What is the solution for y^{''}+y^'=t^2+2t ?

The solution for y^{''}+y^'=t^2+2t is y=c_{1}+c_{2}e^{-t}+(t^3)/3

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y^{''}+y^'=t^2+2t

y^{^{''}} + y^{^{'}} = t^{2} + 2 t

y = c_{1} + c_{2} e^{- t} + \frac{t ^{3}}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + \frac{t ^{3}}{3}

y = c_{1} + c_{2} e^{- t} + \frac{t ^{3}}{3}

y^{^{''}} + 9 y = 10, e^{- t} y (0) y^{^{'}} (0) = 0

y^{^{''}} - 4 y^{^{'}} + 3 y = x cos (x)

y^{^{''}} + y = \frac{cos ( x ) + 1}{sin ( x )}

y^{^{''}} + y^{^{'}} = 1, y^{1} (0) = 1

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

What is the solution for y^{''}+y^'=t^2+2t ?
The solution for y^{''}+y^'=t^2+2t is y=c_{1}+c_{2}e^{-t}+(t^3)/3