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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 2 - lim (\frac{4}{( x - 2 ) ^{5}})

y^{^{'}} - y = 4 sin (3 t)

\frac{1}{s - 1}

n = 4 \sum \infty \frac{1}{( n ) ( n + 1 )}

\int_{0}^{x} \frac{1}{( 1 - x )} d x

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