What is the solution for y^{''}+8y^'+32y=2e^{2x} ?

The solution for y^{''}+8y^'+32y=2e^{2x} is y=e^{-4x}(c_{1}cos(4x)+c_{2}sin(4x))+1/26 e^{2x}

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y^{''}+8y^'+32y=2e^{2x}

y^{^{''}} + 8 y^{^{'}} + 32 y = 2 e^{2 x}

y = e^{- 4 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{26} e^{2 x}

Solution steps

y^{''} (x) + 8 y^{'} (x) + 32 y = 2 e^{2 x}

Solve linear ODE:

y = e^{- 4 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{26} e^{2 x}

y = e^{- 4 x} (c_{1} cos (4 x) + c_{2} sin (4 x)) + \frac{1}{26} e^{2 x}

y^{^{''}} + y - 2 = 5 e^{- 2 x}

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

y^{^{''}} - 6 y^{^{'}} + 9 y = 4 e^{3} t ln (t)

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

y^{^{''}} - 7 y^{^{'}} + 12 y = - e^{x}

What is the solution for y^{''}+8y^'+32y=2e^{2x} ?
The solution for y^{''}+8y^'+32y=2e^{2x} is y=e^{-4x}(c_{1}cos(4x)+c_{2}sin(4x))+1/26 e^{2x}