# [Solved]Another second order DE

1. Oct 10, 2011

### fluidistic

1. The problem statement, all variables and given/known data
I must solve $x^2y''-2y=x$.

2. Relevant equations
Not sure. Reduction of order maybe.

3. The attempt at a solution
I notice that $y_1=x^2$ is solution to the homogeneous DE $x^2y''-2y=0$.
I propose another solution of the form $y_2(x)=v(x)y_1(x)$.
Plugging this y_2 and its second derivative into the original DE, I reach that $v''x^4+4x^3v'=x$.
I called a new variable $u=v'$ to reduce the order of the DE. So the DE reduces to $u'x^4+4x^3u=x$.
I then used the integrating factor method to solve the DE. As integrating factor, I reached $e^{x^4}$. So that $u(x)=\frac{1}{e^{x^4}} \left [ \int{xe^{x^4}dx+C_1} \right ]$. I tried to evaluate this integral via Wolfram Alpha and it returned the imaginary erf function so I prefer to let it under this form.
So now my $v(x)=\int u(x)dx+C_2$ which looks extremely "ugly".

I just tried to solve the original DE into Wolfram and it gives the solution $y(x)=c_1x^2+\frac{c_2}{x}-\frac{x}{2}$. I really don't know what I did wrong. (In http://tutorial.math.lamar.edu/Classes/DE/ReductionofOrder.aspx it seems that the original DE must be homogeneous in order to apply the reduction of order method; however it seems I could do the same even though the DE isn't homogeneous. Could you clarify if I'm right on this?)
Did I do something wrong? How could I reach a "beautiful" result such the one given by Wolfram Alpha?

EDIT: Nevermind, I found out my error. I reach exactly the same answer as Wolfram Alpha. It was when I applied the integrating factor method, I forgot to divide the DE by x⁴ before doing the math. Nice!

Last edited: Oct 10, 2011