# Ordinary differential equations

• Apr 30th 2011, 12:18 PM
Ulysses
Ordinary differential equations
Hi there. Well, I have some doubts about this exercise, I think I've solved it, but I wanted your opinion, which always help. So, it says:

There is no general method for solving the homogeneous general equation of second order
$y''+P(x)y'+Q(x)y=0$ (1)

But if we already know a solution $y_1(x)\neq 0$, then we always can find a second solution linearly independent

$y_2(x)=y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx$
Demonstrate that this functions form a basis for the space of solutions of the differential equation (1).

So, what I did is simple. I know that if the solution is linearly independent, then the Wronskian $w(y_1,y_2,x)$ must be zero. So, I just made the calculus for the wronskian:

$\left| \begin{matrix}{y_1}&{y_2}\\{y_1'}&{y_2'}\end{matri x} \right|=\left| \begin{matrix}{y_1}&{y_1(x) \int \dysplaystyle\frac{e^{-\int P(x)dx}}{y_1^2}dx}\\{y_1'}&{y_1'(x) \int \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2}dx+y_1(x) \displaystyle\frac{e^{- \int P(x)dx}}{y_1^2(x)}dx}\end{matrix} \right|=\displaystyle\frac{e^{- \int P(x)dx}}{y_1(x)}\neq0$

Is this right?

Bye there.
• Apr 30th 2011, 12:24 PM
Ackbeet
Your approach is fine; however I don't think your determinant computation is correct. You've set it up properly, but I get just

$e^{-\int P(x)\,dx}$

for the determinant.
• Apr 30th 2011, 12:25 PM
Ulysses
You're right! thanks.
• Apr 30th 2011, 12:27 PM
Ackbeet
You're welcome!
• Apr 30th 2011, 01:07 PM
Ulysses
This is pretty much like the same. It gives me a differential equation, and one solution and asks me for the entire solution. Did I do it well?

$(1-2x)y''+2y'+(2x-3)y=0,y_1=e^x$
$y_2=e^x \int \displaystyle\frac{e^{-\int 2dx}}{e^{2x}}dx=\displaystyle\frac{e^{-3x}}{-4}$

Then $y(x)=Ae^x-B\displaystyle\frac{e^{-3x}}{4}$

Is that okey? I'm not sure, because I've tried to corroborate the solution with wolfram alpha, but it gives something different.

Here:
&#40;1-2x&#41;y&#39;&#39;&#43;2y&#39;&#43;&#40;2x-3&#41;y&#61;0 - Wolfram|Alpha
• Apr 30th 2011, 02:01 PM
Ackbeet
You haven't normalized the DE to look like (1) in the OP. You have to do that before applying the variation of parameters formula.
• Apr 30th 2011, 03:15 PM
Ulysses
Sorry, I've missed that part, how should I do that?
• Apr 30th 2011, 03:16 PM
Ackbeet
Just divide the entire DE by 1-2x to get the coefficient of y'' equal to 1.