What is the solution for y^{'''}-y=1 ?

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

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y^{'''}-y=1

y^{^{'''}} - y = 1

y = c_{1} e^{t} + e^{- \frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) - 1

Solution steps

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

Solve linear ODE:

y = c_{1} e^{t} + e^{- \frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) - 1

y = c_{1} e^{t} + e^{- \frac{t}{2}} (c_{2} cos (\frac{3 t}{2}) + c_{3} sin (\frac{3 t}{2})) - 1

y^{^{''}} + 7 y^{^{'}} + 12 y = 4 e^{5 t}, y (0) = 2, y^{^{'}} (0) = 0

100 y^{^{''}} - 60 y^{^{'}} + 13 y + 30 = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 2 e^{4 x} + 3 x e^{2 x} + 4 x

y^{^{''}} + 16 y = 4 sin (2 x)

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 15 y = 6 + 4 t

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