What is the solution for y^{''}-6y^'+9y=t^2*e^{3t} ?

The solution for y^{''}-6y^'+9y=t^2*e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^4)/(12)

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y^{''}-6y^'+9y=t^2*e^{3t}

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} \cdot e^{3 t}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = t^{2} \cdot e^{3 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t} t ^{4}}{12}

y^{^{''}} - 36 y = 9 e^{6 x}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 8

y^{^{''}} - 4 y^{^{'}} + 3 y = 4 cos (x) - 3 sin (x)

y^{^{'''}} - 3 y^{^{''}} - y^{^{'}} + 3 y = 3

y^{^{''}} + 4 y^{^{'}} + 3 y = 6 t e^{3 t}

What is the solution for y^{''}-6y^'+9y=t^2*e^{3t} ?
The solution for y^{''}-6y^'+9y=t^2*e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t}t^4)/(12)