An Epilogue to “A Relentless Pursuit”

July 1, 2021January 5, 2023 / Michael Xue / Leave a comment

Let

$p=\frac{dy}{dx},$

differential equation (1) in “A Relentless Pursuit” can be expressed as

$\frac{dp}{dy}-\frac{1}{y}p = \frac{1}{y}p^{-1}.$

This is Bernoulli’s equation

$\frac{dp}{dy} + f(y) p = g(y) p^{\alpha}$

with $f(y) = -\frac{1}{y}, g(y) = \frac{1}{y}$ and $\alpha = -1$ (see “Meeting Mr. Bernoulli“).

Hence,

$p^{1-\alpha} = e^{(\alpha-1)\displaystyle\int f(y)\;dy}\left((1-\alpha)\displaystyle\int e^{-(\alpha-1)\displaystyle\int f(y)\;dy} g(y)\;dy + C\right).\quad\quad\quad(1)$

Substitute $f(y), g(y)$ and $\alpha$ into (1) gives

$p^{1-(-1)} = e^{(-1-1)\displaystyle\int -\frac{1}{y}\; dy}\left((1-(-1))\displaystyle\int e^{-(-1-1)\displaystyle\int -\frac{1}{y}\;dy}\frac{1}{y}\;dy+C\right).$

i.e.,

$p^2 = Cy^2-1$

where it must be true that $C>0$ . Therefore,

$p^2=\frac{1}{C_1^2}y^2-1, C_1>0\implies p^2 = \frac{y^2-C_1^2}{C_1^2} \implies p \overset{C_1>0}{=} \frac{\sqrt{y^2-C_1^2}}{\pm C_1}.$

That is,

$\frac{dy}{dx} = \frac{\sqrt{y^2-C_1^2}}{\pm C_1}.\quad\quad\quad(2)$

There is yet another way to obtain (2):

Since equation (1) in “A Relentless Pursuit“, namely,

$yy''-(y')^2 = 1 \implies 1+(y')^2 = y y''\quad\quad\quad(3)$

and

$(1+(y')^2)'= 2y' y''.\quad\quad\quad(4)$

(4)/(3) yields

$\frac{(1+(y')^2)'}{1+(y')^2} = \frac{2y'}{y}.\quad\quad\quad(5)$

Integrate (5) with respect to $y$ , we have

$\frac{1}{2}\log(1+(y')^2) = \log(y) + C \implies \log(\frac{\sqrt{1+(y')^2}}{y}) = C.$

i.e.,

$\frac{\sqrt{1+(y')^2}}{y}= e^C=\frac{1}{C_1}$

where $C_1>0$ .

It follows that

$\frac{dy}{dx} = \frac{\sqrt{y^2-C_1^2}}{\pm C_1}.$