Differetial Equation with Sine Integral

kathrynmath · #1 $Old$ November 6th, 2009, 08:49 PM

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.

Prove It · #2 $Old$ November 6th, 2009, 11:13 PM

Quote:

Originally Posted by kathrynmath $View Post$

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?

kathrynmath · #3 $Old$ November 7th, 2009, 09:54 AM

y=10x^(-2)Si(x)