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

The solution for y^{''}+4y^'+5y=e^x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^x

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

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

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x}

Solution steps

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

Solve linear ODE:

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x}

y = e^{- 2 x} (c_{1} cos (x) + c_{2} sin (x)) + \frac{1}{10} e^{x}

\frac{d}{d x} (- 8 e^{- 8 x})

\frac{\partial}{\partial r} (r^{2} + s^{2})

x \to 3 - lim (\frac{5 x}{x ^{2} - 9})

\int_{0}^{5} \frac{1}{25 - x ^{2}} d x

x \to - 1 - lim (\frac{9}{x ^{3} - 1})

What is the solution for y^{''}+4y^'+5y=e^x ?
The solution for y^{''}+4y^'+5y=e^x is y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+1/10 e^x