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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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