What is the solution for 1/2 y^{''}+20y=0 ?

The solution for 1/2 y^{''}+20y=0 is y=c_{1}cos(2sqrt(10)t)+c_{2}sin(2sqrt(10)t)

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1/2 y^{''}+20y=0

\frac{1}{2} y^{^{''}} + 20 y = 0

y = c_{1} cos (210 t) + c_{2} sin (210 t)

Solution steps

\frac{1}{2} y^{''} (t) + 20 y = 0

Solve linear ODE:

y = c_{1} cos (210 t) + c_{2} sin (210 t)

y = c_{1} cos (210 t) + c_{2} sin (210 t)

y^{^{''}} + 0 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

- 3 y^{^{''}} + 12 y = 0

y^{^{''}} + 120 y^{^{'}} + 10000 y = 0

y^{^{'''}} - 18 y^{^{''}} + 108 y^{^{'}} - 216 y = 0

y^{^{''}} + 6 \cdot 0 + 16 y = 0

What is the solution for 1/2 y^{''}+20y=0 ?
The solution for 1/2 y^{''}+20y=0 is y=c_{1}cos(2sqrt(10)t)+c_{2}sin(2sqrt(10)t)