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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 5 y^{^{'}} - 6 y = 10 e^{2 x}

y^{^{''}} + 4 y = 9 sin (t)

y^{^{''''}} - y = 6 t + cos (t)

y^{^{''}} + y^{^{'}} + y = x e^{- x}

y^{^{''}} - 6 y^{^{'}} + 13 y = 4 e^{3 x}, y (0) = 2, y^{^{'}} (0) = 4

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