What is the solution for 4y^{''}+4y^'+y=0 ?

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

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4y^{''}+4y^'+y=0

4 y^{^{''}} + 4 y^{^{'}} + y = 0

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

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + y = 0

Solve linear ODE:

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

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

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f (x) = \frac{x ^{3} - 8}{x ^{2} - 4}

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