# Math Help - 2nd ODE trivial solution help.

1. ## 2nd ODE trivial solution help.

The problem is
x''+5x'+6x=3e^(-2t)+e^3t

The homogeneous solution, that I solved is,
x(h)=A(1)e^(-3t)+A(2)e^(-2t)

my problem is when I split it to find the first particular solution, for 3e^(-2t)
I assume
x=B(1)e^-2t
x'=-2B(1)e^-2t
x''=4B(1)e^-2t

I substitute into the original and get
e^(-2t)[4B(1)-10B(1)+6B(1)]=3e^(-2t)

which reduces too
4B-10B+6B=3

This is trivial

any help would be greatly appreciated.

Thank you

2. Originally Posted by Zigy221
The problem is
x''+5x'+6x=3e^(-2t)+e^3t

The homogeneous solution, that I solved is,
x(h)=A(1)e^(-3t)+A(2)e^(-2t)

my problem is when I split it to find the first particular solution, for 3e^(-2t)
I assume
x=B(1)e^-2t
x'=-2B(1)e^-2t
x''=4B(1)e^-2t

I substitute into the original and get
e^(-2t)[4B(1)-10B(1)+6B(1)]=3e^(-2t)

which reduces too
4B-10B+6B=3

This is trivial

any help would be greatly appreciated.

Thank you
Try $y_p = at e^{-2t}$.