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

The solution for y^{''}+2y^'+y=4sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-12/25 sin(2x)-16/25 cos(2x)

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

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

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

y = c_{1} e^{- x} + c_{2} x e^{- x} - \frac{12}{25} sin (2 x) - \frac{16}{25} cos (2 x)

y^{^{''}} - y^{^{'}} - 2 y = 1 - x, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + y = (t - 2 π), y (0) = 10, y^{^{'}} (0) = 0

y^{^{''}} + 16 y = 3.36

y^{^{''}} + 4 y^{^{'}} + 5 y = sin (x)

y^{^{''}} = 2 y + 20 tan^{3} (x), y (0) = 10 tan (x)

What is the solution for y^{''}+2y^'+y=4sin(2x) ?
The solution for y^{''}+2y^'+y=4sin(2x) is y=c_{1}e^{-x}+c_{2}xe^{-x}-12/25 sin(2x)-16/25 cos(2x)