## second order differential equation

I've created a second order homogeneous equation from my orginal data
m(d^2x/dt^2) + kx = 0
how can I turn it into a expression of displacement relevant to time?
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 $$\frac{d^2x}{dt^2} = \frac{-kx}{m}$$ Now, $$\frac{d^2x}{dt^2} = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dv}\frac{dv}{dx} = \frac{d(\frac{1}{2}v^2)}{dx}=\frac{d}{dx}\left(\frac{1}{2}\left(\frac{d x}{dt}\right)^2\right)$$ By applying the chain rule twice. So, $$\frac{d}{dx}\left(\frac{1}{2}\left(\frac{dx}{dt}\right)^2\right) = \frac{-kx}{m}$$ Hopefully you can now solve this as a differential equation
 Isn't it simpler to solve the first one, it being standard? $$x=ACos(\sqrt{\frac{k}{m}}t+\phi_{0})$$

## second order differential equation

Haha yeah, I guess so. But since I haven't learnt the theory for second order differential equations, this is a way using highschool calculus.

If you go through it, it comes out to be the same answer you have