What is the solution for y^{''}+y=3t^2,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=3t^2,y(0)=0,y^'(0)=0 is y=3t^2+6cos(t)-6

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y^{''}+y=3t^2,y(0)=0,y^'(0)=0

y^{^{''}} + y = 3 t^{2}, y (0) = 0, y^{^{'}} (0) = 0

y = 3 t^{2} + 6 cos (t) - 6

Solution steps

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

Solve linear ODE:

y = 3 t^{2} + 6 cos (t) - 6

y = 3 t^{2} + 6 cos (t) - 6

y^{^{''}} + 4 y = cos (4 t)

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

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 3 = e^{- 5 x}

(d^{2} - 9) y = 30 sin^{3} (x)

3 y^{^{''}} - 8 y^{^{'}} - 3 y = 6

What is the solution for y^{''}+y=3t^2,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=3t^2,y(0)=0,y^'(0)=0 is y=3t^2+6cos(t)-6