Sine Integral Function in differential equation

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 4K views
kathrynag
Messages
595
Reaction score
0

Homework Statement



Si(x)=[tex]\int(sint/t)dt[/tex] 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)

Homework Equations





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.
 
Physics news on Phys.org
kathrynag said:

Homework Statement



Si(x)=[tex]\int(sint/t)dt[/tex] 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)

Homework Equations





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.

Divide your original equation by x:

[tex]x^2y'' + 2xy' = 10 \frac {\sin x}{x}[/tex]

which will work for you. You just have an arithmetic mistake.
 
So I have x^2y'+2xy=10sinx/x
(x^2y)'=10sinx/x
x^2y=10Si(x)
y=10x^(-2)Si(x)
 
kathrynag said:
So I have x^2y'+2xy=10sinx/x
(x^2y)'=10sinx/x
x^2y=10Si(x)
y=10x^(-2)Si(x)

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.