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}

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