What is the solution for y^{''}+48/2 y=0 ?

The solution for y^{''}+48/2 y=0 is y=c_{1}cos(2sqrt(6)t)+c_{2}sin(2sqrt(6)t)

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y^{''}+48/2 y=0

y^{^{''}} + \frac{48}{2} y = 0

y = c_{1} cos (26 t) + c_{2} sin (26 t)

Solution steps

y^{''} (t) + \frac{48}{2} y = 0

Solve linear ODE:

y = c_{1} cos (26 t) + c_{2} sin (26 t)

y = c_{1} cos (26 t) + c_{2} sin (26 t)

y^{^{''}} - y^{^{'}} + 0.25 y = 0, y (0) = 2, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + y = 0, y (0) = 0, y^{^{'}} (0) = 1

2 y^{^{''}} + 10 y^{^{'}} + 8 y = 0, y^{^{'}} (0) = 0, y (0) = 5

2 y^{^{''}} - 4 y^{^{'}} + 5 y = 0

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} - 63 y = 0

What is the solution for y^{''}+48/2 y=0 ?
The solution for y^{''}+48/2 y=0 is y=c_{1}cos(2sqrt(6)t)+c_{2}sin(2sqrt(6)t)