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

The solution for y^{'''}-8y^'=0 is y=c_{1}+c_{2}e^{-2^{1+1/2}t}+c_{3}e^{2^{1+1/2}t}

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

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

y = c_{1} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + c_{3} e^{2^{1 + \frac{1}{2}} t}

Solution steps

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

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + c_{3} e^{2^{1 + \frac{1}{2}} t}

y = c_{1} + c_{2} e^{- 2^{1 + \frac{1}{2}} t} + c_{3} e^{2^{1 + \frac{1}{2}} t}

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

y^{^{''}} + 16 y = 0, y (0) = 2, y (0) = - 2

a y^{^{''}} + 2 y^{^{'}} + b y = 0, ab = 1

\frac{d ^{2} x}{d t ^{2}} + 25 x = 0

\frac{d ^{2} y}{d x ^{2}} - 1 \frac{d y}{d x} - 6 y = 0

What is the solution for y^{'''}-8y^'=0 ?
The solution for y^{'''}-8y^'=0 is y=c_{1}+c_{2}e^{-2^{1+1/2}t}+c_{3}e^{2^{1+1/2}t}