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

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

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x^{''}+x^'=sin(t)

x^{^{''}} + x^{^{'}} = sin (t)

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

Solution steps

x^{''} (t) + x^{'} (t) = sin (t)

Solve linear ODE:

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

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

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

y^{^{'''}} - 3 y^{^{'}} + 2 y = 8 e^{2 x}

y^{^{''}} - 4 y^{^{'}} = 8 x e^{3 x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}

y^{^{''}} - 4 y^{^{'}} + 7 y = t e^{t}

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