What is the solution for y^{''}-2y^'+10y=e^{-x} ?

The solution for y^{''}-2y^'+10y=e^{-x} is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+1/13 e^{-x}

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y^{''}-2y^'+10y=e^{-x}

y^{^{''}} - 2 y^{^{'}} + 10 y = e^{- x}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{13} e^{- x}

Solution steps

y^{''} (x) - 2 y^{'} (x) + 10 y = e^{- x}

Solve linear ODE:

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{13} e^{- x}

y = e^{x} (c_{1} cos (3 x) + c_{2} sin (3 x)) + \frac{1}{13} e^{- x}

e^{- x + 1} - e^{- x + 1} \cdot x

\frac{\partial}{\partial x} (y \cdot cos (x - y))

7 x - 4 x^{2}

What is the solution for y^{''}-2y^'+10y=e^{-x} ?
The solution for y^{''}-2y^'+10y=e^{-x} is y=e^x(c_{1}cos(3x)+c_{2}sin(3x))+1/13 e^{-x}