Asymptotes of the integral curve

Question

One of the first tasks in differential equation problem book. Obviously that, it’s separable equation, but integrals can’t be computed. The task is to proof that integral curve have two assymptoties

Answer 1

Suppose that an asymptote exists $y=ax+b$ with $a\neq 0$.

For $x\to\infty$ in this case $y'\to a$ while $y'\sim \sqrt[3]{\frac{y^2}{x^4}}=\sqrt[3]{\frac{a^2}{x^2}}\to 0$ . This is contradictory, thus the supposition is false.

Consequence: If an asymptote exists it can only be an horizontal asymptote $y=c_1$ or a vertical asymptote $x=c_2$.

Suppose that an horizontal asymptote exists : $y=c_1$

For $x\to\infty$ in this case $y'\sim \sqrt[3]{\frac{c_1^2+1}{x^4}}\to 0$ which satisfies $y'=\frac{dc_1}{dx}=0$ .

Conclusion : $y=c_1$ is an asymptote.

Suppose that an horizontal asymptote exists : $x=c_2$

For $y\to\infty$ in this case $x'\sim \sqrt[3]{\frac{c_2^4+1}{y^2}}\to 0$ which satisfies $x'=\frac{dc_2}{dy}=0$ .

Conclusion : $x=c_2$ is an asymptote.

Thus they are two asymptotes, one horizontal and the other vertical.

$c_1$ and $c_2$ depend on some boundary condition not specified in the wording of the problem.

Answer 2

Separation of variables will work: $$f(y) = \int_{y_0}^y \frac {d\tau} {(\tau^2 + 1)^{1/3}} = \int_{x_0}^x \frac {d\tau} {(\tau^4 + 1)^{1/3}} = g(x).$$ Since $f(y)$ is strictly increasing with image $\mathbb R$, the domain of the solution $y(x) = f^{-1}(g(x))$ is $\mathbb R$. Since $g(x)$ has finite limits at $\pm \infty$, the limits of $y(x)$ are also finite.