What is the solution for y^{''}+2y^'+1=4sin(2x) ?

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

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y^{''}+2y^'+1=4sin(2x)

y^{^{''}} + 2 y^{^{'}} + 1 = 4 sin (2 x)

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

Solution steps

y^{''} (x) + 2 y^{'} (x) + 1 = 4 sin (2 x)

Solve linear ODE:

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

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

1.6 q^{^{''}} + 1200 q^{^{'}} + \frac{1}{0.00001} q = 14

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x

\frac{d ^{2} y}{d x ^{2}} + a \frac{d y}{d x} = g

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

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