solutions to these differential equations, intermediate steps

Question

So say I have 2 equations of motion in Cartesian coordinates

$\ddot{x} = -\omega^2x$

and

$\ddot{y} = -\omega^2y.$

How would I get to the solutions:

$x = X\cos[\omega t + A]$

and $y=Y\cos[\omega t + B]?$

Thanks in advance :)

Answer 1

Both equations can be solved the same way $$\ddot{x} = -\omega^2x$$ $$\ddot{x} +\omega^2x=0$$ $$\ddot{x} -i^2\omega^2x=0$$ The characteristic polynomial is $$R^2 -i^2\omega^2=0 \implies R= \pm i \omega$$ Therefore the solution is :

$$x(t)=c_1cos (\omega t)+c_2 \sin( \omega t) \tag {1}$$

Note that $$x = X\cos[\omega t + A]=X( \cos( \omega t)\cos (A)- \sin(\omega t)\sin(A))$$ $$x = X\cos (A) \cos( \omega t)- X\sin(A)\sin(\omega t)$$ Substitute in 1 ( I suppose X is a constant ): $$c_1=X\cos (A) \,\, ; c_2=- X\sin(A)$$

Answer 2

There are multiple ways of doing this, but most of them amount to the same procedure: guess-and-check. The $\sin$ and $\cos$ functions are the ones that get you to their own negatives after two differentiations, so they're a good candidate for the solution. Then you just need to take the $\omega$ into account, which you can do via the Chain Rule. These should be in your standard DE's toolbox. The $X$ and $Y$ amount to constants of integration, which are multiplicative in this case because of an underlying exponential structure.