# Solution of the associated homogeneous problem

1. Apr 22, 2009

### confused88

1. The problem statement, all variables and given/known data

Using the fact that y1 = x-1/2 cos x is a solution of the associated homogeneous problem, obtain the general solution of
x2 y'' + x y' + (x2 - 1/4)y = x3/2

3. The attempt at a solution

Well the first thing i did was divide it by x^2 to obtain:
y'' + 1/x y' + (1 - 1/4x2)y = x-1/2

Then let y'' + 1/x y' + (1 - 1/4x2)y = 0

A second solution will be given as
y2 = y1 $$\int e ^ \int P(x)\frac{}{} (y1)^2$$ eintegral of P(x) dx $$\div$$ (y1)^2

y2 = x-1/2 cos x $$\int$$ - x / ( x-1/2 cos x)^2

2. Apr 22, 2009

### confused88

Re: Integration

Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..

3. Apr 22, 2009

### confused88

Re: Integration

Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..

4. Apr 22, 2009

### confused88

Re: Integration

Umm the first latex image generation is supposed to be an integral,
The second is supposed to be a fraction, so e^integral of P(x) divided by (y1)^2

And the third one is another integral.

Sorry I dunno what happened..