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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 10 y^{^{'}} + 25 y = x cos (5 x)

y^{^{''}} + 9 y = 3 x, y (0) = 1, y^{^{'}} (0) = 3

y^{^{'''}} + 8 y^{^{''}} - 13 y^{^{'}} - 140 y = 4 e^{4} t

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

y^{^{''}} - 35 y^{^{'}} + 9 y = 60 t, y (0) = - 1, y^{^{'}} (0) = 2

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