What is the solution for y^{''}-0*y^'-2y=-2t+4t^2 ?

The solution for y^{''}-0*y^'-2y=-2t+4t^2 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}-2t^2+t-2

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y^{''}-0*y^'-2y=-2t+4t^2

y^{^{''}} - 0 \cdot y^{^{'}} - 2 y = - 2 t + 4 t^{2}

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

Solution steps

y^{''} (t) - 0 \cdot y^{'} (t) - 2 y = - 2 t + 4 t^{2}

Solve linear ODE:

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

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

y^{^{''}} - y = - 20 t + 2

y^{^{'''}} + 2 y^{^{''}} + y^{^{'}} = π e^{x} cos (x)

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

y^{^{''}} + 8 y + 15 = - 18 e^{- 2 t}, y (0) = - 6, y^{^{'}} (0) = 10

y^{^{''}} - 6 y^{^{'}} + 25 y = t^{2} + 2

What is the solution for y^{''}-0*y^'-2y=-2t+4t^2 ?
The solution for y^{''}-0*y^'-2y=-2t+4t^2 is y=c_{1}e^{sqrt(2)t}+c_{2}e^{-sqrt(2)t}-2t^2+t-2