# Differetial Equation with Sine Integral

• Nov 6th 2009, 07:49 PM
kathrynmath
Differetial Equation with Sine Integral
Si(x)=integral(sint/tdt) from 0 to x
integrand is 1 at t=0

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

What I have done:
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.
• Nov 6th 2009, 10:13 PM
Prove It
Quote:

Originally Posted by kathrynmath
Si(x)=integral(sint/tdt) from 0 to x
integrand is 1 at t=0

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

What I have done:
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.

$x^3y' + 2x^2y = 10\sin{x}$

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

$\frac{d}{dx}(x^2y) = 10\cdot\frac{\sin{x}}{x}$

$x^2y = 10\int{\frac{\sin{x}}{x}\,dx}$...

Can you go from here?
• Nov 7th 2009, 08:54 AM
kathrynmath
y=10x^(-2)Si(x)