What is the solution for x^{''}-18x^'+81x=te^{9t} ?

The solution for x^{''}-18x^'+81x=te^{9t} is x=c_{1}e^{9t}+c_{2}te^{9t}+(e^{9t}t^3)/6

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x^{''}-18x^'+81x=te^{9t}

x^{^{''}} - 18 x^{^{'}} + 81 x = t e^{9 t}

x = c_{1} e^{9 t} + c_{2} t e^{9 t} + \frac{e ^{9 t} t ^{3}}{6}

Solution steps

x^{''} (t) - 18 x^{'} (t) + 81 x = t e^{9 t}

Solve linear ODE:

x = c_{1} e^{9 t} + c_{2} t e^{9 t} + \frac{e ^{9 t} t ^{3}}{6}

x = c_{1} e^{9 t} + c_{2} t e^{9 t} + \frac{e ^{9 t} t ^{3}}{6}

x^{^{''}} - 4 x^{^{'}} - 5 x = 2 + e^{- t}

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 9 y = 9 x

y^{^{''}} + 9 y^{^{'}} + 22 y = e^{- 3 x}

(D^{2} - 4 D + 5) y = (x + 1)^{3}

d, d (t) = \frac{1}{2} t^{2} - 10 t

What is the solution for x^{''}-18x^'+81x=te^{9t} ?
The solution for x^{''}-18x^'+81x=te^{9t} is x=c_{1}e^{9t}+c_{2}te^{9t}+(e^{9t}t^3)/6