Below is a collection of some miscellaneous 1D differential equations. See here for more on 2D systems.

Trigonometryโ€ฆ?

ฬ‡๐‘ฅ=1+๐‘ฅ2

The form 1+๐‘ฅ2 is kind of reminiscent of Pythagorean theorem for a right triangle with legs 1 and ๐‘ฅ. 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 ๐‘ฅ=1. However, it rotates at a rate of 1rad๐‘ , and weโ€™d like to track the point of intersection between the lighthouseโ€™s beam and the line ๐‘ฅ=1.

Let the current point of intersection be (1,โ„Ž). Then, weโ€™d like to know how much โ„Ž increases after d๐‘ก seconds.

Our goal is to find ๐ท๐น in terms of ๐ด๐ท. Let ๐‘Ÿ=๐ถ๐ท; then, by Pythagorean theorem, we know that ๐‘Ÿ2=1+โ„Ž2. We also know that ๐ธ๐ท is just a very small arc, so its length is ๐‘Ÿd๐‘ก.

It remains to find ๐น๐ท in terms of ๐ธ๐ท, which can be done by noting that โ–ณ๐ท๐ธ๐นโˆผโ–ณ๐ถ๐ด๐ท . Therefore, ๐น๐ท๐ธ๐ท=๐ท๐ถ๐ด๐ถ=๐‘Ÿ, meaning dโ„Ž=๐น๐ท=๐‘Ÿ2d๐‘ก=(1+โ„Ž2)d๐‘ก, as desired.

Of course, โ„Ž is also simply tan(๐‘ก), therefore the solution to our original differential equation is simply

๐‘ฅ=tan(๐‘ก+๐ถ)

(remember to replace ๐‘ก with ๐‘ก+๐ถ since ๐‘ฅ is a time-invariant system.)

Note that in the general case of ฬ‡๐‘ฅ=๐‘ฅ2+๐‘Ž2 for ๐‘Ž>0, we must be careful to note that the solution is not just ๐‘ฅ=๐‘Žtan(๐‘ก+๐ถ) but rather

๐‘ฅ=๐‘Žtan(๐‘Ž(๐‘ก+๐ถ))

ฬ‡๐‘ฅ=โˆš1โˆ’๐‘ฅ2

We can also make a pretty nice geometric argument here in which โˆš1โˆ’๐‘ฅ2d๐‘ก is interpreted as the y-projection of an arc of length d๐‘ก. Details left to the reader, but solution turns out to be

๐‘ฅ=sin(๐‘ก+๐ถ)

ฬ‡๐‘ฅ=๐‘+๐‘“(๐‘ก)๐‘ฅ

ฬ‡๐‘ฅ=1+๐‘ฅ

Without the 1 on RHS, ๐‘ฅ would just be exp(๐‘ก+๐ถ). Then, note that adding some constant ๐ถ to ๐‘ฅ will leave the LHS unchanged while increasing the RHS by ๐ถ. Therefore, we should just be ๐ถ=โˆ’1 to get

๐‘ฅ=exp(๐‘ก+๐ถ)โˆ’1

ฬ‡๐‘ฅ=1+๐‘๐‘ก๐‘ฅ

Again, letโ€™s try solving it without the 1 on the RHS first. For which functions is differentiation equivalent to multiplication by ๐‘๐‘ก? This is just the class of functions ๐ถ๐‘ก๐‘.

This time, note that we cannot just add a constant ๐ถ to ๐‘ฅ, otherwise the RHS will have a nasty ๐ถ๐‘ก term. However, what if we add a ๐ถ๐‘ก term? Then the LHS will increase by ๐ถ while the RHS increases by ๐‘๐ถ. Therefore we want ๐ถ=1+๐‘๐ถโ‡’๐ถ=11โˆ’๐‘. So, our final solution is

๐‘ฅ=๐‘ก1โˆ’๐‘+๐ถ๐‘ก๐‘