What is the solution for y^{'''}-y^'=5sin(t) ?

The solution for y^{'''}-y^'=5sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+5/2 cos(t)

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y^{'''}-y^'=5sin(t)

y^{^{'''}} - y^{^{'}} = 5 sin (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{5}{2} cos (t)

Solution steps

y^{'''} (t) - y^{'} (t) = 5 sin (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{5}{2} cos (t)

y = c_{1} + c_{2} e^{- t} + c_{3} e^{t} + \frac{5}{2} cos (t)

y^{^{''}} + 4 y^{^{'}} + 8 y = sin (x), y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 3 y^{^{'}} + 33 y = 9 cos (5 x)

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 14 x^{2} - 4 x - 11

4 y^{^{''}} - y = \frac{e x}{2 + 7}

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

What is the solution for y^{'''}-y^'=5sin(t) ?
The solution for y^{'''}-y^'=5sin(t) is y=c_{1}+c_{2}e^{-t}+c_{3}e^t+5/2 cos(t)