What is the solution for y^{''}+y^'=cos(t/2) ?

The solution for y^{''}+y^'=cos(t/2) is y=c_{1}+c_{2}e^{-t}+8/5 sin(t/2)-4/5 cos(t/2)

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y^{''}+y^'=cos(t/2)

y^{^{''}} + y^{^{'}} = cos (\frac{t}{2})

y = c_{1} + c_{2} e^{- t} + \frac{8}{5} sin (\frac{t}{2}) - \frac{4}{5} cos (\frac{t}{2})

Solution steps

y^{''} (t) + y^{'} (t) = cos (\frac{t}{2})

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{8}{5} sin (\frac{t}{2}) - \frac{4}{5} cos (\frac{t}{2})

y = c_{1} + c_{2} e^{- t} + \frac{8}{5} sin (\frac{t}{2}) - \frac{4}{5} cos (\frac{t}{2})

y^{^{''}} - 4 y = t^{2} + 1

5 y^{^{''}} - 10 y^{^{'}} + 10 y = e x sec (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 20 e^{- t}

y^{^{''}} + 9 y = 2 x^{2}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 8 y = cos (x)

What is the solution for y^{''}+y^'=cos(t/2) ?
The solution for y^{''}+y^'=cos(t/2) is y=c_{1}+c_{2}e^{-t}+8/5 sin(t/2)-4/5 cos(t/2)