What is the solution for y^{''}+4y^'+5y=2e^x ?

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

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y^{''}+4y^'+5y=2e^x

y^{^{''}} + 4 y^{^{'}} + 5 y = 2 e^{x}

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

Solution steps

y^{''} (x) + 4 y^{'} (x) + 5 y = 2 e^{x}

Solve linear ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 12 e^{t}, y (0) = - 1, y^{^{'}} (0) = 7

y^{^{''}} + y^{^{'}} = 2 cos (x)

y^{^{''}} + y^{^{'}} - 2 y = 2 x + 3, y (0) = 1, y^{^{'}} (0) = - 4

y^{^{''}} + 5 y^{^{'}} + 4 y = 10 t e^{5 t}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 = 0

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