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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + 3 y^{^{'}} = 1 - e^{- 3 x}

y^{^{''}} + 16 y = 6 x, y (0) = 1, y^{^{'}} (0) = 4

(D^{3} - 2 D^{2} - 3 D) y = 74 e^{- 3 x} cos (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = - 11 t e^{- t}

y^{^{''}} - 4 y^{^{'}} + 8 y = x^{3} - x

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