What is the solution for 64y^{''}+16y^'+y=0 ?

The solution for 64y^{''}+16y^'+y=0 is y=c_{1}e^{-t/8}+c_{2}e^{-t/8}t

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64y^{''}+16y^'+y=0

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

y = c_{1} e^{- \frac{t}{8}} + c_{2} e^{- \frac{t}{8}} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- \frac{t}{8}} + c_{2} e^{- \frac{t}{8}} t

y = c_{1} e^{- \frac{t}{8}} + c_{2} e^{- \frac{t}{8}} t

\int t^{2} cos (1 - t^{3}) d t

\int \frac{1}{4 x ^{3}} d x

t^{3} - 3 t^{2}

\frac{d}{d x} (2 x^{2} - 4 x - 6)

\frac{d}{d x} (arcsec (\frac{x}{2}))

What is the solution for 64y^{''}+16y^'+y=0 ?
The solution for 64y^{''}+16y^'+y=0 is y=c_{1}e^{-t/8}+c_{2}e^{-t/8}t