[SOLVED] integration by parts question

[Monocles]

1. The problem statement, all variables and given/known data
I think my first problem is at integration by parts, but let me know if you see a different error in my work.

Edit: I shoud mention that this is Calculus 2 and we just learned first order differential equations today. So I don't have any fancy methods available.

Find the solution of
x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)
with
y(\pi) = 0


2. Relevant equations
Format of first order differential equations:
y' + P(x) y = q(x)

H(x) = \int P(x) dx


3. The attempt at a solution

x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)

y' + \frac{a y}{x} = \frac{2 x^{-a + 4} \sin(x)}{x^3}

y' + \frac{a y}{x} = 2 x^{-a + 1} \sin(x)

H(x) = \int \frac{a}{x} dx

H(x) = a \ln{x}

Take e^{H(x)} = x^a and multiply it to both sides

x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)

(x^a y)' = 2 x^{-a^2 + a} \sin(x)

x^a y = \int 2 x^{-a^2 + a} \sin(x) dx

This is where I am stuck. The only way I can think of to solve that integral is integration by parts, but the only way I can see that working is by repeating it enough times for the part of the integral before sin(x) to disappear. So I think I went wrong somewhere? But I don't see where.

[Rainbow Child]

x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)

leads to

(x^\alpha\,y)'=2\,x\,\sin x

[Monocles]

Ohh... Well that was easy. I just forgot how to multiply! Thanks.