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?
Largest interval of validity for the solution of a first-order DE.
initial-value-problemsordinary differential equations
Related Solutions
I find it useful to plot the function after solving for the constant from the ICs.
We have:
$$y(x) = -\dfrac{8}{e^{4 x}-2}$$
A plot of $y(x)$ shows:
Of course, analytically, we care about where that denominator is $0$, so we have:
$$e^{4 x}-2 = 0 \rightarrow x = \dfrac{\ln 2}{4} \approx 0.173286795139986$$
From the plot, you can also see what happens as $x \rightarrow \pm \infty$, but it is not bad to derive that analytically either.
One last helpful thing you may have not gotten to yet, is a phase plot, which looks like:
This is a linear differential equation. Therefore it is defined on the largest intervals for which the functions used in the differential equation are all defined.
Hence solutions are defined on intervals for which $\cos x$ doesn't vanish. If you want a solution defined at $0$, the interval is $(-\frac{\pi}{2},\frac{\pi}{2})$
Best Answer
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.