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}

y = x^{2} - 15, y = 1

(x, y) \to (0, 1) lim (\frac{e ^{x y}}{x + 1})

\int cos (y^{3}) d y

f (x) = (x^{3}), at x = 2

x \to 2 lim (\frac{2 - x}{∣ x - 2 ∣})

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}