What is the solution for y^{''}-10y^'+25y=6e^{-5x} ?

The solution for y^{''}-10y^'+25y=6e^{-5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+3/50 e^{-5x}

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y^{''}-10y^'+25y=6e^{-5x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 e^{- 5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3}{50} e^{- 5 x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 6 e^{- 5 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3}{50} e^{- 5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{3}{50} e^{- 5 x}

y^{^{''}} - 5 y^{^{'}} + 6 y = 4, e^{4 x} y (0) = 0, y^{^{'}} (0) = 1

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 24 y = 5 + 2 x

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

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{2 t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 1 + x + x^{2}

What is the solution for y^{''}-10y^'+25y=6e^{-5x} ?
The solution for y^{''}-10y^'+25y=6e^{-5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+3/50 e^{-5x}