What is the solution for y^{'''}-y^'=te^{-t} ?

The solution for y^{'''}-y^'=te^{-t} is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+(2e^{-t}t^2+6e^{-t}t+7e^{-t})/8

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y^{'''}-y^'=te^{-t}

y^{^{'''}} - y^{^{'}} = t e^{- t}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{2 e ^{- t} t ^{2} + 6 e ^{- t} t + 7 e ^{- t}}{8}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{2 e ^{- t} t ^{2} + 6 e ^{- t} t + 7 e ^{- t}}{8}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{2 e ^{- t} t ^{2} + 6 e ^{- t} t + 7 e ^{- t}}{8}

y^{^{''}} - y^{^{'}} = tan (x)

y^{^{''}} + 4 y = \frac{5}{2} cos (2 x)

y^{^{'''}} + 2 y^{^{''}} - y^{^{'}} - 2 y = 1 - 4 x^{3}

y^{^{''''}} + 2 y^{^{''}} + y = 7 + cos (2 t)

x^{^{''}} + 6 x^{^{'}} + 34 x = 68, x (0) = 5, x^{^{'}} (0) = 7

What is the solution for y^{'''}-y^'=te^{-t} ?
The solution for y^{'''}-y^'=te^{-t} is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+(2e^{-t}t^2+6e^{-t}t+7e^{-t})/8