What is the solution for 16y''+8y'+y=0 ?

The solution for 16y''+8y'+y=0 is y=c_{1}e^{-t/4}+c_{2}e^{-t/4}t

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16y''+8y'+y=0

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

y = c_{1} e^{- \frac{t}{4}} + c_{2} e^{- \frac{t}{4}} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- \frac{t}{4}} + c_{2} e^{- \frac{t}{4}} t

y = c_{1} e^{- \frac{t}{4}} + c_{2} e^{- \frac{t}{4}} t

x \to 0 lim (\frac{tan ( 4 x ) cos ( 6 x ) - tan ( 4 x )}{24 x ^{3}})

x \to + 1 lim (\frac{x - 1}{x ^{6} - 1})

x \to 0 lim (\frac{3 x}{sin ( 10 x )})

\frac{\partial}{\partial x} (3 (y - 1)^{3} (x - 6))

\int \frac{x ^{3}}{x ^{2} + 4 x + 4} d x

What is the solution for 16y''+8y'+y=0 ?
The solution for 16y''+8y'+y=0 is y=c_{1}e^{-t/4}+c_{2}e^{-t/4}t