What is the solution for 1/2 x^{''}+25x=0 ?

The solution for 1/2 x^{''}+25x=0 is x=c_{1}cos(sqrt(2)*5t)+c_{2}sin(sqrt(2)*5t)

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1/2 x^{''}+25x=0

\frac{1}{2} x^{^{''}} + 25 x = 0

x = c_{1} cos (2 \cdot 5 t) + c_{2} sin (2 \cdot 5 t)

Solution steps

\frac{1}{2} x^{''} (t) + 25 x = 0

Solve linear ODE:

x = c_{1} cos (2 \cdot 5 t) + c_{2} sin (2 \cdot 5 t)

x = c_{1} cos (2 \cdot 5 t) + c_{2} sin (2 \cdot 5 t)

3 y^{^{''}} - 24 y^{^{'}} = - 48 y

3 y^{^{''}} + 48 y = 0, y (0) = 3, y^{^{'}} (0) = 20

5 \frac{d ^{2} x}{d t ^{2}} - 3 x = 0

y^{^{''''}} - 4 y^{^{'''}} - 5 y^{^{''}} + 36 y^{^{'}} - 36 y = 0

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y^{^{'}} 0 = 10, y^{0} = 2

What is the solution for 1/2 x^{''}+25x=0 ?
The solution for 1/2 x^{''}+25x=0 is x=c_{1}cos(sqrt(2)*5t)+c_{2}sin(sqrt(2)*5t)