# Variation of parameters applied to an ODE

1. Dec 4, 2012

### fluidistic

The ODE to solve via variation of parameters is $(1-x)y''+xy'-y=(1-x)^2$.
Knowing that $e^x$ and $x$ are solutions to the homogeneous ODE.
Now if I call $y_1=x$ and $y_2=e^x$, the Wronskian is $W(y_1,y_2)=e^{x}(x-1)$.
According to http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx, the particular solution of the non homogeneous ODE should be of the form $-y_1 \int \frac{y_2 (1-x)^2}{W(y_1,y_2)} dx+y_2 \int \frac{y_1(1-x)^2}{W(y_1,y_2)}$. This gave me $y_p(x)=-\frac{x^3}{2}-x-1$ I've even checked out with wolfram alpha the evaluation of the integrals, that is the result. However the answer is $y_p=x^2+1$.
I'm clueless on what's going on.

2. Dec 4, 2012

### Mute

If you take a careful look at that page you linked to, you'll notice that although the author writes the ODE as $p(x)y'' + q(x)y' + r(x)y = g(x)$, he eventually states that he will assume p(x) = 1. The wikipedia page for variation of parameters similarly makes this assumption. Try diving your differential equation by (1-x) and then applying the $-y_1 \int \frac{y_2 g(x)}{W(y_1,y_2)} dx+y_2 \int \frac{y_1g(x)}{W(y_1,y_2)}$ formula, where g(x), the non-homogeneous term, is (1-x) rather than (1-x)2.

Last edited: Dec 4, 2012
3. Dec 4, 2012

### fluidistic

Thank you very much. I totally missed this.