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

The solution for y^{''}+4y^'+5y=5x+3e-x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+(4x)/5+(15e-16)/(25)

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{4 x}{5} + \frac{15 e - 16}{25}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{4 x}{5} + \frac{15 e - 16}{25}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{4 x}{5} + \frac{15 e - 16}{25}

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What is the solution for y^{''}+4y^'+5y=5x+3e-x ?
The solution for y^{''}+4y^'+5y=5x+3e-x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+(4x)/5+(15e-16)/(25)