What is the solution for y^{''}+3y=t^3-1 ?

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

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y^{''}+3y=t^3-1

y^{^{''}} + 3 y = t^{3} - 1

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

Solution steps

y^{''} (t) + 3 y = t^{3} - 1

Solve linear ODE:

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

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

\frac{d}{d x} (5 e^{x} - 3 x)

f (x) = (x - 9)^{2}

\int \frac{41 arctan ( x )}{x ^{2}} d x

f (x) = sin (ln (x))

\frac{1}{x + 2}

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