Sine Integral Function in differential equation

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Homework Help Overview

The discussion revolves around expressing the solution to a differential equation involving the sine integral function, Si(x). The initial value problem presented is x^3y'+2x^2y=10sinx, with the condition y(0)=0.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the manipulation of the differential equation and the integration process. There is a focus on how to incorporate the sine integral function into the solution while addressing potential integration challenges.

Discussion Status

Some participants have provided algebraic manipulations and suggested alternative forms of the equation. There is an acknowledgment of the need for clarity in the integration steps and the application of the initial condition, but no consensus on the final solution has been reached.

Contextual Notes

Participants note the importance of the initial condition y(0)=0 and discuss the implications of the sine integral function being a definite integral, which may affect the integration process.

kathrynag
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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.
 
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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.
 

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