An Epilogue to “A Relentless Pursuit”

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

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}.

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