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Chris L T521 · #11 August 23rd, 2009, 06:48 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.

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