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

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

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x^{''}+2x=0

x^{^{''}} + 2 x = 0

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

Solution steps

x^{''} (t) + 2 x = 0

Solve linear ODE:

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

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

y = x^{3} - 3 x + 1

\frac{\partial}{\partial x} (sin (x) + x^{2} e^{y} - 1)

f (x) = \frac{ln ( x )}{x ^{3}}

x \to 0.01 lim (a x^{2} - x + b)

\int cos (x) (9 + 5 sin^{2} (x)) d x

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