What is the solution for y^{''}+y^'=sin(t),y(0)=1,y^'(0)=2 ?

The solution for y^{''}+y^'=sin(t),y(0)=1,y^'(0)=2 is y= 5/2+3/2-5/2 e^{-t}+(-cos(t)-sin(t))/2

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y^{''}+y^'=sin(t),y(0)=1,y^'(0)=2

y^{^{''}} + y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 2

y = \frac{5}{2} + \frac{3}{2} - \frac{5}{2} e^{- t} + \frac{- cos ( t ) - sin ( t )}{2}

Solution steps

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

Solve linear ODE:

y = \frac{5}{2} + \frac{3}{2} - \frac{5}{2} e^{- t} + \frac{- cos ( t ) - sin ( t )}{2}

y = \frac{5}{2} + \frac{3}{2} - \frac{5}{2} e^{- t} + \frac{- cos ( t ) - sin ( t )}{2}

\int - 8 x d x

\int \frac{x ^{4} cos ( x )}{8} d x

\int_{0}^{2} (4 x + 5) e^{- x} d x

\frac{d}{d x} (\frac{2 x ^{2} + 2 x + 4}{x})

y^{^{'}} = \frac{( x y + 2 y )}{( x + 3 )}

What is the solution for y^{''}+y^'=sin(t),y(0)=1,y^'(0)=2 ?
The solution for y^{''}+y^'=sin(t),y(0)=1,y^'(0)=2 is y= 5/2+3/2-5/2 e^{-t}+(-cos(t)-sin(t))/2