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

The solution for y^{''}+y^'-2y=sin(t) is y=c_{1}e^t+c_{2}e^{-2t}-3/10 sin(t)-1/10 cos(t)

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

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

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3}{10} sin (t) - \frac{1}{10} cos (t)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3}{10} sin (t) - \frac{1}{10} cos (t)

y = c_{1} e^{t} + c_{2} e^{- 2 t} - \frac{3}{10} sin (t) - \frac{1}{10} cos (t)

What is the solution for y^{''}+y^'-2y=sin(t) ?
The solution for y^{''}+y^'-2y=sin(t) is y=c_{1}e^t+c_{2}e^{-2t}-3/10 sin(t)-1/10 cos(t)