What is the solution for y^{''}+y^'=sin(x) ?

The solution for y^{''}+y^'=sin(x) is y=c_{1}+c_{2}e^{-x}+(-cos(x)-sin(x))/2

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y^{''}+y^'=sin(x)

y^{^{''}} + y^{^{'}} = sin (x)

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

Solution steps

y^{''} (x) + y^{'} (x) = sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 4 e^{- x} cos (2 x)

y^{^{''}} - y^{^{'}} - 2 y = 2 e^{- x}

y^{^{''}} + 5 y^{^{'}} - 6 y = x e^{- 5 x}

y^{^{''}} + 7 y^{^{'}} + 12 y = t^{2} e^{- 3 t}

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} tan (x)

What is the solution for y^{''}+y^'=sin(x) ?
The solution for y^{''}+y^'=sin(x) is y=c_{1}+c_{2}e^{-x}+(-cos(x)-sin(x))/2