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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

f^{^{'}} (x) = (\frac{1 - sec ( x )}{tan ( x )})

\frac{\partial}{\partial x} (1 - x^{2})

\frac{\partial}{\partial x} (r cos (t) sin (x))

\int \frac{x}{( x + 3 )} d x

\frac{\partial}{\partial y} (4 x^{2} y)

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