What is the solution for 2y^{''}+16y=0 ?

The solution for 2y^{''}+16y=0 is y=c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t)

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2y^{''}+16y=0

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

y^{^{''}} - 15 y^{^{'}} + 54 y = 0

2 y^{^{''}} + 4 y^{^{'}} + 2 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = \frac{2}{3}, y^{^{'}} (0) = - \frac{5}{3}

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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