What is the solution for y^{'''}-y^'=e^t(1+t) ?

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

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y^{'''}-y^'=e^t(1+t)

y^{^{'''}} - y^{^{'}} = e^{t} (1 + t)

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

Solution steps

y^{'''} (t) - y^{'} (t) = e^{t} (1 + t)

Solve linear ODE:

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

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

(d^{2} - 2 d + 5) y = e^{2 x} \cdot sin (x)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{3} + 2 x^{2} + 5 x + 2

y^{^{''}} + 9 y = 2 x^{2} e^{3 x} + 5

3 y^{^{''}} - 5 y^{^{'}} + 6 y = e^{x} cos (2 x)

2 y^{^{'''}} - 3 y^{^{''}} + y^{^{'}} = x e^{2 x}

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