What is the solution for y^{''}-y=-20t+2 ?

The solution for y^{''}-y=-20t+2 is y=c_{1}e^t+c_{2}e^{-t}+20t-2

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

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

y = c_{1} e^{t} + c_{2} e^{- t} + 20 t - 2

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} e^{- t} + 20 t - 2

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

y^{^{''}} + 2 y^{^{'}} + 5 y = e^{- t} cos (2 t)

What is the solution for y^{''}-y=-20t+2 ?
The solution for y^{''}-y=-20t+2 is y=c_{1}e^t+c_{2}e^{-t}+20t-2