What is the solution for y^{''}-10y^'+25y=2e^{-3x} ?

The solution for y^{''}-10y^'+25y=2e^{-3x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+1/32 e^{-3x}

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y^{''}-10y^'+25y=2e^{-3x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 2 e^{- 3 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{1}{32} e^{- 3 x}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 2 e^{- 3 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{1}{32} e^{- 3 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{1}{32} e^{- 3 x}

What is the solution for y^{''}-10y^'+25y=2e^{-3x} ?
The solution for y^{''}-10y^'+25y=2e^{-3x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+1/32 e^{-3x}