What is the solution for y^{''}+4y^'+5y=2e^{3x} ?

The solution for y^{''}+4y^'+5y=2e^{3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/13 e^{3x}

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y^{''}+4y^'+5y=2e^{3x}

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{13} e^{3 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 2 e^{3 x}

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{13} e^{3 x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{13} e^{3 x}

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = 6 e^{2 x} + 16 x^{2}

y^{^{''}} + 4 y = \frac{3}{sin ( 2 x )}

a y^{^{''}} + 2 y^{^{'}} = 4 e^{2 t}, y (0) = 0, y^{^{'}} (0) = 0

(D^{2} + 3 D) y = - 18 x

y^{^{''}} - 2 y^{^{'}} - 8 y = 4 e^{2} - 21 e^{- 3 x}

What is the solution for y^{''}+4y^'+5y=2e^{3x} ?
The solution for y^{''}+4y^'+5y=2e^{3x} is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/13 e^{3x}