x^2 y'' - 5x y' - 7y = x^3

The question asks me to solve x^2 y'' - 5x y' - 7y = x^3 as function of x. In other words, it's asking for y(x). I am given the initial conditions y(1) = 10 and y'(1) = 8.

Someone who wasn't very sure told me to use characteristic equations and the method of undetermined coefficients but is that correct? If so, how do I set it up mechanically? I only know how to use characteristic equations when there are no x variables in the differential equation. I am really, really stuck. I only divided both sides by x^2.

The answer is:

http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3;+y(1)+%3D+10,+y'(1)+%3D+8

If someone could show me how to set up the problem (tedious steps aren't necessary), I would greatly appreciate it!

Thanks in advance!

Re: x^2 y'' - 5x y' - 7y = x^3

Re: x^2 y'' - 5x y' - 7y = x^3

Sorry for the dumb question but, is this process just a convenient trick or is it how I am supposed to solve most differential equations with x variables in them? Also, is there a name for this process other than just "substitution?"

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Re: x^2 y'' - 5x y' - 7y = x^3

Also, I think I almost get it but I get the wrong answer. I'm stuck at evaluating the new initial conditions as well so it would be great if you could help me with that too but I have the answer without the initial conditions plugged in (http://www.wolframalpha.com/input/?i=Solve+x^2+y''+-+5xy'+-+7y+%3D+x^3) and it differs from the answer I get.

I'm attaching my work.

Re: x^2 y'' - 5x y' - 7y = x^3

The equation is of the type Cauchy- Euler. A Google search will turn up alot of info.

I was a little careless with my notation above and I think it has confused you. When we do this substitution we are chainging the independant variable from x to u. So we need to sub out the x's in the original equation. Using the work above you should get the equation.

Simplifying will reduce out all of the x's and gives the equation.

Note: here the primes represent differentation with respect to u.

I hope this clears it up a bit.

TES