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

The solution for y^{''}-y^'=t^2,y(0)=0,y^'(0)=0 is y=2e^t-2-(t^3)/3-t^2-2t

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

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

y = 2 e^{t} - 2 - \frac{t ^{3}}{3} - t^{2} - 2 t

Solution steps

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

Solve linear ODE:

y = 2 e^{t} - 2 - \frac{t ^{3}}{3} - t^{2} - 2 t

y = 2 e^{t} - 2 - \frac{t ^{3}}{3} - t^{2} - 2 t

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 x

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 7 e^{- t} + 2

y^{^{'''}} + y^{^{''}} = e^{t} + t + 1

y^{^{''}} - y^{^{'}} = - 4

y^{^{''}} + b^{2} y = x cos (a x)

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