MHB -242.17.8 Solve y''-10y'+25y&=2e^{5x} by variation of parameters.

[karush]

$\tiny{242.17.8}$
2000
$\textrm{Solve the given equation by variation of parameters.}$
\begin{align*}\displaystyle
y''-10y'+25y&=2e^{5x}\\
\end{align*}
$\textrm{the homogeneous equation:}$
\begin{align*}\displaystyle
x^2-10x+25&=0\\
(x-5)^2&=0\\
x&=5\\
y_h&=c_1 e^{5x}+c_2x^{5x}
\end{align*}
$\textit{now what}$

 

[MarkFL]

Your fundamental solution set for the corresponding homogeneous equation is:

$$\left\{e^{5x},xe^{5x}\right\}$$

So now we take as our particular solution:

$$y_p(x)=v_1(x)e^{5x}+v_2(x)xe^{5x}$$

We may determine $v_1(x)$ and $v_2(x)$ by solving the system:

$$e^{5x}v_1'+xe^{5x}v_2'=0$$

$$\left(e^{5x}\right)'v_1'+\left(xe^{5x}\right)'v_2'=2e^{5x}$$

for $v_1'(x)$ and $v_2'(x)$ and integrating. :D

 

[MarkFL]

Let's first solve this ODE using the method of undetermined coefficients. We know the homogeneous solution is:

$$y_h(x)=c_1e^{5x}+c_2xe^{5x}$$

And so upon inspection of the RHS, we see that the particular solution will take the form:

$$y_p(x)=Ax^2e^{5x}$$

Differentiating, we obtain:

$$y_p'(x)=Ax(5x+2)e^{5x}$$

$$y_p''(x)=A\left(25x^2+20x+2\right)e^{5x}$$

Substituting into the ODE, we have:

$$\left(A\left(25x^2+20x+2\right)e^{5x}\right)-10\left(Ax(5x+2)e^{5x}\right)+25\left(Ax^2e^{5x}\right)=2e^{5x}$$

Distributing, and combining like terms, we get:

$$2Ae^{5x}=2e^{5x}$$

From this, we conclude $A=1$ and so:

$$y_p(x)=x^2e^{5x}$$

And so the solution is:

$$y(x)=y_h(x)+y_p(x)=c_1e^{5x}+c_2xe^{5x}+x^2e^{5x}$$

Okay, now back to variation of parameters. We need to solve the system:

$$e^{5x}v_1'+xe^{5x}v_2'=0$$

$$5e^{5x}v_1'+(5x+1)e^{5x}v_2'=2e^{5x}$$

First, let's divide through both equations by $e^{5x}\ne0$ to obtain:

$$v_1'+xv_2'=0$$

$$5v_1'+(5x+1)v_2'=2$$

Let's multiply the first equation by -5 and then add:

$$v_2'=2\implies v_1'=-2x$$

And so integrating (we need only 1 particular solution, so for simplicity, we take both constants of integration to be zero for simplicity...can you explain why not doing so would in fact not yield anything useful to our particular solution?), we obtain:

$$v_1=-x^2,\,v^2=2x$$

And so, our particular solution is:

$$y_p(x)=\left(-x^2\right)e^{5x}+\left(2x\right)xe^{5x}=x^2e^{5x}$$

This checks with what we found using the method of undetermined coefficients, and so we likewise conclude:

$$y(x)=y_h(x)+y_p(x)=c_1e^{5x}+c_2xe^{5x}+x^2e^{5x}$$

 

[karush]

just want to thank you again for clearing up the fog.

btw

how do you manage to crank out the mountains of LaTEX you do to help everbody?

or do you have huge cut and paste library?

 

[MarkFL]

btw

how do you manage to crank out the mountains of LaTEX you do to help everbody?

or do you have huge cut and paste library?

After almost 7 years of helping with math online, I think in $\LaTeX$ now. (Nerd)

It really takes little time...and when I am posting a series of equations, I typically just copy the last line and then modify it. :D