What is the solution for y^{''}+4y=-16e^{-2x}+2sin(x) ?

The solution for y^{''}+4y=-16e^{-2x}+2sin(x) is y=c_{1}cos(2x)+c_{2}sin(2x)-2e^{-2x}+2/3 sin(x)

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y^{''}+4y=-16e^{-2x}+2sin(x)

y^{^{''}} + 4 y = - 16 e^{- 2 x} + 2 sin (x)

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

Solution steps

y^{''} (x) + 4 y = - 16 e^{- 2 x} + 2 sin (x)

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} = 8 x + 1

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

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

y^{^{''}} + 2 y^{^{'}} + y = 3 cos (t)

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (x) - sin (2 x)

What is the solution for y^{''}+4y=-16e^{-2x}+2sin(x) ?
The solution for y^{''}+4y=-16e^{-2x}+2sin(x) is y=c_{1}cos(2x)+c_{2}sin(2x)-2e^{-2x}+2/3 sin(x)