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

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

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

y^{^{''}} + 10 y^{^{'}} + 25 y = 14 \cdot 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 \cdot 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}

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 9 e^{- t} + 6

y^{^{''}} = - 4 y^{^{'}} - 5 y + 2

16 \frac{d ^{4} y}{d x ^{4}} + 24 \frac{d ^{2} y}{d x ^{2}} + 9 = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 2 sin (4 x)

y^{^{''}} + 7 y^{^{'}} + 10 y = e^{3 x}, y (0) = 0, y^{^{'}} (0) = 0

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