What is the solution for y^{''}+4y^'+4y=6sin(x) ?

The solution for y^{''}+4y^'+4y=6sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+18/25 sin(x)-24/25 cos(x)

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y^{''}+4y^'+4y=6sin(x)

y^{^{''}} + 4 y^{^{'}} + 4 y = 6 sin (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{18}{25} sin (x) - \frac{24}{25} cos (x)

Solution steps

y^{''} (x) + 4 y^{'} (x) + 4 y = 6 sin (x)

Solve linear ODE:

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{18}{25} sin (x) - \frac{24}{25} cos (x)

y = c_{1} e^{- 2 x} + c_{2} x e^{- 2 x} + \frac{18}{25} sin (x) - \frac{24}{25} cos (x)

y^{^{'''}} - y^{^{'}} = sin^{2} (x)

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

y^{^{''}} + 9 y = 2, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + 2 y^{^{'}} + y = t

y^{^{''}} + y = cos (x) + sec (x)

What is the solution for y^{''}+4y^'+4y=6sin(x) ?
The solution for y^{''}+4y^'+4y=6sin(x) is y=c_{1}e^{-2x}+c_{2}xe^{-2x}+18/25 sin(x)-24/25 cos(x)