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

The solution for y^{''}+9y=t^2e^3t+6 is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^3t^3)/9-(2e^3t)/(27)+2/3

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

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

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3} t ^{3}}{9} - \frac{2 e ^{3} t}{27} + \frac{2}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3} t ^{3}}{9} - \frac{2 e ^{3} t}{27} + \frac{2}{3}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{e ^{3} t ^{3}}{9} - \frac{2 e ^{3} t}{27} + \frac{2}{3}

\frac{1}{( s - 3 ) ( s + 1 )}

\int \frac{3 x - y}{9} d y

y^{^{''}} - 14 y^{^{'}} + 49 y = 3 t^{- 1} e^{7 t}

\frac{d}{d x} (2 csc (x) + 4 cot (x))

(- 3, 4), (2, - 9)

What is the solution for y^{''}+9y=t^2e^3t+6 ?
The solution for y^{''}+9y=t^2e^3t+6 is y=c_{1}cos(3t)+c_{2}sin(3t)+(e^3t^3)/9-(2e^3t)/(27)+2/3