Some fun ones

Trigonometry…?

̇𝑥=1+𝑥2

The form 1+𝑥2 is kind of reminiscent of Pythagorean theorem for a right triangle with legs 1 and 𝑥. How can we use this to our advantage?

Consider the following setup: a lighthouse is situated at the origin and currently aims directly at the line 𝑥=1. However, it rotates at a rate of 1rad𝑠, and we’d like to track the point of intersection between the lighthouse’s beam and the line 𝑥=1.

Let the current point of intersection be (1,). Then, we’d like to know how much increases after d𝑡 seconds.

Our goal is to find 𝐷𝐹 in terms of 𝐴𝐷. Let 𝑟=𝐶𝐷; then, by Pythagorean theorem, we know that 𝑟2=1+2. We also know that 𝐸𝐷 is just a very small arc, so its length is 𝑟d𝑡.

It remains to find 𝐹𝐷 in terms of 𝐸𝐷, which can be done by noting that 𝐷𝐸𝐹𝐶𝐴𝐷 . Therefore, 𝐹𝐷𝐸𝐷=𝐷𝐶𝐴𝐶=𝑟, meaning d=𝐹𝐷=𝑟2d𝑡=(1+2)d𝑡, as desired.

Of course, is also simply tan(𝑡), therefore the solution to our original differential equation is simply

𝑥=tan(𝑡+𝐶)

(remember to replace 𝑡 with 𝑡+𝐶 since 𝑥 is a time-invariant system.)

Note that in the general case of ̇𝑥=𝑥2+𝑎2 for 𝑎>0, we must be careful to note that the solution is not just 𝑥=𝑎tan(𝑡+𝐶) but rather

𝑥=𝑎tan(𝑎(𝑡+𝐶))

̇𝑥=1𝑥2

We can also make a pretty nice geometric argument here in which 1𝑥2d𝑡 is interpreted as the y-projection of an arc of length d𝑡. Details left to the reader, but solution turns out to be

𝑥=sin(𝑡+𝐶)

̇𝑥=𝑐+𝑓(𝑡)𝑥

̇𝑥=1+𝑥

Without the 1 on RHS, 𝑥 would just be exp(𝑡+𝐶). Then, note that adding some constant 𝐶 to 𝑥 will leave the LHS unchanged while increasing the RHS by 𝐶. Therefore, we should just be 𝐶=1 to get

𝑥=exp(𝑡+𝐶)1

̇𝑥=1+𝑐𝑡𝑥

Again, let’s try solving it without the 1 on the RHS first. For which functions is differentiation equivalent to multiplication by 𝑐𝑡? This is just the class of functions 𝐶𝑡𝑐.

This time, note that we cannot just add a constant 𝐶 to 𝑥, otherwise the RHS will have a nasty 𝐶𝑡 term. However, what if we add a 𝐶𝑡 term? Then the LHS will increase by 𝐶 while the RHS increases by 𝑐𝐶. Therefore we want 𝐶=1+𝑐𝐶𝐶=11𝑐. So, our final solution is

𝑥=𝑡1𝑐+𝐶𝑡𝑐