What is the solution for y^{''}-10y^'+25y=4xe^{5x} ?

The solution for y^{''}-10y^'+25y=4xe^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+(2e^{5x}x^3)/3

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y^{''}-10y^'+25y=4xe^{5x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 4 x e^{5 x}

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

Solution steps

y^{''} (x) - 10 y^{'} (x) + 25 y = 4 x e^{5 x}

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (9 x e^{2 x y})

\frac{s ^{2} + 2 s + 5}{s ^{2} ( s + 1 )}

e^{i θ}

\frac{d}{d z} (e^{i z})

\frac{d y}{d t} = 1.7 y

What is the solution for y^{''}-10y^'+25y=4xe^{5x} ?
The solution for y^{''}-10y^'+25y=4xe^{5x} is y=c_{1}e^{5x}+c_{2}xe^{5x}+(2e^{5x}x^3)/3