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 am presenting these two problems together because I feel they are essentially the same. Of course, regardless, problem 24 is more complex. In problem 24, let \dot{x}=y, then \ddot{x}=y\frac{dy}{dx}, so the original equation becomes: \frac{dy}{dx}=x^2+\frac{x^5}{y} This makes it similar in form to problem 23.
First, let’s solve problem 23. When I solve differential equations, I usually first analyze whether there are solutions of the form y=cx^n. This type of solution is generally the simplest. Substituting it in gives: cnx^{n-1}=x+\frac{1}{c}x^{3-n} We can find that such a solution indeed exists. Setting n-1=3-n gives n=2, and then we have 2c=1+\frac{1}{c}, which yields c=1 or c=-\frac{1}{2}. Thus, we have obtained two particular solutions.
However, this did not provide any special help to us. After reaching this point, I stopped for a long time. Two days ago, while rethinking this problem in class, I came up with an idea. The left side of the equation is in the form of a quotient, while the right side is in the form of a sum. To perform separation of variables, I intended to turn the left side into a product form, i.e., transform it into the form f(x')f(y') through a transformation. Based on the characteristics of the right side, we can set a new variable: x=t y^k Substituting this in, we have x+\frac{x^3}{y}=t y^k+t^3 y^{3k-1}. To combine like terms, let k=3k-1, which solves to k=\frac{1}{2}. That is, we can consider the transformation x=t\sqrt{y}. At this time, dx=\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy. The original equation can then become: \frac{dy}{\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy}=(t+t^3)\sqrt{y}
Note that until now, we have been exploring, and I did not know if it would ultimately succeed. Taking the reciprocal: \sqrt{y}\times \frac{\sqrt{y}dt+\frac{t}{2\sqrt{y}}dy}{dy}=\frac{1}{t+t^3}
Rearranging gives: y\frac{dt}{dy}=\frac{1}{t+t^3}-\frac{t}{2}, \quad \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, indicating that our initial guess was very successful. Moreover, this integral is not difficult. The result of the integration is: \begin{aligned} \text{Const.} + \ln y = -\frac{1}{3}\ln[(2+t^2)(t^2-1)^2] \\ y\sqrt[3]{(2+t^2)(t^2-1)^2} = \text{Const.} \end{aligned}
Substituting x=t\sqrt{y}, we get (2y+x^2)(x^2-y)^2=C
This can be regarded as the final answer. When C=0, we obtain the two particular solutions we solved for earlier. Therefore, we have successfully solved problem 23. As for problem 24, it is similar, and the answer will be analyzed in another article.
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