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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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