ODE ( 2nd order nonhomogeneous equation)

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Homework Help Overview

The discussion revolves around finding a particular solution to a second-order nonhomogeneous ordinary differential equation (ODE) of the form y'' + y' + y = (sin x)^2 using the method of undetermined coefficients.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the challenge of determining the particular solution for (sin x)^2, with one suggesting a form similar to that used for sin x, while another proposes using trigonometric identities to rewrite (sin x)^2 for a suitable solution form.

Discussion Status

The discussion includes various approaches to the problem, with some participants exploring the use of trigonometric identities to simplify the expression. Guidance has been offered regarding the form of the particular solution, but no consensus has been reached on the final approach.

Contextual Notes

Participants note the limitations of assuming a solution based on standard forms for sine and cosine functions, highlighting the need to consider identities for squaring these functions.

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Homework Statement


By using the method of undetermined coefficients,find the particular solution of
y''+y'+y=(sin x)^2


Homework Equations


i know how to determine the particular solution IF it is sin x.
Ex: sin x ====> Asin x + B cos x (particular)

but i wonder how to determine the (sin x)^2


The Attempt at a Solution

 
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i haven't actually seen a problem like this come up, but i think similarly to finding the particular solution to something like lhs=t^2 is A+Bt+Ct^2, it'd be something like:

Asin(x)+Bcos(x)+Ccos^2(x)+Dsin^2(x)

i'm not 100% sure, but i'd try something like that and see if it works out. good luck!
 
You can't assume a solution of the form cos2 x or sin2 x because sine or cosine squared are not of the form that gets as solutions to a homogeneous equation with constant coefficients. However, you CAN use a trigonometric identity. Since cos(a+ b)= cos(a)cos(b)- sin(a)sin(b), taking a= b= x, cos(2x)= cos2(x)- sin2(x). Replacing cos2(x) by 1- sin2(x), cos(2x)= 1- 2sin2(x) so sin2(x)= (1/2)(1- cos(2x)). Look for a particular solution of the form A+ Bsin(2x)+ Ccos(2x).
 
Thx for helping ! i did solve the Q:!)
 

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