What is the solution for y^{''}+4sqrt(2)y^'+8y=0 ?

The solution for y^{''}+4sqrt(2)y^'+8y=0 is y=c_{1}e^{-2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}t

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y^{''}+4sqrt(2)y^'+8y=0

y^{^{''}} + 42 y^{^{'}} + 8 y = 0

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

Solution steps

y^{''} (t) + 42 y^{'} (t) + 8 y = 0

Solve linear ODE:

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

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

\frac{d ^{3} y}{d x ^{3}} - 21 \frac{d y}{d x} - 20 y = 0

10 y^{^{''}} - 3 y^{^{'}} = 0

y^{^{''}} + 3 y^{^{'}} + 8 y = 0

\frac{d ^{2} x}{d t ^{2}} = x

y^{^{''}} - 6 y^{^{'}} + 8 y = 0, y (0) = 2, y^{^{'}} (0) = 2

What is the solution for y^{''}+4sqrt(2)y^'+8y=0 ?
The solution for y^{''}+4sqrt(2)y^'+8y=0 is y=c_{1}e^{-2^{1+1/2}t}+c_{2}e^{-2^{1+1/2}t}t