Solving Differential Equations with Variation of Parameters

[Illusionist]

Homework Statement

Given that y=x^2 is a solution to the differential equation:
(x^2)y'' + 2xy' - 6y = 0 <--- Eq.(1)
find the general solution of the differential equation
(x^2)y'' + 2xy' - 6y = 10(x^7) + 15(x^2) <--- Eq.(2)
Hence write down a second linear dependent solution of equation (1) and a particular solution of equation (2).

Homework Equations

I've basically concluded that variation of parameters is necessary. I don't think I completely understand what is being asked.

The Attempt at a Solution

I tried letting y= V(y1) = V(X^2)
hence y'= (x^2)V' + 2xV and
y''= (x^2)V'' + 2xV' + 2V
Here is where I think I'm getting confused, sub. back into (1) I get:
V''(x^4)+2(x^3)V'+2V(x^2)+2(x^3)V'+4(x^2)V-6V(x^2)=0
which equals V''(x^4)+4(x^3)V"=0
This is where I come to a dead end, any help or advice would be greatly appreciated, thank you.

 

[Hurkyl]

I assume you meant V'' x^4 + 4 x^3 V' = 0? Doesn't that simplify into a form that's very easy to solve?


Anyways, aren't you trying to solve equation (2)? You should be substituting into that.

 

[ice109]

isn't there another variation of parameters that can be used to find another linearly independent solution of a DE where you take the solution you find from the characterstic, multiplying it by u(x), then differentiating it and plugging it into and solving for u(x)?

 

[Hurkyl]

That's what he did, except he called it V instead of u(x).

 

[Illusionist]

Ok so I'm at V''(x^4) + 4(x^3)V'=10(x^7) + 15(x^2). I then just devided all by (x^4) to get V'' alone, hence"
V'' + V'(4/x) = 10(x^3) + 15/(x^2)
I then let u=V' and u'=V'', therefore:
u' +u(4/x) = 10(x^3) + 15/(x^2), I then let P(x)=4/x and hence I(x)=x^4 after integration.
Now I have (d/dx) (x^4)u = integral of [10(x^7) + 15(x^2).dx],
hence (x^4)u= (10x^8 / 8) + 5(x^3) + C,
V'= (5x^4 / 4) + 5/x + C/(x^4) and finally
V= (x^5)/4 + 5log(x) - C/(3x^3) +D.

This is wrong and I just don't know where I went wrong. Thanks for the help so far guys.

 

[ice109]

That's what he did, except he called it V instead of u(x).

yea but there's also the use of the wronskian which is called variation of parameters

 

[Illusionist]

Would anyone know where I went wrong above?

 

[Illusionist]

Please, anyone?

 

[ice109]

Please, anyone?

write it out in tex and i'll help but i really can't decypher what you've written