Sine Integral Function in differential equation

1. Nov 6, 2009

kathrynag

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

Si(x)=$$\int(sint/t)dt$$ from 0 to x
integrand is 1 at t=0

express the solutiony(x) of the initial value problem x^3y'+2x^2y=10sinx, y(0)=0 in terms of Si(x)

2. Relevant equations

3. The attempt at a solution
y'+2/xy=10x^-3sinx
multiply by x^2
x^2y'+2xy=10x^-2sinx
(x^2y)'=10x^-2sinx
Now I get stuck because I feel like this can't be integrated and I don't think i can just insert Si(x) because I need to account for 10x^-2.

2. Nov 6, 2009

LCKurtz

Divide your original equation by x:

$$x^2y'' + 2xy' = 10 \frac {\sin x}{x}$$

which will work for you. You just have an arithmetic mistake.

3. Nov 7, 2009

kathrynag

So I have x^2y'+2xy=10sinx/x
(x^2y)'=10sinx/x
x^2y=10Si(x)
y=10x^(-2)Si(x)

4. Nov 7, 2009

LCKurtz

Your answer may be correct, but you need to show more explicitly the integral and how you used the initial condition y(0) = 1. Your second step isn't clear because the Si function is a definite integral.