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}

y^{^{'''}} + 3 y^{^{''}} - y^{^{'}} - 3 y = 3 x - 14

y^{^{''}} - 16 y^{^{'}} + 64 y = - \frac{2}{3 x} e^{8 x}

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 4 (x - 1)

y^{^{''}} - 6 y^{^{'}} - 3 + cos (x) = 0

\frac{d ^{2} y}{d x ^{2}} + y - 1 = 0, y (1) = 0

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