What is the solution for y^{''}+9y=5t^4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/9-(20t^2)/(27)-40/243 cos(3t)+40/243

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y^{''}+9y=5t^4,y(0)=0,y^'(0)=0

y^{^{''}} + 9 y = 5 t^{4}, y (0) = 0, y^{^{'}} (0) = 0

y = \frac{5 t ^{4}}{9} - \frac{20 t ^{2}}{27} - \frac{40}{243} cos (3 t) + \frac{40}{243}

Solution steps

y^{''} (t) + 9 y = 5 t^{4}

Solve linear ODE:

y = \frac{5 t ^{4}}{9} - \frac{20 t ^{2}}{27} - \frac{40}{243} cos (3 t) + \frac{40}{243}

y = \frac{5 t ^{4}}{9} - \frac{20 t ^{2}}{27} - \frac{40}{243} cos (3 t) + \frac{40}{243}

\frac{d}{d x} (arcsech (x))

\frac{\partial}{\partial x} (6 x^{\frac{1}{2}} + 2 y)

x \to 0 lim (sin (x) ln (4 x))

\int \frac{I n ( x + 1 )}{x + 1} d x

What is the solution for y^{''}+9y=5t^4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y=5t^4,y(0)=0,y^'(0)=0 is y=(5t^4)/9-(20t^2)/(27)-40/243 cos(3t)+40/243