What is the solution for y^{''}-10y^'+25y=6x^2e^{5x} ?

The solution for y^{''}-10y^'+25y=6x^2e^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+(e^{5x}x^4)/2

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y^{''}-10y^'+25y=6x^2e^{5x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 x^{2} e^{5 x}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{5 x} x ^{4}}{2}

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 6 x^{2} e^{5 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{5 x} x ^{4}}{2}

y = c_{1} e^{5 x} + c_{2} x e^{5 x} + \frac{e ^{5 x} x ^{4}}{2}

y^{^{''}} - y^{^{'}} - 2 y = 2 x - 1

x^{^{''}} + 4 x = 2 e^{t}

y^{^{'''}} + y = 4 e^{x}

y^{^{''}} + 4 y = 2 e^{- x}

y^{^{''}} + 2 y^{^{'}} + 10 y = 50 x, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}-10y^'+25y=6x^2e^{5x} ?
The solution for y^{''}-10y^'+25y=6x^2e^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+(e^{5x}x^4)/2