# 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 $b.$

• What happens if you differentiate twice directly? – infinitylord Oct 20 '17 at 7:40

Differentiate twice:

$$x=a\sin(t+b)\\ x' = a\cos(t+b) \\ x'' = -a \sin(t+b)$$

Thus the required equation is $x'' = -x$. Ha reminds me of simple harmonic motion.

• that's it? simpler than I made it look like – user463026 Oct 20 '17 at 7:46
• I think thats about it. – SJ. Oct 20 '17 at 7:59

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.

$x''(t)=-a\sin(t+b)$

so $x''(t)+x(t)=0$

is what you are looking for