What is the solution for y^{''}+10y^'+25y=14e^{-5x} ?

The solution for y^{''}+10y^'+25y=14e^{-5x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+7x^2e^{-5x}

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y^{''}+10y^'+25y=14e^{-5x}

y^{^{''}} + 10 y^{^{'}} + 25 y = 14 e^{- 5 x}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + 7 x^{2} e^{- 5 x}

Solution steps

y^{''} (x) + 10 y^{'} (x) + 25 y = 14 e^{- 5 x}

Solve linear ODE:

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + 7 x^{2} e^{- 5 x}

y = c_{1} e^{- 5 x} + c_{2} x e^{- 5 x} + 7 x^{2} e^{- 5 x}

x \to \infty lim (- \frac{1}{x})

x \to 1 lim (cos (\frac{π}{3}))

\frac{d}{d x} (\frac{2}{3} π + 1)

\frac{\partial}{\partial x} (\frac{( 2 x + y )}{z})

\int cos (4 x) cos (5 x) d x

What is the solution for y^{''}+10y^'+25y=14e^{-5x} ?
The solution for y^{''}+10y^'+25y=14e^{-5x} is y=c_{1}e^{-5x}+c_{2}xe^{-5x}+7x^2e^{-5x}