What is the solution for 1/2 y^{''}+2y^'+6y=0 ?

The solution for 1/2 y^{''}+2y^'+6y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

Popular >

1/2 y^{''}+2y^'+6y=0

\frac{1}{2} y^{^{''}} + 2 y^{^{'}} + 6 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

\frac{1}{2} y^{''} (t) + 2 y^{'} (t) + 6 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))

y^{^{''}} + 16 y = 0, y (0) = 0, y (1) = 1

\frac{d ^{2}}{d t ^{2}} (θ (t)) + g θ (t) = 0

15 y^{^{''}} + 90 y^{^{'}} + 135 y = 0, y (0) = 3, y^{^{'}} (0) = 0

12 y^{^{''}} + 64 y = 0, y (0) = 4, y^{^{'}} (0) = 7

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 0, y^{^{'}} (0) = 0.9

What is the solution for 1/2 y^{''}+2y^'+6y=0 ?
The solution for 1/2 y^{''}+2y^'+6y=0 is y=e^{-2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))