What is the solution for y^{''}+2y^'+17y=0 ?

The solution for y^{''}+2y^'+17y=0 is y=e^{-t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}+2y^'+17y=0

y^{^{''}} + 2 y^{^{'}} + 17 y = 0

y = e^{- t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\frac{d}{d x} (\frac{1}{x ^{2} + 9})

\int \frac{5 x ^{2}}{x ^{2} + 7} d x

\int sin^{3} (a x) d x

sin (x), sin (2 x), [0, \frac{π}{3}]

f (x) = \frac{1}{x} - 2 x

What is the solution for y^{''}+2y^'+17y=0 ?
The solution for y^{''}+2y^'+17y=0 is y=e^{-t}(c_{1}cos(4t)+c_{2}sin(4t))