Question about differential equations?(nonhomogeneous)

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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.
 
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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