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

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

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y^{''}+2y+y=2e^{-t}

y^{^{''}} + 2 y + y = 2 e^{- t}

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

Solution steps

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

Solve linear ODE:

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

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

\frac{x}{x - 2}

\int_{2}^{5} x^{4} ln (2 x) d x

\int_{0}^{18} t^{2} + 2 t + 1 d t

\frac{\partial}{\partial y} (3 y - e^{s i n (x)})

x \to 3 - lim (\frac{6 x}{2 x - 5})

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