Basic Skills in Mathematics 23, 24 (Part I)

数学基本技艺之23、24（上） 苏剑林 · 2013-09-26 · Mathematics competition differential equation art

23. Solve the quasi-homogeneous equation $\frac{dy}{dx}=x+\frac{x^3}{y}$
24. Solve the quasi-homogeneous equation $\ddot{x}=x^5+x^2\dot{x}$

I'm putting these two problems together because I feel that they are essentially the same problem — though of course problem 24 is a bit more complicated. In problem 24, let $\dot{x}=y$, so that $\ddot{x}=y\frac{dy}{dx}$, and the original equation becomes:

$$\frac{dy}{dx}=x^2+\frac{x^5}{y}$$

which puts it in roughly the same form as problem 23. more

Let's start by solving problem 23. When I solve a differential equation, I generally first check whether there's a solution of the form $y=cx^n$, since such solutions are usually the simplest. Substituting it in gives:

$$cnx^{n-1}=x+\frac{1}{c}x^{3-n}$$

We can see that such a solution does indeed exist: setting n-1=3-n gives n=2, and then we get $2c=1+\frac{1}{c}$, from which $c=1$ or $c=-\frac{1}{2}$. This gives us two particular solutions.

However, this doesn't really help us much. After getting this far I set the problem aside for quite a while, but a couple of days ago, while thinking about it again during class, I came up with an approach. The left-hand side of the equation is in the form of a quotient, while the right-hand side is a sum; in order to separate variables, I wanted to turn the left-hand side into a product, i.e., transform it into the form $f(x')f(y')$. Based on the structure of the right-hand side, I introduced a new variable:

$$x=t y^k$$

Substituting this in gives $x+\frac{x^3}{y}=t y^k+t^3 y^{3k-1}$. To combine like terms, I set $k=3k-1$, which gives $k=\frac{1}{2}$, so we can consider the transformation $x=t\sqrt{y}$. At this point $dx=\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy$. The original equation then becomes:

$$\frac{dy}{\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy}=(t+t^3)\sqrt{y}$$

It should be noted that up to this point we've been exploring blindly — I had no idea whether this would ultimately work out. Flipping it over:

$$\sqrt{y}\times \frac{\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy}{dy}=\frac{1}{t+t^3}$$

and rearranging gives:

$$y\frac{dt}{dy}=\frac{1}{t+t^3}-\frac{t}{2},\frac{dy}{y}=\frac{2t(1+t^2)dt}{2-t^2-t^4}=\frac{(1+t^2)d(t^2)}{2-t^2-t^4}$$

At this point we have successfully separated the variables, which shows that our initial guess paid off. Moreover, this integral isn't too hard; the result of the integration is:

$$\begin{eqnarray*}Const.+ln y=-\frac{1}{3}ln[(2+t^2)(t^2-1)^2] \\ y\sqrt[3]{(2+t^2)(t^2-1)^2}=Const.\end{eqnarray*}$$

Substituting back $x=t\sqrt{y}$, we obtain

$$(2y+x^2)(x^2-y)^2=C$$

and this can be taken as the final answer. When C=0, we recover the two particular solutions found earlier. So we have successfully solved problem 23. Problem 24 is handled in a similar way, and the answer will be analyzed in another post.