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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int \frac{1}{( x \cdot x ^{2} - 16 )} d x

y = e^{- x} cos (x)

\int \frac{1}{x ^{3} - 4 x ^{2}} d x

\frac{\partial}{\partial x} (\frac{9 x}{7 y + 8 z})

\frac{x d y}{d x} + 6 y = x^{3} - x

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