What is the solution for x^{''}-10x^'+25x=4te^{5t} ?

The solution for x^{''}-10x^'+25x=4te^{5t} is x=c_{1}e^{5t}+c_{2}te^{5t}+(2e^{5t}t^3)/3

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x^{''}-10x^'+25x=4te^{5t}

x^{^{''}} - 10 x^{^{'}} + 25 x = 4 t e^{5 t}

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

Solution steps

x^{''} (t) - 10 x^{'} (t) + 25 x = 4 t e^{5 t}

Solve linear ODE:

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

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

x^{^{''}} + 9 x = 2 sin (t)

y^{^{''}} + 2 y^{^{'}} = 2

y^{^{''}} + 2 y^{^{'}} = 7

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = (1 + e^{- x})

y^{^{''}} - 4 y = \frac{e ^{2 x} - e ^{- 2 x}}{2}

What is the solution for x^{''}-10x^'+25x=4te^{5t} ?
The solution for x^{''}-10x^'+25x=4te^{5t} is x=c_{1}e^{5t}+c_{2}te^{5t}+(2e^{5t}t^3)/3