What is the solution for y^{''}+9y=2t^2-4,y(0)=0,y^'(0)=-4 ?

The solution for y^{''}+9y=2t^2-4,y(0)=0,y^'(0)=-4 is y= 40/81 cos(3t)-4/3 sin(3t)+(2t^2)/9-40/81

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y^{''}+9y=2t^2-4,y(0)=0,y^'(0)=-4

y^{^{''}} + 9 y = 2 t^{2} - 4, y (0) = 0, y^{^{'}} (0) = - 4

y = \frac{40}{81} cos (3 t) - \frac{4}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{40}{81}

Solution steps

y^{''} (t) + 9 y = 2 t^{2} - 4

Solve linear ODE:

y = \frac{40}{81} cos (3 t) - \frac{4}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{40}{81}

y = \frac{40}{81} cos (3 t) - \frac{4}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{40}{81}

y^{^{''}} + 3 y^{^{'}} + 4 y = 2 sin (3 t)

y^{^{''}} + y^{^{'}} = 2 x + x

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 e^{- 2 x} - e^{- x}

y^{^{''''}} - 16 y = 2 e^{- 2 x} + 3 e^{- 3 x} + cos (2 x) - 1

y^{^{''}} + 19 y = 4 e^{8 x}

What is the solution for y^{''}+9y=2t^2-4,y(0)=0,y^'(0)=-4 ?
The solution for y^{''}+9y=2t^2-4,y(0)=0,y^'(0)=-4 is y= 40/81 cos(3t)-4/3 sin(3t)+(2t^2)/9-40/81