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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to 2 lim (3 x^{2} + 2 x - 1)

\int \frac{1}{x ^{2} + a} d x

\int \frac{y ^{4} + y ^{2} - 1}{y ^{3} + y} d y

n = 0 \sum \infty \frac{5}{2 ^{n}}

\frac{5}{( s - 2 ) ( s - 3 ) ( s - 6 )}

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