What is the solution for y^{''}+9y=2t^2-2,y(0)=0,y^'(0)=-8 ?

The solution for y^{''}+9y=2t^2-2,y(0)=0,y^'(0)=-8 is y= 22/81 cos(3t)-8/3 sin(3t)+(2t^2)/9-22/81

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y^{''}+9y=2t^2-2,y(0)=0,y^'(0)=-8

y^{^{''}} + 9 y = 2 t^{2} - 2, y (0) = 0, y^{^{'}} (0) = - 8

y = \frac{22}{81} cos (3 t) - \frac{8}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{22}{81}

Solution steps

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

Solve linear ODE:

y = \frac{22}{81} cos (3 t) - \frac{8}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{22}{81}

y = \frac{22}{81} cos (3 t) - \frac{8}{3} sin (3 t) + \frac{2 t ^{2}}{9} - \frac{22}{81}

What is the solution for y^{''}+9y=2t^2-2,y(0)=0,y^'(0)=-8 ?
The solution for y^{''}+9y=2t^2-2,y(0)=0,y^'(0)=-8 is y= 22/81 cos(3t)-8/3 sin(3t)+(2t^2)/9-22/81