Largest interval of validity for the solution of a first-order DE.

Question

I have been trying to solve for the largest interval for which the particular solution of the differential equation \begin{equation*} 9x^2 \frac{dy}{dx} = y^2+3xy-36x^2, \; \; y(-1) = 0 \end{equation*} is defined. I've already solved for the particular solution which turned out to be \begin{equation*} \ln|x| + \frac{3}{2\sqrt{5}} \ln \left( \frac{3+\sqrt{5}}{2} \right) = \frac{3}{2\sqrt{5}} \ln \left| \frac{y/x - 3 - 3\sqrt{5}}{y/x - 3 + 3\sqrt{5}} \right| \end{equation*} I cannot seem to express this as an explicit function so I can determine the interval of validity. How can I solve for the interval if the particular solution is implicitly defined?

Answer 1

This equation is also a Riccati equation, so you could parametrize $y(x)=-4\dfrac{u(x)}{u'(x)}$ with $u(-1)=0$, $u'(-1)\ne 0$. Inserting this gives $$ y'=-4+4\frac{uu''}{u'^2} \\~\\ 9x^2y'=-36x^2+36x^2\frac{uu''}{u'^2} =16\frac{u^2}{u'^2}-12x\frac{u}{u'}-36x^2 \\~\\ \implies 0=9x^2u''+3xu'-4u $$ This now is an Euler-Cauchy equation with characteristic polynomial $$ 0=9m(m-1)+3m-4=9m^2-6m-4=(3m-1)^2-5 $$ The solution has thus the form $$ u(x)=A[(-x)^{(\sqrt5+1)/3}-(-x)^{-(\sqrt5-1)/3}] $$ This is defined on all of $(-\infty,0)$. The derivative in the denominator $$ u'(x)=\frac{A}3x^{-1}[(\sqrt5+1)(-x)^{(\sqrt5+1)/3}+(\sqrt5-1)(-x)^{-(\sqrt5-1)/3}] $$ has no roots on that interval.