What is the solution for y^{''}+4y^'+12y=0 ?

The solution for y^{''}+4y^'+12y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

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y^{''}+4y^'+12y=0

y^{^{''}} + 4 y^{^{'}} + 12 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

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

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

\int_{0}^{4} (13 - t) t d t

y = t^{2} + 5 cos (t)

\frac{d y}{d x} sin (x + y) = 0

\int \frac{sec ( x ) + cos ( x )}{4 cos ( x )} d x

x \to 0 lim (x \cdot lo g_{2} (x))

What is the solution for y^{''}+4y^'+12y=0 ?
The solution for y^{''}+4y^'+12y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))