TITLE: Unsolvable Differential Equation?
QUESTION [14 upvotes]: Is there an analytic solution to the following equation?
$$ \frac{dx}{dy} = \frac{x^2 - y^2}{x^2 + y^2}  $$
I believe the answer is 'no'. It isn't separable or exact
I've had trouble finding any documentation that can help me explore this problem further. I have solved the problem numerically to explore the solution space, and it seems to make sense as the answer, but I also need to show some work in attempting to solve it analytically.
Is there a substitution that can be done to solve this? Or is this, in fact, an unsolvable problem (analytically)?

REPLY [5 votes]: Given homogenius ODE is
$$\dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{y^2+x^2}{y^2-x^2}.$$
Can be applied substitution
$$y= (u-1)x,\tag1$$
$$x\dfrac{\mathrm du}{\mathrm dx} = -u+1-\dfrac{(u-1)^2+1}{(u-1)^2-1},$$
$$x\dfrac{\mathrm du}{\mathrm dx}=-u-\dfrac{2}{(u-1)^2-1},$$
$$\dfrac{\mathrm dx}{x}=\dfrac{2u-u^2}{u^3-2u^2+2}{\mathrm du}.\tag2$$
Since the denominator of RHS(2) has a root
$$r=\dfrac13(2-\sqrt[3]{19+\sqrt{297}}-\sqrt[3]{19-\sqrt{297}})\approx-0.83928\,67552\,14161,\tag3$$
\begin{align}
&27(r^3-2r^2+2) = (3r-2)^3 - 12(3r-2) + 38\\
&=  -\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)^3
+12\left(\sqrt[3]{19+\sqrt{297}}+\sqrt[3]{19-\sqrt{297}}\right)+38\\
&= \left(-3\sqrt[3]{19+\sqrt{297}\,}\,\sqrt[3]{19-\sqrt{297}\,}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right)\\
&= \left(-3\sqrt[3]{64}+12\right)\left(\sqrt[3]{19+\sqrt{297}\,}\,+\sqrt[3]{19-\sqrt{297}\,}\,\right) = 0,
\end{align}
then
$$u^3-2u^2+2 = (u-r)(u^2-(2-r)u+r^2-2r),$$
$$R(u)=\dfrac{2u-u^2}{u^3-2u^2+2}=\dfrac A{u-r}+\dfrac{B(2u-2+r)}{u^2-(2-r)u+r^2-2r}
+\dfrac{C}{u^2-(2-r)u+r^2-2r},$$
where
$$A = \lim_{u\to r}(u-r)R(u) = \dfrac{2r-r^2}{3r^2-4r} = -\dfrac{2-r}{4-3r},$$
$$A+2B = \lim_{u\to \infty}uR(u) = -1,\quad B=-\dfrac{1-r}{4-3r},$$
$$\dfrac {B-A}r+\dfrac C{r^2-2r} = R(0)=0,\quad C=\dfrac{2-r}{4-3r},$$
$$u^2-(2-r)u+r^2-2r = \left(u-\dfrac{2-r}2\right)^2+\dfrac{(2-r)(-3r-2)}4.$$
Then from $(2)$ should
$$(4-3r)\ln \mathrm{const}|x| = -(2-r)\ln|u-r|-(1-r)\ln|u^2-(2-r)u+r^2-2r|\\
+2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}},$$
$$\color{green}{\mathbf{\small C_1|x|^{4-3r}|u-r||u^3-2u^2-2|^{1-r} = \exp\left(2\sqrt{\dfrac{2-r}{-3r-2}}\arctan\dfrac{2u-2+r}{\sqrt{(2-r)(-3r-2)}}\right)}},\tag4$$
where $C_1$ is the arbitrary constant.
Formulas $(1),(3),(4)$ present the common solution of the given ODE.