# Solution of the associated homogeneous problem

• 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)^2And the third one is another integral.In summary, the conversation discusses finding the general solution of a differential equation using the fact that one of the solutions is known. The attempt at a solution involves dividing the equation and using a second solution formula, but there is a problem with integrating. The person asks for help with integrating the equation.

## Homework Statement

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

## 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

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..

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..

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..