What is the solution for y^{'''}+8y^'=0 ?

The solution for y^{'''}+8y^'=0 is y=c_{1}+c_{2}cos(2^{1+1/2}t)+c_{3}sin(2^{1+1/2}t)

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y^{'''}+8y^'=0

y^{^{'''}} + 8 y^{^{'}} = 0

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

Solution steps

y^{'''} (t) + 8 y^{'} (t) = 0

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} = 0

y^{^{''}} - 3 y^{^{'}} = 0, y (0) = 2 - 2 e^{3}, y (1) = 0

y^{^{''}} - 19 y^{^{'}} + 90 y = 0

y^{^{''}} + 5 y^{^{'}} + \frac{29}{4} y = 0, y (0) = 1, y^{^{'}} (0) = - \frac{7}{2}

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

What is the solution for y^{'''}+8y^'=0 ?
The solution for y^{'''}+8y^'=0 is y=c_{1}+c_{2}cos(2^{1+1/2}t)+c_{3}sin(2^{1+1/2}t)