $y_2$ . Due to the linear independence of $y_1$ and $y_2$ , $W(y_1,y_2)\neq 0$ .

We can now find $u_1$ and $u_2$ .

Example 19:

Solve $y''+y=\tan x$ .

Solve the homogeneous equation:

$y''+y=0 \implies r^2+1=0 \implies r=\pm i$

Thus, $y_c=c_1\cos(x)+c_2\sin(x)$ .

Now that we have $y_1=\cos(x)$ and $y_2=\sin(x)$ we can use variation of parameters.

$W_1=\det \left[ {\begin{array}{*{20}c} 0 & {\sin x } \\ {\tan x} & {\cos x} \\ \end{array} } \right]=-\frac{\sin^2(x)}{cos(x)}=\frac{\cos^2(x)-1}{cos(x)}=\cos(x)-\sec(x)$

$W_2=\det \left[ {\begin{array}{*{20}c} {cos(x) } & 0 \\ {-sin(x) } & {\tan(x)} \\ \end{array} } \right]=\cos(x)\tan(x)=\sin(x)$

$W = \det \left[ {\begin{array}{*{20}c} {\cos(x)} & {\sin(x)} \\ {-\sin(x)} & {\cos(x)} \\ \end{array} } \right]=\cos^2(x)-\left(-\sin^2(x)\right)=cos^2(x)+sin^2(x)=1$

Therefore $u_1^/=\frac{W_1}{W}=\cos(x)-\sec(x)$ and $u_2^/=\frac{W_2}{W}=\sin(x)$ .

Now find $u_1$ and $u_2$ .

$u_1=\int \left(\cos(x)-\sec(x)\right)\,dx=\sin(x)-\ln\left|\sec(x)+\tan(x)\right|$

$u_2=\int \sin(x)\,dx=-\cos(x)$

Therefore, our particular solution will be:

$\color{red}\boxed{y=\left[\sin(x)-\ln\left|\sec(x)+\tan(x)\right|\right]\cos(x)-\frac{1}{2}\sin(2x)}$

My next couple posts will be on applications of non-homogeneous differential equations (in particular, the spring mass systems)...

Chris L T521 · #9 $Old$ August 22nd, 2008, 11:36 PM

It's been a while since I've last posted. I finally found the time right now to do this, since it has been on my mind for a while. I put this together a little quickly, but I will come back later and make changes [if necessary].

Applications of Homogeneous and Non Homogeneous Differential Equations (Spring Mass Systems)

A useful application to the real world would be spring mass systems. We prefer to use homogeneous or non-homogeneous differential equations to model the systems.

There are two types of systems. One models damped motion [due to a dashpot of some sort or due to friction], and the other models undamped motion [sometimes known as free motion].

Undamped Spring Mass Systems

Consider a spring of length $l$ hanging from the ceiling. The spring has a mass $m$ attached to its end. The system comes to rest at its equilibrium position.

When the system is place in motion, the spring varies in length about the equilibrium position. We will denote this distance from the equilibrium position as $x$ .

When a free body analysis is done, we see two different forces:

$F=ma$ [Newton's 2nd Law] and $F=-kx$ [Hooke's Law].

Thus, we see that $\sum F_y=0$ when $-kx-ma=0\implies ma+kx=0$

Since $a=\frac{d^2x}{dt^2}$ , our equation becomes the differential equation $m\frac{d^2x}{dt^2}+kx=0\implies x''+\frac{k}{m}x=0$ .

Letting $\frac{k}{m}=\omega^2$ , the DE becomes $x''+\omega^2x=0$ .

However, we are dealing with systems that are being acted upon my external forces, so we make a modification to our DE. If an external force is applied, then the DE takes on the form $x''+\omega^2x=F(t)$ , which is non-homogeneous. When no external force is applied, $F(t)=0$ , which then makes the equation homogeneous.

Let's take a look at an example:

Example 20:

A mass weighing 2 pounds stretches a spring 6 inches. At $t=0$ , the mass is released from a point
8 inches below the equilibrium position with an upward velocity of $\tfrac{4}{3}~ft/s$ . Determine the equation of free motion.

This can be seen as an initial value problem.

Since the weight of the mass is 2 pounds, we can determine the mass, because $w=mg\implies m=\frac{w}{g}\implies m=\frac{2~lbs}{32~ft/s^2}\implies m=\frac{1}{16}~slug$

Since the string is displaced by 6 inches = 1/2 a foot, we can determine k:

$F=kx\implies 2~lbs=k\left(\tfrac{1}{2}~ft\right)\implies k=4~lbs/ft$

At $t=0$ , the mass is released 8 inches below the equilibrium position. This implies that the condition is $x(0)=\tfrac{2}{3}~ft$ .

At $t=0$ , the mass is released with an upward velocity of $\tfrac{4}{3}~ft/s$ . This implies that the condition is $x'(0)=-\tfrac{4}{3}~ft/s$ .

Keep in mind the signs of these conditions. If the mass starts off above the equilibrium position, the sign applied to the condition is negative, and if the mass starts off below the equilibrium poistion, the sign applied to the condition is positive. The same idea is applied to upward and downward velocities.

Let's start to set up our DE:

Since $\frac{k}{m}=\omega^2$ , we see that $\omega^2=64$

So our IVP is:

$x''+64x=0;~~x(0)=\tfrac{2}{3},~~x'(0)=-\tfrac{4}{3}$

We see that the characteristic equation has the form

$r^2+64=0$

This implies that we have complex conjugate roots. Thus, we see that $r=\pm8i\implies x(t)=A\cos(8t)+B\sin(8t)$

Applying the condition $x(0)=\tfrac{2}{3}$ , we see that $A=\tfrac{2}{3}$ .

Before we apply the second condition, let's find $x'(t)$

$x'(t)=-8A\sin(8t)+8B\cos(8t)$

Applying the condition $x'(0)=-\tfrac{4}{3}$ , we see that $-\tfrac{4}{3}=8B\implies B=-\tfrac{1}{6}$ .

Thus, the equation of free motion is $\color{red}\boxed{x(t)=\tfrac{2}{3}\cos(8t)-\tfrac{1}{6}\sin(8t)}$

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Damped Motion

Let's go back to the equation we came up with for undamped motion:

$mx''+kx=F(t)$

For damped motion, we need to introduce a damping term. The term is $\beta x'$ .

When introduced into the system, we see from a free body diagram that $ma=-kx-\beta x'$

Thus, we see that $ma+\beta x'+kx=0$ .

Since $a=x''$ the DE becomes $mx''+\beta x'+kx=0\implies x''+\frac{\beta}{m}x'+\frac{k}{m}x=0$ .

Letting $\frac{k}{m}=\omega^2$ and $\frac{\beta}{m}=2\lambda$ , the DE is then transformed into

$x''+2\lambda x'+\omega^2 x=0$

Given a particular external force, F(T), we can say the DE takes on the form

$x''+2\lambda x'+\omega^2 x=F(t)$

In the case that there is no external force, $F(t)=0$ .

Three Cases for Damped Motion

Solving $x''+2\lambda x'+\omega^2 x=0$ , we see that we have the auxiliary equation $r^2+2\lambda r+\omega^2=0$

Using the quadratic formula, we see that $r=\frac{-2\lambda\pm\sqrt{4\lambda^2-4\omega^2}}{2}\implies r=\frac{-2\lambda\pm2\sqrt{\lambda^2-\omega^2}}{2}\implies r=-\lambda\pm\sqrt{\lambda^2-\omega^2}$

We have three cases, depending on the value of $r=-\lambda\pm\sqrt{\lambda^2-\omega^2}$

Case 1 : Overdamped

We have overdamped motion when r consists of real and distinct roots.

Other factors for determining the overdamped case are when $\beta>>k\implies \lambda^2>\omega^2$

Thus, the general solution for overdamped motion has the form $\color{red}\boxed{x(t)=c_1e^{(-\lambda+\sqrt{\lambda^2-\omega^2})t}+c_2e^{(-\lambda-\sqrt{\lambda^2-\omega^2})t}}$

Case 2 : Critically damped

We have critically damped motion when r consists of real, repeated roots.

One major factor that reveals this is when $\lambda^2=\omega^2$ .

Thus, the general solution for critically damped motion has the form $\color{red}\boxed{x(t)=c_1e^{-\lambda t}+c_2te^{-\lambda t}}$

Case 3 : Underdamped

We have underdamped motion when r consists of complex, conjugate roots.

Major factors that reveal this is when $k>>\beta$ and when $\lambda^2<\omega^2$ .

This then implies that the solution to the characteristic equation has the form $r=-\lambda\pm\sqrt{\omega^2-\lambda^2}i$ .

Thus, the general solution for underdamped motion has the form $\color{red}\boxed{x(t)=e^{-\lambda t}\left[c_1\cos\left(\sqrt{\omega^2-\lambda^2}t\right)+c_2\sin\left(\sqrt{\omega^2-\lambda^2}t\right)\right]}$

Let us take a look at another example.

Example 21:

A 4-foot spring measures 8 feet long after an 8-pound weight is attatched
to it. The medium through with the weight moves offers a resistance
numerically equal to $\sqrt{2}$ times the instantaneous velocity. Find the equation
of motion if the weight is released from the equilibrium postion with a downward
velocity of 5 ft/s. Find the time at which the weight attains its extreme
displacement from the equilibrium position.
Identify the system as overdamped, underdamped, or critically damped.

$w=mg\implies m=\frac{w}{g}$

Since $w=8~lbs$ and $g=32~ft/s^2$ , $m=\tfrac{1}{4}~slug$

$F=kx\implies k=\frac{F}{x}$

Since $F=w=8~lbs$ , and $x=4~ft$ , $k=2~lbs/ft$

Since the medium through with the weight moves offers a resistance numerically equal to $\sqrt{2}$ times the instantaneous velocity, this implies that $\beta=\sqrt{2}$ .

Now, let us determine the conditions:

Since the mass is being released from the equilibrium position, we see that $x(0)=0$

Since the mass will have a downward velocity of $5~ft/s$ , this implies that $x'(0)=5$

Since $\frac{k}{m}=\omega^2$ and $\frac{\beta}{m}=2\lambda$ , we see that $\omega^2=8$ and $2\lambda=4\sqrt{2}$

Now, we have the IVP:

$x''+4\sqrt{2}x'+8x=0;~~x(0)=0,~~x'(0)=5$

Solving, we see that the DE has the characteristic equation $r^2+4\sqrt{2}r+8=0$ . It turns out to be a perfect square:

$r^2+4\sqrt{2}r+8=0\implies\left(r+2\sqrt{2}\right)^2=0\implies r=-2\sqrt{2}$ with multiplicity 2.

Thus, $x(t)=c_1e^{-2\sqrt{2}t}+c_2te^{-2\sqrt{2}t}\implies x(t)=(c_1+c_2t)e^{-2\sqrt{2}t}$

When $x(0)=0$ , $c_1=0$ .

Now we need to find $x'(t)$

$x'(t)=-2\sqrt{2}(c_1+c_2t)e^{-2\sqrt{2}t}+c_2e^{-2\sqrt{2}t}$

When, $x'(0)=5$ , $5=-2\sqrt{2}c_1+c_2$

But $c_1=0$ , so $c_2=5$

Therefore, the equation of motion is $\color{red}\boxed{x(t)=5te^{-2\sqrt{2}t}}$ .

Now, we need to find the time when the greatest displacement is achieved.

This is the case when $x'(t)=0$

Since $x'(t)=5e^{-2\sqrt{2}t}-10\sqrt{2}te^{-2\sqrt{2}t}$ , the time when the greatest displacement is achieved when $0=5-10\sqrt{2}t\implies t=\frac{1}{2\sqrt{2}}\implies \color{red}\boxed{t=\frac{\sqrt{2}}{4}}$ .

Since we saw that the auxiliary equation had real repeated roots, we can tell that the equation of motion will be critically damped.

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My next post may be on another application of second order DE's [Electronic Circuits], or I may dive into systems of DEs and matrix methods.

Chris L T521 · #10 $Old$ December 27th, 2008, 10:56 AM

Something told me now is the best time to start updating this tutorial, especially since I'm currently on vacation from school. This post will be on electrical circuits.

Electrical Circuits

First, let me start off with a diagram.

The type of circuit we will analyze will be a RLC circuit.

In a RLC circuit, there is a(n):

- resistor with resistance $R$ ohms ( $\Omega$ )
- inductor with inductance $L$ henries
- capacitor with capacitance $C$ farads.
- Source with voltage $E$ volts.

There is a nice relationship between current, $I$ , and charge, $Q$ . Current is the rate of change of the flow of charges. Thus, we can say $I=\frac{\,dQ}{\,dt}$ .

According to the fundamental elementary principles of electricity, we see that the voltage drop across the three elements are as follows:

- Across a resistor, the voltage drop is resistance times current, or $RI$ .

- Across an inductor, the voltage drop is the inductance times the rate of change in the current, or $L\frac{\,dI}{\,dt}$ .

- Across a capacitor, the voltage drop is the charge divided by capacitance, or $\frac{1}{C}Q$ .

Now, we can analyze the behavior of the circuit by using one of Kirchoff's Laws:

The (Algebraic) sum of the voltage drops across the elements in a simple loop of an electrical circuit is equal to the applied voltage.

Thus, we see that if $E\left(t\right)$ is the applied voltage from the source, we get the equation $L\frac{\,dI}{\,dt}+RI+\frac{1}{C}Q=E\left(t\right)$ .

Using the relationship between current and charge, we can rewrite this as the second order non-homogeneous DE $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$ .

The solution to this differential equation, of course, is the amount of charge, $Q$ , at any given time, $t$ .

However, we usually want to solve for current. You can solve for current in one of two ways:

1) If you differentiate both sides of the differential equation, we get $L\frac{\,d^3Q}{\,dt^3}+R\frac{\,d^2Q}{\,dt^2}+\frac{1}{C}\frac{\,dQ}{\,dt}=E^{\prime}\left(t\right)\implies L\frac{\,d^2I}{\,dt^2}+R\frac{\,dI}{\,dt}+\frac{1}{C}I=E^{\prime}\left(t\right)$ . Now the solution to the DE is current, $I$ .

2) Once you solve $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$ , differentiate the solution to get current, $I$ .

Now, what happens if we are not dealing with a RLC Circuit?!?! We make slight modifications.

- If we have a RL Circuit, we solve the differential equation $L\frac{\,d^2Q}{\,dt^2}+R\frac{\,dQ}{\,dt}=E\left(t\right)$

- If we have a RC Circuit, we solve the differential equation $R\frac{\,dQ}{\,dt}+\frac{1}{C}Q=E\left(t\right)$

- If we have a LC Circuit, we solve the differential equation $L\frac{\,d^2Q}{\,dt^2}+\frac{1}{C}Q=E\left(t\right)$

Let us go through a couple of examples:

Example 22

Suppose that in an RLC Circuit, we have a resistance of 60 $\mathit{\Omega}$ , an inductance of 2 henries, and a capacitance of 0.0025 farads. Now, let the circuit have an emf of $\mathit{E\left(t\right)=100e^{-10t}}$ volts. Find the current in the circuit, given that the inital current in the circuit is zero, and the charge on the capacitor is one coloumb.

This is another initial value problem.

Here, we are to solve the DE $2\frac{\,d^2Q}{\,dt^2}+60\frac{\,dQ}{\,dt}+\frac{1}{0.0025}Q=100e^{-10t}$ , where $I\left(0\right)=0,~Q\left(0\right)=1$ .

The DE can be rewritten as $\frac{\,d^2Q}{\,dt^2}+30\frac{\,dQ}{\,dt}+200Q=50e^{-10t}$

In solving the homogeneous equation, we get the characteristic equation $r^2+30r+200=0\implies \left(r+20\right)\left(r+10\right)=0$ . We now see that this gives us $r_1=-20$ and $r_2=-10$ .

Thus, our complimentary solution is $Q_c=c_1e^{-20t}+c_2e^{-10t}$

Now, to find the particular solution, I will apply the method of the annihilator (in a sense, the annihilator method leads to the method of undetermined coefficients).

Rewriting the DE in differential operator notation, we get $\left(D^2+30D+200\right)\left(y\right)=50e^{-10t}$

The term that annihilates $50e^{-10t}$ is $\left(D+10\right)$

Thus, applying the annihilator to both sides of the DE, we can then convert the DE to the characteristic equation $(r+10)\left(r^2+30r+200\right)=0$

The particular solution is $r_p=-10$

Thus, it will take on the form $Q_p=Ate^{-10t}$

Now, substituting this into the original DE, we get $10Ae^{-10t}=50e^{-10t}$ (Verify)

You now get that $A=5$

Thus, the general solution to the DE is $Q\left(t\right)=c_1e^{-20t}+c_2e^{-10t}+5te^{-10t}$

Let us now apply the initial conditions:

$Q\left(0\right)=1\implies 1=c_1+c_2$

To apply the second condition, find $I\left(t\right)$

$I\left(t\right)=\frac{\,dQ}{\,dt}=-20c_1e^{-20t}-10c_2e^{-10t}-50te^{-10t}+5e^{-10t}$

Thus, $I\left(0\right)=0\implies 0=-20c_1-10c_2+5$ .

Solving these two equations for $c_1$ and $c_2$ , we get $c_1=-\tfrac{1}{2}$ and $c_2=-\tfrac{3}{2}$ (Verify)

Thus, the current is $I\left(t\right)=-\tfrac{1}{2}\left(-20\right)e^{-20t}-\left[\tfrac{3}{2}\left(-10\right)+5\right]e^{-10t}-50te^{-10t}=\color{red}\boxed{10e^{-20t}+10e^{-10t}-50te^{-10t}}$

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I will post another example when I find the time later today

Chris L T521 · #11 $Old$ August 23rd, 2009, 07:48 PM

Its been about 8 months since I've updated this. This post will probably be the first of two on systems of differential equations.

Systems of Differential Equations (Part I)

In all the previous posts, we dealt with differential equations that had one dependent variable. Now, we introduce the idea of a system of differential equations that have two or more dependent variables. For now, we consider first order systems of two (or three) differential equations.

When we construct our system, we consider the following:

$\begin{aligned}f\!\left(t,x,y,x^{\prime},y^{\prime}\right) & = 0\\g\!\left(t,x,y,x^{\prime},y^{\prime}\right) & = 0\end{aligned}$

where $t$ is the independent variable. A solution to this system would be a pair of functions $x\!\left(t\right)$ and $y\!\left(t\right)$ such that both equations were satisfied.

Let's go through the following example to introduce us to solving techniques.

Example 23

Find a general solution to the following system of differential equations:
$\left\{\begin{aligned}x^{\prime} & = y\\ y^{\prime} & = 2x+y\end{aligned}\right.$

To solve this, we will use techniques in solving second order differential equations.

Since $x^{\prime}=y$ , we see that when we differentiate the equation wrt x, we have $x^{\prime\prime}=y^{\prime}$ . Now take notice that $y^{\prime}$ was defined in the second equation. So it follows that $x^{\prime\prime}=y^{\prime}=2x+y$ . Also, since $x^{\prime}=y$ , it now follows that we have $x^{\prime\prime}=2x+x^{\prime}$ , which becomes the second order equation $x^{\prime\prime}-x^{\prime}-2x=0$ .

From here, its a walk in the park...

The characteristic equation is $r^2-r-2=0\implies \left(r+1\right)\left(r-2\right)=0$ . Thus, $r_1=-1$ and $r_2=2$ . Therefore, $\color{red}\boxed{x\!\left(t\right)=c_1e^{-t}+c_2e^{2t}}$ .

Now that we have a solution for x, we can find the solution for y, since $x^{\prime}=y$ . It now follows that $\color{red}\boxed{y\!\left(t\right)=-c_1e^{-t}+2c_2e^{2t}}$ .

These two functions form the solution to this system of differential equations.

Let's go through another simple example:

Example 24

Find a particular solution to the system of differential equations

$\left\{\begin{aligned}x^{\prime}&=-y\\y^{\prime}&=13x+4y\end{aligned}\right.$

given that $x(0)=0$ and $y(0)=3$ .

Again, we note that $x^{\prime}=-y\implies -x^{\prime\prime}=y^{\prime}$ .

We then substitute this value into the second equation to get

$-x^{\prime\prime}=13x+4y$ .

Now, substitute the first equation into the second to obtain the second order equation

$-x^{\prime\prime}=13x+4\left(-x^{\prime}\right)\implies x^{\prime\prime}-4x^{\prime}+13x=0$

The characteristic equation is $r^2-4r+13=0\implies r=\frac{4\pm\sqrt{16-52}}{2}\implies r=2\pm 3i$

Thus, $x(t)=e^{2t}\left[c_1\cos\!\left(3t\right)+c_2\sin\!\left(3t\right)\right]$

Since $-x^{\prime}=y$ , it follows that

$y(t)=-2e^{2t}\left[c_1\cos\!\left(3t\right)+c_2\sin\!\left(3t\right)\right]-e^{2t}\left[-3c_1\sin\!\left(3t\right)+3c_2\cos\!\left(3t\right)\right]$ $=e^{2t}\left[\left(-3c_2-2c_1\right)\cos\!\left(3t\right)+\left(3c_1-2c_2\right)\sin\!\left(3t\right)\right]$

We now apply the initial conditions:

$x(0)=0\implies 0=c_1$

$y(0)=3\implies 3=-3c_2-2c_1\implies c_2=-1$

Therefore, our pair of solutions to the system of differential equations is

$\color{red}\boxed{x(t)=-e^{2t}\sin\!\left(3t\right)}$ and $\color{red}\boxed{y(t)=e^{2t}\left[3\cos\!\left(3t\right)+2\sin\!\left(3t\right)\right]}$

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Let us now move on to a technique that is good for solving small systems of differential equations. (We will resort to matrix methods when we have 4 or more equations -- that will be the next post.)

The Method of Elimination

As the title suggests, we will use elimination techniques to help us reduce the system of equations into a differential equation with one unknown variable.

Let us consider a nth order linear differential operator

$L=a_nD^n+a_{n-1}D^{n-1}+\dots+a_1D+a_0$

where $D$ represents differentiation with respect to $t$ .

Let's now consider a system of differential equations defined by

$\left\{\begin{aligned}L_1x+L_2y &= f_1\!\left(t\right)\\L_3x+L_4y &= f_2\!\left(t\right)\end{aligned}\right.$

where $L_1$ , $L_2$ , $L_3$ and $L_4$ are (different) linear differential operators.

Let's say we wanted to eliminate the independent variable $x$ . Multiplying the first equation by $L_3$ and the second equation by $L_1$ , we have the system

$\left\{\begin{aligned}L_3L_1x+L_3L_2y &= L_3f_1\!\left(t\right)\\L_1L_3x+L_1L_4y &= L_1f_2\!\left(t\right)\end{aligned}\right.$

Since the linear differential operators multiply like regular polynomials, it follows that $L_1L_2=L_2L_1$ . Now we can subtract the two equations to get

$L_3L_2y-L_1L_4y=L_3f_1\!\left(t\right)-L_1f_2\!\left(t\right)\implies\left(L_3L_2-L_1L_4\right)y=L_3f_1\!\left(t\right)-L_1f_2\!\left(t\right)$

With minor manipulations, we end up with $\left(L_1L_4-L_2L_3\right)y=L_1f_2\!\left(t\right)-L_3f_1\!\left(t\right)\implies\begin{vmatrix}L_1 & L_2 \\ L_3 & L_4\end{vmatrix}y=\begin{vmatrix} L_1 & f_1\!\left(t\right)\\ L_3 & f_2\!\left(t\right)\end{vmatrix}$

Once we know what $y(t)$ is, we can then substitute it into either equation in the original system.

Similarly, if we eliminate $y$ , we end up with $\begin{vmatrix}L_1 & L_2 \\ L_3 & L_4\end{vmatrix}x=\begin{vmatrix} f_1\!\left(t\right) & L_2\\ f_2\!\left(t\right) & L_4\end{vmatrix}$

Let us go through a couple examples.

Example 25

Find the general solution for the system

$\left\{\begin{aligned}(D-4)x+3y &= 0\\-6x+(D+7)y&=0\end{aligned}\right.$

Let us first eliminate $x$ .

Then it follows that we have the equation

$\begin{vmatrix}D-4 & 3 \\ -6 & D+7\end{vmatrix}y=0\implies\left[(D-4)(D+7)-(-3)(6)\right]y=0$ $\implies \left(D^2+3D-10\right)y=0$ .

Now the characteristic equation is $r^2+3r-10=0$ . It follows that $r=-5$ or $r=2$ .

Thus, $y=b_1e^{2t}+b_2e^{-5t}$ .

If we choose to eliminate $y$ instead, we get

$\begin{vmatrix}D-4 & 3 \\ -6 & D+7\end{vmatrix}x=0\implies\left[(D-4)(D+7)-(-3)(6)\right]x=0$ $\implies \left(D^2+3D-10\right)x=0$ .

Thus, it follows that $x=a_1e^{2t}+a_2e^{-5t}$ .

However, there is a slight dilemma. It appears that our solution set contains four different arbitrary constants. However, by the Theorem for Existence and Uniqueness of Linear Systems, since we have two equations in our system, we should only have exactly two different arbitrary constants. So what now? The solution is simple: Substitute both functions into one of the equations in the original system.

If we substitute them into the first equation $(D-4)x+3y=0\implies x^{\prime}-4x+3y=0$ , we see that

$0=\left(2a_1e^{2t}-5a_2e^{-5t}\right)-4\left(a_1e^{2t}+a_2e^{-5t}\right)+3\left(b_1e^{2t}+b_2e^{-5t}\right)$ $=\left(-2a_1+3b_1\right)e^{2t}+\left(-9a_2+3b_2\right)e^{-5t}$ .

We now use the fact that $e^{2t}$ and $e^{-5t}$ are linearly independent. Thus, it follows that $-2a_1+3b_1=0\implies a_1=\tfrac{3}{2}b_1$ and $-9a_2+3b_2=0\implies a_2=\tfrac{1}{3}b_2$ .

Therefore, the general solution to our system is

$\color{red}\boxed{x(t)=\tfrac{3}{2}b_1e^{2t}+\tfrac{1}{3}b_1e^{-5t}}$ and $\color{red}\boxed{y=b_1e^{2t}+b_2e^{-5t}}$

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The next post in the tutorial will be on matrix methods to solving systems of differential equations. I will try to post that in the next couple days.

Chris L T521 · #12 $Old$ August 24th, 2009, 12:51 AM

I'm in such the mood to post Part II....so I'll do it now. XD

Systems of Differential Equations (Part II - Matrix Methods)

In part one, we covered basic techniques on how to solve first order system of two (or three) differential equations. What we will discuss in this post are techniques used in solving systems with a larger number of equations, and look at some non-linear systems.

Matrix-Valued Functions

A matrix-valued function is of the form

$\mathbf{x}(t)=\begin{bmatrix}x_(t)\\ x_2(t)\\ \vdots\\ x_n(t)\end{bmatrix}$ or $\mathbf{A}(t)=\begin{bmatrix}a_{11}(t) & a_{12}(t) & \dots & a_{1n}(t)\\ a_{21}(t) & a_{22}(t) & \dots & a_{2n}(t)\\ \vdots & \vdots & \phantom{x}& \vdots\\ a_{m1}(t) & a_{m2}(t) & \dots & a_{mn}(t)\end{bmatrix}$

where each entry is a function of $t$ . Now, $\mathbf{x}(t)$ or $\mathbf{A}(t)$ is differentiable if each entry is differentiable. Thus, we define $\frac{\,d\mathbf{A}}{\,dt}=\left[\frac{\,da_{ij}}{\,dt}\right]$

Let us now look into a popular method (which we will spend the rest of the post discussing) -- the Eigenvalue Method of Homogeneous Systems.

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Eigenvalue Method of Homogeneous Systems

Let us consider the following first order system of $n$ differential equations

$\left\{\begin{aligned}x_1^{\prime} &= a_{11}x_1+a_{12}x_2+\dots+a_{1n}x_n\\x_2^{\prime} &= a_{21}x_1+a_{22}x_2+\dots+a_{2n}x_n\\ &\vdots\\ x_n^{\prime} &= a_{n1}x_1+a_{n2}x_2+\dots+a_{nn}x_n\\\end{aligned}\right.$

It suffices to find $n$ linearly independent solution vectors $\mathbf{x}_1,\mathbf{x}_2,\dots,\mathbf{x}_n$ such that

$\mathbf{x}(t)=c_1\mathbf{x}_1+c_2\mathbf{x}_2+\dots+c_n\mathbf{x}_n$

is a solution to the general system.

We anticipate the solution vectors to be of the form

$\mathbf{x}(t)=\begin{bmatrix}x_1\\x_2\\x_3\\\vdots\\x_n\end{bmatrix}=\begin{bmatrix}v_1e^{\lambda t}\\v_2e^{\lambda t}\\v_3e^{\lambda t}\\\vdots\\v_ne^{\lambda t}\end{bmatrix}=\begin{bmatrix}v_1\\v_2\\v_3\\\vdots\\v_n\end{bmatrix}e^{\lambda t}=\mathbf{v}e^{\lambda t}$

where $\lambda,v_1,v_2,v_3,\dots,v_n$ are appropriate scalar constants.

To expand on this, let us rewrite our general system in matrix form:

$\mathbf{x}^{\prime}=\mathbf{Ax}$

Now, let us substitute the anticipated solution into the differential equation to get

$\left(\mathbf{v}e^{\lambda t}\right)^{\prime}=\mathbf{A}\left(\mathbf{v}e^{\lambda t}\right)\implies \lambda\mathbf{v}e^{\lambda t}=\mathbf{Av}e^{\lambda t}$

Cancelling out $e^{\lambda t}$ , we now have

$\lambda\mathbf{v}=\mathbf{Av}$ .

From this, we see that $\mathbf{x}=\mathbf{v}e^{\lambda t}$ will be a nontrivial solution of $\mathbf{x}^{\prime}=\mathbf{Ax}$ given that $\mathbf{v}\neq\mathbf{0}$ and such that $\mathbf{Av}$ is a scalar multiple of $\mathbf{v}$ .

So ... How do we find $\mathbf{v}$ and $\lambda$ ??

First, we rewrite $\lambda\mathbf{v}=\mathbf{Av}$ as $\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0}$ .

Now we recall from linear algebra, this equation has a nontrivial solution iff

$\det\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$ .

Thus, $\lambda$ is referred to the eigenvalue of $\mathbf{A}$ , and $\mathbf{v}$ is the associated eigenvector.

We also define $\det\left(\mathbf{A}-\lambda\mathbf{I}\right)=0$ to be the characteristic equation of $\mathbf{A}$ .

Now, we lay out the steps of the eigenvalue method:

1. First solve the characteristic equation for the eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ of the matrix $\mathbf{A}$ .

2. Attempt to find $n$ linearly independent eigenvectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_n$ associated with the eigenvalues.

3. If step 2 is possible (it may not always be!), we have $n$ linearly independent solutions $\mathbf{x}_1=\mathbf{v}_1e^{\lambda_1t}, \mathbf{x}_2=\mathbf{v}_2e^{\lambda_2t},\dots,\mathbf{x}_n=\mathbf{v}_ne^{\lambda_nt}$ . Thus, $\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)+\dots+c_n\mathbf{x}_n(t)$ is the general solution of $\mathbf{x}^{\prime}=\mathbf{Ax}$

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Let us now go through two special cases (each illustrated by an example):

Case I: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real and distinct.

Let us start with an example.

Example 26

Find a general solution for the system

$\left\{\begin{aligned}x_1^{\prime} & = 4x_1 + 2x_2\\ x_2^{\prime} &= 3x_1-x_2\end{aligned}\right.$

To solve this, let us rewrite the system in matrix form:

$\mathbf{x}^{\prime}=\begin{bmatrix}4 & 2\\3 & -1\end{bmatrix}\mathbf{x}$

It follows that the characteristic equation is

$\begin{vmatrix}4-\lambda & 2 \\ 3 & -1-\lambda\end{vmatrix}=-\left(4-\lambda\right)\left(1+\lambda\right)-6=\lambda^2-3\lambda-10=0$

Thus, $\lambda^2-3\lambda-10=0\implies\left(\lambda-5\right)\left(\lambda+2\right)=0\implies \lambda_1=-2$ and $\lambda_2=5$ .

Now that we have the eigenvalues, let us try to find the eigenvectors.

Note that the eigenvector equation in this case is

$\begin{bmatrix}4-\lambda & 2 \\ 3 & -1-\lambda\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

Case I: $\lambda=-2$ .

Here, the eigenvector equation becomes

$\begin{bmatrix}6& 2 \\ 3 & 1\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

This gives us the linear system

$\left\{\begin{aligned}6v_1+2v_2 & =0\\ 3v_1 + v_2 &= 0\end{aligned}\right.$ .

It is evident that there are infinitely many solutions. So what now? What we usually do is pick a simple value. So for example, if $v_1=1$ , we have $v_2=-3$ .

Therefore, $\mathbf{v}_1=\begin{bmatrix}1\\-3\end{bmatrix}$ is the eigenvector associated to $\lambda_1=-2$ . Thus, $\mathbf{x}_1(t)=\begin{bmatrix}1\\-3\end{bmatrix}e^{-2t}$ is a solution to the general equation.

Case II: $\lambda=5$ .

Here, the eigenvector equation becomes

$\begin{bmatrix}-1& 2 \\ 3 & -6\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

This gives us the linear system

$\left\{\begin{aligned}-v_1+2v_2 & =0\\ 3v_1 - 6v_2 &= 0\end{aligned}\right.$ .

It is evident that there are infinitely many solutions. So what now? What we usually do is pick a simple value. So for example, if $v_2=1$ , we have $v_1=2$ .

Therefore, $\mathbf{v}_2=\begin{bmatrix}2\\1\end{bmatrix}$ is the eigenvector associated to $\lambda_2=5$ . Thus, $\mathbf{x}_2(t)=\begin{bmatrix}2\\1\end{bmatrix}e^{5t}$ is a solution to the general equation.

It is easy to show that $e^{-2t}$ and $e^{5t}$ are linearly independent (via Wronskian).

Now, by the principle of superposition, it follows that

$\color{red}\boxed{\mathbf{x}(t)=c_1\begin{bmatrix}1\\-3\end{bmatrix}e^{-2t}+c_2\begin{bmatrix}2\\1\end{bmatrix}e^{5t}}$

satisfies $\mathbf{x}^{\prime}=\begin{bmatrix}4&2\\3&-1\end{bmatrix}\mathbf{x}$

(Written in scalar form, the solutions would be $\color{red}\boxed{\mathbf{x}_1(t)=c_1e^{-2t}+2c_2e^{5t}}$ and $\color{red}\boxed{\mathbf{x}_2(t)=-3c_1e^{-2t}+c_2e^{5t}}$ )

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Case II: $\lambda_1,\lambda_2,\dots,\lambda_n$ are complex.

Prelim Theory

We are after real valued solutions (it will turn out to be real and imaginary parts of the general solution). When complex eigenvalues pop up, they always appear in conjugate pairs (i.e. $\lambda=p+qi$ and $\bar{\lambda}=p-qi$ ).

Now, if $\mathbf{v}$ is an eigenvector associated with $\lambda$ , such that

$\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}=\mathbf{0}$ ,

then taking complex conjugates in the equation gives us

$\left(\mathbf{A}-\bar{\lambda}\mathbf{I}\right)\overline{\mathbf{v}}=\mathbf{0}$

If we take

$\mathbf{v}=\begin{bmatrix}a_1+b_1i\\a_2+b_2i\\\vdots\\a_n+b_ni\end{bmatrix}=\begin{bmatrix}a_1\\a_2\\\vdots\\a_n\end{bmatrix}+\begin{bmatrix}b_1\\b_2\\\vdots\\b_n\end{bmatrix}i=\mathbf{a}+\mathbf{b}i$ ,

then $\overline{\mathbf{v}}=\mathbf{a}-\mathbf{b}i$

Therefore, the complex-valued solution associated with $\lambda$ and $\mathbf{v}$ is

$\mathbf{x}(t)=\mathbf{v}e^{\lambda t}=\mathbf{v}e^{\left(p+qi\right)t}=\left(\mathbf{a}+\mathbf{b}i\right)e^{pt}\left[\cos\!\left(qt\right)+\sin\!\left(qt\right)\right]$

Rearranging, we have

$\mathbf{x}(t)=e^{pt}\left[\mathbf{a}\cos\!\left(qt\right)-\mathbf{b}\sin\!\left(qt\right)\right]+ie^{pt}\left[\mathbf{b}\cos\!\left(qt\right)+\mathbf{a}\sin\!\left(qt\right)\right]$ .

Therefore,

$\begin{aligned}\mathbf{x}_1(t)&=\Re\left(\mathbf{x}(t)\right)=e^{pt}\left[\mathbf{a}\cos\!\left(qt\right)-\mathbf{b}\sin\!\left(qt\right)\right]\\\mathbf{x}_2(t)&=\Im\left(\mathbf{x}(t)\right)=e^{pt}\left[\mathbf{b}\cos\!\left(qt\right)+\mathbf{a}\sin\!\left(qt\right)\right]\end{aligned}$

I leave it for you to verify we get the same set of solutions when we check the real and imaginary parts of $\overline{\mathbf{v}}e^{\bar{\lambda}t}$ .

Example 27

Find the general solution of the system

$\begin{aligned}x_1^{\prime} &= 4x_1-3x_2\\ x_2^{\prime}&= 3x_1+4x_2\end{aligned}$

Our coefficient matrix $\mathbf{A}=\begin{bmatrix}4&-3\\3&4\end{bmatrix}$ has the characteristic equation

$\begin{bmatrix}4-\lambda & -3 \\ 3 & 4-\lambda\end{bmatrix}=\left(4-\lambda\right)^2+9=0\implies \lambda=4-3i$ and $\bar{\lambda}=4+3i$ .

Substituting $\lambda=4-3i$ into the eigenvector equation, we have

$\begin{bmatrix}3i & -3\\ 3 & 3i\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

Thus, we have the linear system

$\left\{\begin{aligned}iv_1-v_2 & = 0\\ v_1 + iv_2 & = 0\end{aligned}\right.$

If we take $v_1=1$ , $v_2=i$ . Thus, $\mathbf{v}=\begin{bmatrix}1\\i\end{bmatrix}$ is complex eigenvector associated with $\lambda=4-3i$ .

Now, the corresponding complex solution is

$\mathbf{x}(t)=\begin{bmatrix}1\\i\end{bmatrix}e^{\left(4-3i\right)t}=\begin{bmatrix}1\\i\end{bmatrix}e^{4t}\left(\cos\!\left(3t\right)-i\sin\!\left(3t\right)\right)=e^{4t}\begin{bmatrix}\cos\!\left(3t\right)-i\sin\!\left(3t\right)\\i\cos\!\left(3t\right)+\sin\!\left(3t\right)\end{bmatrix}$

Thus,

$\mathbf{x}_1(t)=\Re\left(\mathbf{x}(t)\right)=e^{4t}\begin{bmatrix}\cos\!\left(3t\right)\\\sin\!\left(3t\right)\end{bmatrix}$ and $\mathbf{x}_2(t)=\Im\left(\mathbf{x}(t)\right)=e^{4t}\begin{bmatrix}-\sin\!\left(3t\right)\\\cos\!\left(3t\right)\end{bmatrix}$

Therefore, a real-valued general solution to $\mathbf{x}^{\prime}=\mathbf{Ax}$ is

$\color{red}\boxed{\mathbf{x}(t)=c_1\mathbf{x}_1(t)+c_2\mathbf{x}_2(t)=e^{4t}\begin{bmatrix}c_1\cos\!\left(3t\right)-c_2\sin\!\left(3t\right)\\c_1\sin\!\left(3t\right)+c_2\cos\!\left(3t\right)\end{bmatrix}}$ .

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I will have to post a Part III for Case III: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real, but not distinct.

I will have that posted sometime tomorrow or the next day.

Chris L T521 · #13 $Old$ August 29th, 2009, 11:03 PM

System of Differential Equations (Part III - Matrix Methods (cont.))

In Part II, we ended with two special cases for the eigenvalues of an n x n matrix system. We now devote an entire post
to the last special case.

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Case III: $\lambda_1,\lambda_2,\dots,\lambda_n$ are real but not distinct.

When $\lambda_1,\lambda_2,\dots,\lambda_n$ were distinct (real or complex), then the general solution of
$\mathbf{x}^{\prime}=\mathbf{Ax}$ took on the form

$\mathbf{x}(t)=c_1\mathbf{v}_1e^{\lambda_1t}+c_2\mathbf{v}_2e^{\lambda_2t}+\dots+c_n\mathbf{v}_ne^{\lambda_nt}$ .

We now consider when the characteristic equation $\left|\mathbf{A}-\lambda\mathbf{I}\right|=0$ doesn't have n
distinct root --> the characteristic equation has at least one repeated root.

In that case, we refer to the eigenvalue as having multiplicity. An eigenvalue is of multiplicity k if it is a k-fold
root of the characteristic equation. If $\lambda$ is of multiplicity k, then there is at least one
eigenvector $\mathbf{v}$ associated with it. However, we may not always be able to find k linearly
independent eigenvectors associated with $\lambda$ (this is referred to as a defect of $\lambda$ ,
which will be discussed later). If we can find k linearly independent eigenvectors associated with $\lambda$ ,
we say that $\lambda$ is complete.

Example 28

Find a general solution of the system

$\mathbf{x}^{\prime}=\begin{bmatrix}9 & 4 & 0\\-6 & -1 & 0\\6 & 4 & 3\end{bmatrix}\mathbf{x}$

The characteristic equation of $\mathbf{A}=\begin{bmatrix}9 & 4 & 0\\-6 & -1 & 0\\6 & 4 & 3\end{bmatrix}$ is

$\begin{vmatrix}9-\lambda & 4 & 0\\-6 & -1-\lambda & 0\\6 & 4 & 3-\lambda\end{vmatrix}=(3-\lambda)\begin{vmatrix}9 -\lambda & 4\\ -6 & -1-\lambda\end{vmatrix}$ $=(3-\lambda)(\lambda^2-8\lambda+15)=(5-\lambda)(3-\lambda)^2=0$

Here, we see that $\lambda_1=5$ and $\lambda_2=3$ with multiplicity 2.

Case I: $\lambda=5$

The eigenvector equation is

$\begin{bmatrix}4 & 4 & 0\\-6 & -6 & 0\\6 & 4 & -2\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin {bmatrix}0\\0\\0\end{bmatrix}$

Thus, we have the following system of equations:

$\left\{\begin{aligned}4v_1+4v_2 & = 0\\-6v_1-6v_2&=0\\6v_1+4v_2-2v_3&=0\end{aligned}\right.$

The first two deduce to $v_2=-v_1$

Now, it follows the third equation can be written as $2v_1-2v_3=0\implies v_3=v_1$ . Thus, if we pick $v_1=1$ , we have the eigenvector $\mathbf{v}_1=\begin{bmatrix}1\\-1\\1\end{bmatrix}$ associated with
$\lambda=5$ .

Case II: $\lambda=3$

The eigenvector equation is

$\begin{bmatrix}6 & 4 & 0\\-6 & -6 & 0\\6 & 4 & 0\end{bmatrix}\begin{bmatrix}v_1\\v_2\\v_3\end{bmatrix}=\begin {bmatrix}0\\0\\0\end{bmatrix}$

Thus, we have a nonzero eigenvector iff $6v_1+4v_2=0\implies v_2=-\tfrac{3}{2}v_1$ . Thus, $v_3$
is arbitrary. So if we pick $v_3=1$ , we can let $v_1=v_2=0$ . Thus, $\mathbf{v}_2=\begin {bmatrix}0\\0\\1\end{bmatrix}$ is an associated eigenvector to $\lambda=3$ . However, there is one more eigenvector!

If we pick $v_3=0$ , we can pick $v_1$ and $v_2$ such that we don't have the zero vector. So if we take $v_2=2$ , we see that $v_3=-3$ . Thus, $\mathbf{v}_3=\begin{bmatrix}2\\-3\\0\end{bmatrix}$ is the eigenvector associated with $\lambda=3$ .

Therefore, the general solution is

$\color{red}\boxed{\mathbf{x}(t)=c_1\begin{bmatrix}1\\-1\\1\end{bmatrix}e^{5t}+c_2\begin{bmatrix}0\\0\\1\end{bmatrix}e^{3t}+c_3\begin{bmatrix}2\\-3\\0\end{bmatrix}e^{3t}}$

Remark: With regards to the two eigenvectors for $\lambda=3$ , the fact that $v_2=-\tfrac{3}{2}v_1$ is worth taking note of. The eigenvector can be rewritten as

$\mathbf{v}=\begin{bmatrix}v_1\\-\frac{3}{2}v_1\\v_3\end{bmatrix}=v_3\begin{bmatrix}0\\0\\1\end{bmatrix}+\tfrac{1}{2}v_1\begin{bmatrix}2\\-3\\0\end{bmatrix}=v_3\mathbf{v}_2-\tfrac{1}{2}v_1\mathbf{v}_3$

Thus, we could replace $\mathbf{v}$ for the eigenvector and still get the same answer we did when considering both eigenvectors. This tells us that we don't have to worry about making the right choice -- its just advisible that we pick the simplest one.

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Defective Eigenvalues

We start this section with an example.

Example 29

Consider the coefficient matrix $\mathbf{A}=\begin{bmatrix}1 &-3\\3 & 7\end{bmatrix}$ .

The characteristic equation is $\begin{vmatrix}1-\lambda & -3\\ 3 & 7-\lambda\end{vmatrix}=\lambda^2-8\lambda+16=\left(\lambda-4\right)^2=0$ .

Thus, $\lambda=4$ is an eigenvalue of multiplicity two.

Now, the eigenvector equation is

$\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}v_1\\v_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$ .

Thus, it follows that our system of equation is

$\left\{\begin{aligned}-3v_1-3v_2 & = 0\\3v_1 + 3v_2 & = 0\end{aligned}\right.$

Thus, $v_2=-v_1$ .

Thus the eigenvector is of the form $\mathbf{v}=\begin{bmatrix}v_1\\-v_1\end{bmatrix}=v_1\begin{bmatrix}1\\-1\end{bmatrix}$ .

This implies that all eigenvectors associated with $\lambda=4$ will be a constant multiple of $\begin{bmatrix}1\\-1\end{bmatrix}$ . Therefore, there is only one linearly independent eigenvector associated with $\lambda=4$ , making $\lambda=4$ incomplete.

The eigenvalue in the above example is incomplete, or defective.

Now, if an eigenvalue $\lambda$ has $p<k$ linearly independent eigenvectors, then $d=k-p$ is the number of missing eigenvectors - the defect of the defective eigenvalue $\lambda$ .

In Example 29, the defect would be $d=2-1=1$ .

What we do now is consider a way to solve a system of differential equations given the defect $d=1$ .

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Case IV: $\lambda$ has multiplicity two and is defective.

Suppose that $\lambda$ has one linearly independent eigenvector, implying that $\mathbf{x}_1(t)=\mathbf{v}_1e^{\lambda t}$ is the only solution (that we know of) to $\mathbf{x}^{\prime}=\mathbf{Ax}$ .

However, we hope to find a second solution of the form $\mathbf{x}_2(t)=\mathbf{v}_2te^{\lambda t}$ . Substituting it into the system, we have

$\mathbf{v}_2e^{\lambda t}+\lambda\mathbf{v}_2te^{\lambda t}=\mathbf{Av}_2te^{\lambda t}$ .

Since the coefficients of $e^{\lambda t}$ and $te^{\lambda t}$ need to balance, it follows from the above equation that $\mathbf{v}_2=\mathbf{0}$ and consequently, $\mathbf{x}_2(t)\equiv\mathbf{0}$ .

Since that didn't work, let us extend our original idea and replace $\mathbf{v}_2t$ with $\mathbf{v}_1t+\mathbf{v}_2$ . So we suppose now that the second solution will take on the form

$\mathbf{x}_2(t)=\mathbf{v}_1te^{\lambda t}+\mathbf{v_2}e^{\lambda t}$ .

Substituting this into $\mathbf{x}^{\prime}=\mathbf{Ax}$ , we get

$\left(\mathbf{v}_1+\lambda\mathbf{v}_2\right)e^{\lambda t}+\lambda\mathbf{v}_1te^{\lambda t}=\mathbf{Av}_1te^{\lambda t}+\mathbf{Av}_2e^{\lambda t}$

Comparing coefficents of $e^{\lambda t}$ and $te^{\lambda t}$ , we see that

$\mathbf{v}_1+\lambda\mathbf{v}_2=A\mathbf{v}_2\implies \left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v_2}=\mathbf{v_1}$

and

$\lambda\mathbf{v}_1=\mathbf{Av}_1\implies \left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_1=\mathbf{0}$

The second equation confirms that $\mathbf{v}_1$ is an eigenvector for $\lambda$ . Now, it follows that $\mathbf{v}_2$ satisfies the equation

$\left(\mathbf{A}-\lambda\mathbf{I}\right)^2\mathbf{v}_2=\left(\mathbf{A}-\lambda\mathbf{I}\right)\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_2=\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_1=\mathbf{0}$

This tells us that it suffices to find a single solution $\mathbf{v}_2$ to the equation $\left(\mathbf{A}-\lambda\mathbf{I}\right)^2\mathbf{v}_2=\mathbf{0}$ such that $\mathbf{v}_1=\left(\mathbf{A}-\lambda\mathbf{I}\right)\mathbf{v}_2\neq\mathbf{0}$ .

It is always possible to find a solution when the defective eigenvalue $\lambda$ has multiplicity two.

Let us go through an example that illustrates this process.

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Example 30

Find the general solution to the system

$\mathbf{x}^{\prime}=\begin{bmatrix}1 & -3\\ 3 & 7\end{bmatrix}\mathbf{x}$

In example 29, we showed that the characteristic equation produced a defective eigenvalue $\lambda=4$ of multiplicity two.

We now start by calculation $\left(\mathbf{A}-4\mathbf{I}\right)^2$ :

$\left(\mathbf{A}-4\mathbf{I}\right)^2=\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$

Thus, $\left(\mathbf{A}-4\mathbf{I}\right)^2\mathbf{v}_2=\mathbf{0}\implies \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\mathbf{v}_2=\mathbf{0}$ implies that $\mathbf{v}_2$ can be of any (nonzero) form.

So if we take $\mathbf{v}_2=\begin{bmatrix}1\\0\end{bmatrix}$ , then we see that

$\left(\mathbf{A}-4\mathbf{I}\right)\mathbf{v}_2=\begin{bmatrix}-3 & -3\\3 & 3\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}-3\\3\end{bmatrix}=\mathbf{v}_1$ .

This eigenvector is nonzero, and thus associated with the eigenvalue $\lambda=4$ (Note that this is -3 times the eigenvector we found in example 29).

Therefore, the two solutions to the system are

$\mathbf{x}_1(t)=\mathbf{v}_1e^{4t}=\begin{bmatrix}-3\\3\end{bmatrix}e^{4t}$

and

$\mathbf{x}_2(t)=\left(\mathbf{v}_1t+\mathbf{v}_2\right)e^{4t}=\begin{bmatrix}-3t+1\\3t\end{bmatrix}e^{4t}$

Therefore, the general solution to the system is

$\color{red}\boxed{\mathbf{x}(t)=c_1\begin{bmatrix}-3\\3\end{bmatrix}e^{4t}+c_2\begin{bmatrix}-3t+1\\3t\end{bmatrix}e^{4t}=\begin{bmatrix}-3c_1-3c_2t+c_2\\3c_1+3c_2t\end{bmatrix}e^{4t}}$

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This will conclude the systems of differential equations section of the tutorial.

I will start working on the first of three (or maybe four) posts on Laplace Transforms and their use in IVPs.

Chris L T521 · #14 $Old$ August 30th, 2009, 11:15 PM

Laplace Transforms (Part I - Introduction, IVPs and Partial Fraction Techniques)

There are many types of transformations out there. For example, differentiation and integration are types of linear transformations. However, there is one particular transform that we would like to analyze. This transform is of the form:

$\int_0^{\infty}K\!\left(s,t\right)f\!\left(t\right)\,dt$

where $K\!\left(s,t\right)$ is called the kernel of the transformation.

In this case, we are interested in the transform with a kernel of $K\!\left(s,t\right)=e^{-st}$ . With this kernel, we take $f\!\left(t\right)$ and transform it into another function $F\!\left(s\right)$ . This transformation described by $\int_0^{\infty}e^{-st}f\!\left(t\right)\,dt=F\!\left(s\right)$ is called the Laplace Transform. It is denoted by $\mathcal{L}\left\{f\!\left(t\right)\right\}$ .

Before we go and derive all the common Laplace Transforms (we will derive many more as we get futher into later posts), let us take a look at a familar function to some of us (this may also be totally knew to some of you out there).

Given $x\in\mathbb{R}$ , where $x>0$ , we define the Gamma Function $\Gamma\!\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$ . It has the property $\Gamma\!\left(1\right)=1$ and $\Gamma\!\left(x+1\right)=x\Gamma\!\left(x\right)$ .

Now, if $n\in\mathbb{N}$ , then it follows by a similar idea that $\Gamma\left(n+1\right)=n\Gamma\!\left(n\right)$ . If we continue simplifying, we have

$\begin{aligned}n\Gamma\!\left(n\right) & = n\left(n-1\right)\Gamma\!\left(n-1\right)\\ &= n\left(n-1\right)\left(n-2\right)\Gamma\!\left(n-2\right)\\ &\vdots\\ &=n\left(n-1\right)\left(n-2\right)\dots2\cdot\Gamma\!\left(2\right)\\ &= n\left(n-1\right)\left(n-2\right)\dots2\cdot1\cdot\Gamma\!\left(1\right)\end{aligned}$

This implies that when $n\in\mathbb{N}$ , $\Gamma\!\left(n+1\right)=n!$ .

(Thus it is interesting to point out that since $1\in\mathbb{N},\,\Gamma\!\left(1\right)=\color{red}\boxed{1=0!}$ , an identity for factorials.)

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Common Laplace Transforms

In this part, I will list the common Laplace Transforms, and leave the derivation of each in a spoiler for you to look at if you decide too.

$\mathcal{L}\left\{1\right\}=\frac{1}{s},\,s>0$

Spoiler:

$\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a},\,s>a$ (This will pop up again, when we talk about translation theorems)

Spoiler:

$\mathcal{L}\left\{t^a\right\}=\frac{\Gamma\left(a+1\right)}{s^{a+1}},\,s>0$ ; If $n\in\mathbb{N},\,\mathcal{L}\left\{t^n\right\}=\frac{\Gamma\!\left(n+1\right)}{s^{n+1}}=\frac{n!}{s^{n+1}}$

Spoiler:

Given