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Chris L T521 · #3 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...

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