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

\int_{1}^{\infty} x e^{- 10 x} d x

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x^{2} - 2, 7

\frac{d ^{2}}{d x ^{2}} (arccos (3 x))

\int \frac{1}{x ln ( x )} 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