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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 2 y = x + sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

4 y^{^{''}} - 16 y^{^{'}} + 64 y = - \frac{2 e ^{8 x}}{3 x}

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

y^{^{''}} + 5 y^{^{'}} + 6 y = x^{2}

y^{^{''}} + 9 y = e^{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}e^{-t}-1/2 sin(t)