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

The solution for y^{''}+2y^'+y=sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-3/25 sin(2x)-4/25 cos(2x)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{3}{25} sin (2 x) - \frac{4}{25} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{3}{25} sin (2 x) - \frac{4}{25} cos (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{3}{25} sin (2 x) - \frac{4}{25} cos (2 x)

\frac{\partial}{\partial x} (x - 2 y^{2})

\frac{d}{d x} (r (θ, x) cos (θ))

\int (2 x + 3)^{2} d x

sin (u)

y = 1 + e^{x}

What is the solution for y^{''}+2y^'+y=sin(2x) ?
The solution for y^{''}+2y^'+y=sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-3/25 sin(2x)-4/25 cos(2x)