Solution of the associated homogeneous problem

[confused88]

Homework Statement

Please help me with this, I would really appreciate it:

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 = x^3/2

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 $$\int e ^ \int P(x)\frac{}{} (y1)^2$$ e^{integral of P(x) dx} $$\div$$ (y1)^2

So here is where i have a problem, can someone please help me integrate

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

please help me..?

 

[confused88]

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 don't know what happened..

please help

 

[confused88]

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 don't know what happened..

please help me

 

[confused88]

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 don't know what happened..

please help me