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

The solution for y^{'''}-y^{''}=t^2+e^t is y=c_{1}e^t+c_{2}+c_{3}t-(t^4}{12}-\frac{t^3)/3-t^2+e^tt

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

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

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{t ^{4}}{12} - \frac{t ^{3}}{3} - t^{2} + e^{t} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{t ^{4}}{12} - \frac{t ^{3}}{3} - t^{2} + e^{t} t

y = c_{1} e^{t} + c_{2} + c_{3} t - \frac{t ^{4}}{12} - \frac{t ^{3}}{3} - t^{2} + e^{t} t

y^{^{''}} - 2 y^{^{'}} - 15 y = 480 e^{- t}

y^{^{''}} - 4 y^{^{'}} + 4 y = (24 x^{2} + 2) e^{2 x}

y^{^{''}} - 3 y^{^{'}} - 4 y = cos (x)

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 7 e^{- t} + 4

y^{^{''}} + y^{^{'}} + 2 y = sin (t)

What is the solution for y^{'''}-y^{''}=t^2+e^t ?
The solution for y^{'''}-y^{''}=t^2+e^t is y=c_{1}e^t+c_{2}+c_{3}t-(t^4}{12}-\frac{t^3)/3-t^2+e^tt