What is the solution for y^{''}+4y^'+16y=0 ?

The solution for y^{''}+4y^'+16y=0 is y=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))

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y^{''}+4y^'+16y=0

y^{^{''}} + 4 y^{^{'}} + 16 y = 0

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 16 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

y = e^{- 2 t} (c_{1} cos (23 t) + c_{2} sin (23 t))

6 ln (x)

27 sin (3 x)

\frac{d y}{d x} + \frac{6}{x} y = 0

u^{\frac{7}{2}}

n = 1 \sum \infty \frac{2}{n ^{2} + 2 n}

What is the solution for y^{''}+4y^'+16y=0 ?
The solution for y^{''}+4y^'+16y=0 is y=e^{-2t}(c_{1}cos(2sqrt(3)t)+c_{2}sin(2sqrt(3)t))