Question about differential equations?(nonhomogeneous)

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The discussion centers on finding the particular solution for the nonhomogeneous differential equation y'' + 9y' = 162sin(9t) + 1296cos(9t). The correct particular solution is identified as y_P = Asin(9t) + Bcos(9t). The roots of the complementary function are confirmed as 0 and -9, indicating distinct roots. The initial setup presented by the user is deemed incorrect, and the correct approach is emphasized, focusing on substituting y_P into the differential equation to solve for constants A and B.

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hard_assteel
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For this problem I want to find the particular solution;
but i am not sure if it is right setup.


y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

Thank You
 
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hard_assteel said:
For this problem I want to find the particular solution;
but i am not sure if it is right setup.


y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

Thank You
No. Here's your particular solution: yp = Asin(9t)+Bcos(9t)
This would be your particular solution for a right hand side of any of the following:
c1sin(9t)
c2cos(9t)
c1sin(9t) + c2cos(9t)
 
hard_assteel said:
y''+9y'=162sin(9t)+1296cos(9t)

the roots for complementary function are 0,-9
so it has distinct roots.

That's true, but it's of no use in finding the particular solution.

my setup is;
[Asin(9t)+Bcos(9t)]+[Asin(9t)+Bcos(9t)]
Axsin(9t)+Bxcos(9t)

am I going about it the right way?

To be honest, I can't figure out what you're thinking. But no, what you have written is not right. All you need is y_P=A\sin(9t)+B\cos(9t). Plug it into your differential equation and solve for A and B.
 

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