What is the solution for x^{''}-x^'-2x=5cos(2t) ?

The solution for x^{''}-x^'-2x=5cos(2t) is x=c_{1}e^{2t}+c_{2}e^{-t}-1/4 sin(2t)-3/4 cos(2t)

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x^{''}-x^'-2x=5cos(2t)

x^{^{''}} - x^{^{'}} - 2 x = 5 cos (2 t)

x = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} sin (2 t) - \frac{3}{4} cos (2 t)

Solution steps

x^{''} (t) - x^{'} (t) - 2 x = 5 cos (2 t)

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} sin (2 t) - \frac{3}{4} cos (2 t)

x = c_{1} e^{2 t} + c_{2} e^{- t} - \frac{1}{4} sin (2 t) - \frac{3}{4} cos (2 t)

What is the solution for x^{''}-x^'-2x=5cos(2t) ?
The solution for x^{''}-x^'-2x=5cos(2t) is x=c_{1}e^{2t}+c_{2}e^{-t}-1/4 sin(2t)-3/4 cos(2t)