1. ## variation of parameters

I have to use variation of parameters to find the general solution of:

x^2y'' + xy' - y = x^2 e^x

but all the examples I've got don't have x's in and tell you what the fundamental set of solutions are...so could anyone start me off or tell me in steps how I would go about doing this question?

2. This is a form of Euler Equation:

$

x^2 y'' + xy' - y = x^2 e^x

$

--------------------------------------------------------------------------
$
\\
\\
$

I'll show you how to find the homogeneous solution. and you'll go from there:

The trick is to use the following substitution:

$
t = \ln x
\\
$

thus :

$
\begin{array}{l}
x = e^t \\
\frac{d}{{dx}}y = \frac{{dy}}{{dt}}\frac{{dt}}{{dx}} = \frac{{dy}}{{dt}}\frac{1}{x} \\
\end{array}

$

$

\frac{{d^2 }}{{dx^2 }}y = \frac{d}{{dx}}\left( {\frac{{dy}}{{dt}}\frac{1}{x}} \right) = - \frac{1}{{x^2 }}\frac{{dy}}{{dt}} + \frac{1}{x}\frac{{d^2 y}}{{dt^2 }}\frac{{dt}}{{dx}} = \frac{1}{{x^2 }}\left( {\frac{{d^2 y}}{{dt^2 }} - \frac{{dy}}{{dt}}} \right)

$

after substituting these results into the ode we get a regular constant coefficients ODE:

$

\begin{array}{l}
x^2 \left[ {\frac{1}{{x^2 }}\left( {\frac{{d^2 y}}{{dt^2 }} - \frac{{dy}}{{dt}}} \right)} \right] + x\left( {\frac{{dy}}{{dt}}\frac{1}{x}} \right) - y = 0 \\
\\
y'' - y' + y' - y = 0 \\
\\
y'' - y = 0 \\
\end{array}
$

the characteristic polynomial is:

$

\begin{array}{l}
\lambda ^2 - 1 = 0 \\
\lambda = \pm 1 \\
\end{array}
$

so the general solution is:

$

y = Ae^{ - t} + Be^t
$

or in terms of x:

$

y = A\frac{1}{x} + Bx

$

now you can find the particular solution using the method of variation of parameters...

[The latex came kinda massed up, don't know how to fix it]

3. OK I've already found one of the solutions to be x, but can't get the other one...I've been working on this for hours!

Thanks for replying, but I don't understand how to use that substitution?? I only need to get the other solution and I'll be OK from there!

EDIT: wait you've just explained it for me

4. Ohhhhhhh so the solutions are just y1 = x and y2 = x^-1?

Thanks so much for explaining that so clearly!

5. but I don't understand how to use that substitution?
you change the dependent variable from x to t and use the chain rule to express the derivatives of y in the new variable t.

6. Originally Posted by hunkydory19
Ohhhhhhh so the solutions are just y1 = x and y2 = x^-1?

Thanks so much for explaining that so clearly!
No. $y_1=|x|$ and $y_2 = |x|^{-1}$ are the linearly independent solution which form a basis for the equation, $x^2 y'' + xy' - y = 0$ on $(-\infty,0)\cup (0,\infty)$. But to solve $x^2y'' + xy' - y = x^2 e^x$ you need to use Variation of Parameters and find a particular solution $p(x)$ in that case the solution will be $k_1y_1+k_2y_2 + p(x)$ for $k_1,k_2\in \mathbb{R}$.