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

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

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

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

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{3} e^{x} - \frac{5 x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{3} e^{x} - \frac{5 x}{2}

y = c_{1} + c_{2} e^{- 2 x} + \frac{1}{3} e^{x} - \frac{5 x}{2}

y^{^{'''}} + y^{^{'}} = x + sin (x) + cos (x)

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

y^{^{''}} + y^{^{'}} - 2 y = 2 t, y (0) = 0

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

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

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