What is the solution for 1/4 y^{''}+sqrt(2)y^'+2y=0 ?

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

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1/4 y^{''}+sqrt(2)y^'+2y=0

\frac{1}{4} y^{^{''}} + 2 y^{^{'}} + 2 y = 0

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

Solution steps

\frac{1}{4} y^{''} (t) + 2 y^{'} (t) + 2 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 ^{2} y}{d t ^{2}} + 25 y = 0

y^{^{'''}} + 9 y^{^{''}} = 0

y^{^{''}} + 36 y = 0, y (π) = 0, y (0) = 0

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

y^{^{'''}} - y^{^{''}} - 56 y^{^{'}} = 0

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