What is the solution for x^{''}+x^'-2x=8sin(2t) ?

The solution for x^{''}+x^'-2x=8sin(2t) is x=c_{1}e^t+c_{2}e^{-2t}-6/5 sin(2t)-2/5 cos(2t)

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x^{''}+x^'-2x=8sin(2t)

x^{^{''}} + x^{^{'}} - 2 x = 8 sin (2 t)

x = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{6}{5} sin (2 t) - \frac{2}{5} cos (2 t)

Solution steps

x^{''} (t) + x^{'} (t) - 2 x = 8 sin (2 t)

Solve linear ODE:

x = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{6}{5} sin (2 t) - \frac{2}{5} cos (2 t)

x = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{6}{5} sin (2 t) - \frac{2}{5} cos (2 t)

What is the solution for x^{''}+x^'-2x=8sin(2t) ?
The solution for x^{''}+x^'-2x=8sin(2t) is x=c_{1}e^t+c_{2}e^{-2t}-6/5 sin(2t)-2/5 cos(2t)