What is the solution for y^{'''}-y^'=9t ?

The solution for y^{'''}-y^'=9t is y=c_{1}+c_{2}e^{-t}+c_{3}e^t-(9t^2)/2

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y^{'''}-y^'=9t

y^{^{'''}} - y^{^{'}} = 9 t

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - \frac{9 t ^{2}}{2}

Solution steps

y^{'''} (t) - y^{'} (t) = 9 t

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - \frac{9 t ^{2}}{2}

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} - \frac{9 t ^{2}}{2}

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

y^{^{''}} + 4 y = sin^{2} (2 x)

x^{^{''}} + 36 x = 33 cos (5 t)

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} - 12 y = 24

y^{^{''}} + y = 3 x sin (x)

What is the solution for y^{'''}-y^'=9t ?
The solution for y^{'''}-y^'=9t is y=c_{1}+c_{2}e^{-t}+c_{3}e^t-(9t^2)/2