What is the solution for y^{''}-y=sin(x) ?

The solution for y^{''}-y=sin(x) is y=c_{1}e^x+c_{2}e^{-x}-1/2 sin(x)

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y^{''}-y=sin(x)

y^{^{''}} - y = sin (x)

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

Solution steps

y^{''} (x) - y = sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} + 2 y = e^{- t}, y (4) = 1, y^{^{'}} (4) = 3

y^{^{''}} + 25 y = 2 x cos (5 x)

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} - 12 y = 3 cos (x)

4 y^{^{''}} - 16 y = \frac{1}{e ^{x}}

y^{^{''}} + y^{^{'}} - 2 y = x e^{x}

What is the solution for y^{''}-y=sin(x) ?
The solution for y^{''}-y=sin(x) is y=c_{1}e^x+c_{2}e^{-x}-1/2 sin(x)