What is the solution for y^{''}+2y^'+y=sin(8x) ?

The solution for y^{''}+2y^'+y=sin(8x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-63/4225 sin(8x)-16/4225 cos(8x)

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y^{''}+2y^'+y=sin(8x)

y^{^{''}} + 2 y^{^{'}} + y = sin (8 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{63}{4225} sin (8 x) - \frac{16}{4225} cos (8 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) + y = sin (8 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{63}{4225} sin (8 x) - \frac{16}{4225} cos (8 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{63}{4225} sin (8 x) - \frac{16}{4225} cos (8 x)

y^{^{''}} - 6 y^{^{'}} + 5 y = t e^{2 t}, y (0) = 2, y^{^{'}} (0) = 3

y^{^{'''}} + 12 y^{^{''}} - 13 y = x e^{x} + 5 x

y^{^{''}} + y = 2, y (0) = 0

y^{^{''}} + 5 y^{^{'}} - 24 y = - 92 t + 48 t^{2}

y^{^{''}} - 5 y^{^{'}} = 5 e^{5 x}

What is the solution for y^{''}+2y^'+y=sin(8x) ?
The solution for y^{''}+2y^'+y=sin(8x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-63/4225 sin(8x)-16/4225 cos(8x)