What is the solution for y^{''}+4y^'=sin(2x) ?

The solution for y^{''}+4y^'=sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/20 sin(2x)-1/10 cos(2x)

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y^{''}+4y^'=sin(2x)

y^{^{''}} + 4 y^{^{'}} = sin (2 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{20} sin (2 x) - \frac{1}{10} cos (2 x)

Solution steps

y^{''} (x) + 4 y^{'} (x) = sin (2 x)

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{20} sin (2 x) - \frac{1}{10} cos (2 x)

y = c_{1} + c_{2} e^{- 4 x} - \frac{1}{20} sin (2 x) - \frac{1}{10} cos (2 x)

(D^{2} + 13 D + 13) x = 12 t^{2} + 2 t - cos (t)

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - y = x e^{\frac{x}{2}}

y^{^{''}} + 9 y = e^{2 x} + x^{2} sin (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x^{- 1} e^{x}

What is the solution for y^{''}+4y^'=sin(2x) ?
The solution for y^{''}+4y^'=sin(2x) is y=c_{1}+c_{2}e^{-4x}-1/20 sin(2x)-1/10 cos(2x)