What is the solution for 1/4 y^{''}+y^'+y=sin(x) ?

The solution for 1/4 y^{''}+y^'+y=sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+12/25 sin(x)-16/25 cos(x)

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1/4 y^{''}+y^'+y=sin(x)

\frac{1}{4} y^{^{''}} + y^{^{'}} + y = sin (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{12}{25} sin (x) - \frac{16}{25} cos (x)

Solution steps

\frac{1}{4} y^{''} (x) + y^{'} (x) + y = sin (x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{12}{25} sin (x) - \frac{16}{25} cos (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{12}{25} sin (x) - \frac{16}{25} cos (x)

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

y^{^{''}} + y = sin (t), y (0) = (0), y^{^{'}} (0) = 1

y^{^{''}} - 7 y^{^{'}} + 12 y = - 288 t^{3}

y^{^{''}} + y = - 3

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

What is the solution for 1/4 y^{''}+y^'+y=sin(x) ?
The solution for 1/4 y^{''}+y^'+y=sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+12/25 sin(x)-16/25 cos(x)