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

The solution for y^{''}+2y^'+5y=2+e^x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+2/5+1/8 e^x

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

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

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2}{5} + \frac{1}{8} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2}{5} + \frac{1}{8} e^{x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{2}{5} + \frac{1}{8} e^{x}

What is the solution for y^{''}+2y^'+5y=2+e^x ?
The solution for y^{''}+2y^'+5y=2+e^x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+2/5+1/8 e^x