ODE ( 2nd order nonhomogeneous equation)

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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:!)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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