Obtain the differential equation of the family of hyperbolas with center at the origin and transverse axis along the x-axis.

Step-by-step solution

1. 1

   Count the arbitrary constants to predict the order. The family carries two independent parameters, $a$ and $b$. Eliminating $n$ constants requires $n$ differentiations, so the answer will be a second-order differential equation. Knowing this in advance tells you when to stop differentiating.

2. 2

   Differentiate once with respect to $x$. Treating $y$ as a function of $x$ and using the chain rule on $y^2$:

   $\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0 \quad\Longrightarrow\quad \frac{x}{a^2}=\frac{y\,y'}{b^2}$

   The constant $1$ on the right of the original equation has already vanished — one relation between $a^2$ and $b^2$ is now available.

3. 3

   Differentiate a second time. Differentiating $\dfrac{x}{a^2}=\dfrac{y\,y'}{b^2}$ and using the product rule on $y\,y'$:

   $\frac{1}{a^2}=\frac{(y')^2+y\,y''}{b^2}$

4. 4

   Divide the two relations to kill both constants at once. Both equations have the form (something)$/a^2$ = (something)$/b^2$, so dividing the first by the second makes the ratio $b^2/a^2$ cancel:

   $\frac{x/a^2}{1/a^2}=\frac{y\,y'/b^2}{\left((y')^2+y\,y''\right)/b^2}\quad\Longrightarrow\quad x=\frac{y\,y'}{(y')^2+y\,y''}$

   This is the crucial economy of the method: one division removes $a$ and $b$ together, instead of solving for each separately.

5. 5

   Clear the denominator to get the differential equation.

   $x\left[\left(\frac{dy}{dx}\right)^2+y\,\frac{d^2y}{dx^2}\right]=y\,\frac{dy}{dx}$

   or, written with everything on one side,

   $x\,y\,y''+x\,(y')^2-y\,y'=0$

   No $a$ or $b$ appears, so it holds for every member of the family — which is exactly what "the differential equation of the family" means.

6. 6

   Verify on a concrete member. Take $a=2$, $b=3$, so $y=3\sqrt{x^2/4-1}$, and evaluate at $x=3$: there $y=3.354102$, $y'=2.012461$ and $y''=-0.536656$. Then

   $x\left[(y')^2+y\,y''\right]=3\left[4.050+(-1.800)\right]=6.750,\qquad y\,y'=3.354102\times 2.012461=6.750$

   Both sides agree, and repeating the check with different $a,b$ gives the same agreement — confirming the constants really were eliminated.