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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{2} 144 x + 289 d x

16 4 4 x^{4} + 4

t^{3} - \frac{1}{4 t ^{7}}

\frac{\partial}{\partial y} (y^{2} sin (x))

f (x) = \frac{1}{x - 1}, at x = 2

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