What is the solution for y^{'''}+y^'+2=0 ?

The solution for y^{'''}+y^'+2=0 is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)-2t

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y^{'''}+y^'+2=0

y^{^{'''}} + y^{^{'}} + 2 = 0

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) - 2 t

Solution steps

y^{'''} (t) + y^{'} (t) + 2 = 0

Solve linear ODE:

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) - 2 t

y = c_{1} + c_{2} cos (t) + c_{3} sin (t) - 2 t

(d^{3} - d^{2} - 6 d) y = x^{2} + 1

(d^{2} + 1) \cdot y = e^{- x} + cos (x) + x^{3} + e^{x} \cdot cos (x)

4 y^{^{''}} + 4 y = 2 sec (\frac{t}{2})

(d^{2} + 4 d + 5) y = 50 x + 13 e^{3} x

(d^{2} - 2 d + 2) y = e^{- x} + sin (x)

What is the solution for y^{'''}+y^'+2=0 ?
The solution for y^{'''}+y^'+2=0 is y=c_{1}+c_{2}cos(t)+c_{3}sin(t)-2t