What is the solution for y^{''}-12y^'+36y=te^{6t} ?

The solution for y^{''}-12y^'+36y=te^{6t} is y=c_{1}e^{6t}+c_{2}te^{6t}+(e^{6t}t^3)/6

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y^{''}-12y^'+36y=te^{6t}

y^{^{''}} - 12 y^{^{'}} + 36 y = t e^{6 t}

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

Solution steps

y^{''} (t) - 12 y^{'} (t) + 36 y = t e^{6 t}

Solve linear ODE:

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

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

\int sin (\frac{nπ x}{4}) d x

\frac{d}{d x} (\frac{2 + sin ( x )}{2 x + cos ( x )})

y = ∣ x ∣, y = x^{2} - 1, x = - 1, x = 1

5 t \frac{d y}{d t} - 5 y = t

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

What is the solution for y^{''}-12y^'+36y=te^{6t} ?
The solution for y^{''}-12y^'+36y=te^{6t} is y=c_{1}e^{6t}+c_{2}te^{6t}+(e^{6t}t^3)/6