Produce a differential equation from the family of curves $x=a\cdot \sin(t+b)$ I'm trying to produce a differential equation from the family of curves:$$x=a\cdot \sin(t+b), ~a,b\in \mathbb{R}$$
I differentiated once with respect to $t$, here $x$ is a function of $t$:
$$x'=a\cos(t+b) \Rightarrow a=\frac{x'}{\cos(t+b)}$$
and rewrote the equation as $$x=\frac{x'}{\cos(t+b)}\cdot \sin(t+b)=x' \tan(t+b)$$
differentiating again gives me:
$$x'=x'' \tan(t+b)+\sec^2(t+b)x'$$
Im not sure how to get rid of $b.$

If you settle for first order then you have an energy equation with only one retained constant $a$.
$$ x^2 + (\dot x)^2 =a^2 $$
For the second order by differenting this once more you have both constants vanishing resulting in the well known simple harmonic motion differential equation.
