What is the solution for 16y^{''}=-49y^' ?

The solution for 16y^{''}=-49y^' is y=c_{1}+c_{2}e^{-(49t)/(16)}

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16y^{''}=-49y^'

16 y^{^{''}} = - 49 y^{^{'}}

y = c_{1} + c_{2} e^{- \frac{49 t}{16}}

Solution steps

16 y^{''} (t) = - 49 y^{'} (t)

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{49 t}{16}}

y = c_{1} + c_{2} e^{- \frac{49 t}{16}}

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

y^{^{''}} + 4 y = 0, y (0) = 0

\frac{d ^{2} q}{d t ^{2}} + \frac{d q}{d t} + q = 0

y^{^{''}} - y^{^{'}} - 2 y = 0, y (0) = - 8, y^{^{'}} (0) = - 1

y^{^{''}} + 18 y = 0, y (0) = - 3, y^{^{'}} (0) = - 1

What is the solution for 16y^{''}=-49y^' ?
The solution for 16y^{''}=-49y^' is y=c_{1}+c_{2}e^{-(49t)/(16)}