easy but hard differential equation

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 10u = 26e^{-3t}\)

 

[khansaheb]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 10u = 26e^{-3t}\)

Please show us what you have tried and exactly where you are stuck.

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Please share your work/thoughts about this problem

 

[logistic_guy]

\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)

\(\displaystyle u'' + 2u' + 10u = 26e^{-3t}\)

The homogeneous solution is:

\(\displaystyle u(t) = c_1e^{-t}\cos 3t + c_2e^{-t}\sin 3t\)

 

[logistic_guy]

A guess for the particular solution will be:

\(\displaystyle u_p = Ae^{-3t}\)

When we substitute this solution in the original differential equation, we get:

\(\displaystyle A = 2\)

Then, the general solution to the OP differential equation is:

\(\displaystyle u(t) = c_1e^{-t}\cos 3t + c_2e^{-t}\sin 3t + 2e^{-3t}\)

I was a little bit lazy to show all the details to get this solution. But you are free to ask me about any part you did not understand!