What is the solution for y^{''}+4y^'+5y=e^{-x}+15x ?

The solution for y^{''}+4y^'+5y=e^{-x}+15x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/2 e^{-x}+3x-12/5

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y^{''}+4y^'+5y=e^{-x}+15x

y^{^{''}} + 4 y^{^{'}} + 5 y = e^{- x} + 15 x

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} + 3 x - \frac{12}{5}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = e^{- x} + 15 x

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} + 3 x - \frac{12}{5}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{2} e^{- x} + 3 x - \frac{12}{5}

x^{^{''}} + 9.8 x = cos (2 t)

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

y^{^{''}} - 6 y^{^{'}} + 8 y = 2, y (0) = 0, y^{^{'}} (0) = 0

x^{^{''}} + 49 x = 52 cos (6 t)

y^{^{''}} - 3 y^{^{'}} + 2 y = - 2 x

What is the solution for y^{''}+4y^'+5y=e^{-x}+15x ?
The solution for y^{''}+4y^'+5y=e^{-x}+15x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/2 e^{-x}+3x-12/5