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)

y^{^{''}} - 5 y^{^{'}} + 6 y = (12 x - 7) e^{- x}

y^{^{''}} + 25 y = 20 sin (5 x) + x cos (x)

2 y^{^{''}} - 13 y^{^{'}} + 20 y = 40 x^{2} + 8 x + 10

y^{^{''}} - 2 y^{^{'}} + 10 y = 20 x, y (0) = 0, y^{^{'}} (0) = 0

- a y^{^{''}} + b y - 1 = 0

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)