What is the solution for y^{''}+6y^'+34y=25e^{-3x} ?

The solution for y^{''}+6y^'+34y=25e^{-3x} is y=e^{-3x}(c_{1}cos(5x)+c_{2}sin(5x))+e^{-3x}

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y^{''}+6y^'+34y=25e^{-3x}

y^{^{''}} + 6 y^{^{'}} + 34 y = 25 e^{- 3 x}

y = e^{- 3 x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + e^{- 3 x}

Solution steps

y^{''} (x) + 6 y^{'} (x) + 34 y = 25 e^{- 3 x}

Solve linear ODE:

y = e^{- 3 x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + e^{- 3 x}

y = e^{- 3 x} (c_{1} cos (5 x) + c_{2} sin (5 x)) + e^{- 3 x}

\frac{d}{d x} (4 x - 5 x^{\frac{5}{6}})

f (x) = x cos (x) + 2 tan (x)

\int coth (\frac{x}{3}) d x

\frac{d}{d x} (a \cdot e^{x})

x \to - 5 - lim (\frac{2}{x ^{2} - 25})

What is the solution for y^{''}+6y^'+34y=25e^{-3x} ?
The solution for y^{''}+6y^'+34y=25e^{-3x} is y=e^{-3x}(c_{1}cos(5x)+c_{2}sin(5x))+e^{-3x}