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)

y^{^{''}} + c \cdot y = g (x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = a

y^{^{''}} - 36 y = 24 sin (6 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{x} (cos (x) - 3 sin (x))

y^{^{''}} + 7 y^{^{'}} + 11 y = 11 x^{3} + 10 x^{2} - 8 x + 9

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)