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

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

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y^{''}+9y=t^3

y^{^{''}} + 9 y = t^{3}

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

Solution steps

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

Solve linear ODE:

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

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

n = 1 \sum \infty \frac{6}{n ( n + 3 )}

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

\int sin (\frac{nπ x}{L}) d x

f (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 1)^{3}

\int \frac{1}{sec ( x ) tan ( x )} d x

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