# A Relentless Pursuit

Problem: Solving differential equation $y \frac{d^2y}{dx^2} - (\frac{dy}{dx})^2 = 1.\quad\quad\quad(1)$

Solution:

Let $p = \frac{dy}{dx}.\quad\quad\quad(2)$

Then by chain rule, $\frac{d^2 y}{dx^2} = \frac{dp}{dx} =\frac{dp}{dy}\cdot\frac{dy}{dx} = p\frac{dp}{dy}.$

Rewrite (1) as $y\cdot p\frac{dp}{dy} - p^2=1.$

Equivalently, $\frac{p}{1+p^2}\cdot\frac{dp}{dy} = \frac{1}{y}.\quad\quad\quad(3)$

Integrate (3) with respect to $y$ gives $\displaystyle\int \frac{1}{2}\cdot\frac{2p}{1+p^2}\cdot\frac{dp}{dy}\;dy =\displaystyle \int \frac{1}{y}\;dy\implies\frac{1}{2}\log(1+p^2)= \log(y) + C.$

Hence $\log(\sqrt{1+p^2}) - \log(y) = C.$

Or, $\frac{\sqrt{1+p^2}}{y} = e^{C}=\frac{1}{C_1}\quad\quad\quad(4)$

where $C_1 > 0$ since $e^C>0$.

Square (4) gives $\frac{1+p^2}{y^2} = \frac{1}{C_1^2}.\quad\quad\quad(5)$

Solving for $p$ from (5), we obtain $p^2 = \frac{y^2-C_1^2}{C_1^2} \overset{(4)}{\implies} p = \pm \frac{\sqrt{y^2-C_1^2}}{C_1}.$

And so, $\frac{dy}{dx} \overset{(2)}{=} \frac{\sqrt{y^2-C_1^2}}{\pm C_1}.$

Or, $\pm C_1\cdot\displaystyle\frac{1}{\sqrt{y^2-C_1^2}}\cdot\frac{dy}{dx} = 1.\quad\quad\quad(6)$

Integrate (6) with respect to $x$ yields $\pm C_1\cdot\displaystyle\int\frac{1}{\sqrt{y^2-C_1^2}}\cdot\frac{dy}{dx}\;dx= \int\;dx\overset{(\star)}{\implies}\pm C_1\cdot\cosh^{-1}\left(\frac{y}{C_1}\right)=x+C_2.$

i.e., $\cosh^{-1}(\frac{y}{C_1}) = \frac{x+C_2}{\pm C_1}.$

Therefore, $y = C_1\cdot\cosh\left(\frac{x+C_2}{\pm C_1}\right)=C_1\cdot\cosh\left(\frac{x+C_2}{C_1}\right)$

since $\cosh$ is an even function.

Let $y=C_1z \implies \frac{dy}{dz}=C_1.$ $\displaystyle\int \frac{1}{\sqrt{y^2-C_1^2}} \;dy=\int\frac{C_1}{\sqrt{(C_1z)^2-C_1^2}}\;dz\overset{C_1>0}{=}\int\frac{1}{\sqrt{z^2-1}}\;dz\overset{(\star\star)}{=}\cosh^{-1}(z)$ $=\cosh^{-1}(\frac{y}{C_1}).$

i.e., $\displaystyle\int \frac{1}{\sqrt{y^2-C_1^2}} \;dy = \cos^{-1}\left(\frac{y}{C_1}\right).\quad\quad\quad(\star)$

Let $z = \sec(\theta) \implies \frac{dz}{d\theta} = \sec(\theta)\tan(\theta).$ $\displaystyle\int \frac{1}{\sqrt{z^2-1}}\;dz=\displaystyle\int \frac{\sec(\theta)\tan(\theta)}{\sqrt{\sec(\theta)^2-1}}\;d\theta$ $= \displaystyle\int \sec{\theta}\;d\theta \overset{(\star\star\star)}{=} \log(\sec(\theta)+\tan(\theta)) = \log(z + \sqrt{z^2-1})$

By definition (see “Deriving Two Inverse Functions“), $= \cosh^{-1}(z).$

i.e., $\displaystyle\int \frac{1}{\sqrt{z^2-1}} \;dz= \cosh^{-1}(z).\quad\quad\quad(\star\star)$

Without using a CAS, $\displaystyle \int \sec(x)\;dx = \displaystyle\int\frac{1}{\cos(x)}\;dx=\displaystyle\int \frac{\cos(x)}{\cos(x)^2}\;dx= \displaystyle\int\frac{\cos(x)}{1-\sin(x)^2}\;dx$

Let $t = \sin(x) \implies 1 = \cos(x)\frac{dx}{dt}\implies \frac{dx}{dt} = \sec(x)$, $=\displaystyle\int \frac{1}{1-t^2} \;dt = \displaystyle\int\frac{1}{2}\left(\frac{1}{1-t} + \frac{1}{1+t}\right)\;dt=\frac{1}{2}(-\log(1-t) + \log(1+t))$ $= \frac{1}{2}\log\left(\frac{1+t}{1-t}\right)=\frac{1}{2}\log\left(\frac{(1+t)(1+t)}{(1-t)(1+t)}\right)=\frac{1}{2}\log\left(\frac{(1+t)^2}{1-t^2}\right)=\frac{1}{2}\log\left(\frac{(1+\sin(x))^2}{1-\sin(x)^2}\right)$ $=\frac{1}{2}\log\left(\frac{(1+\sin(x))^2}{\cos(x)^2}\right)=\log\left(\frac{1+\sin(x)}{\cos(x)}\right)=\log(\sec(x) + \tan(x)).$

i.e., $\displaystyle\int \sec(x)\;dx = \log(\sec(x) + \tan(x)).\quad\quad\quad(\star\star\star).$

Exercise-1 What is the derivative of $\cosh^{-1}(x)?$ hint: $(\star\star)$

# Deriving Generalized Leibniz’s Integral Rule

The general form of Leibniz’s Integral Rule with variable limits states:

Suppose $f(x, t)$ satisfies the condition stated previously for the basic form of Leibniz’s Rule (LR-1, see “A Semi-Rigorous Derivation of Leibniz’s Rule“) . In addition, $a(t), b(t)$ are defined and have continuous derivatives for $t_1\le t\le t_2.$ Then for $t_1\le t \le t_2,$ $\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1)$

(1) can be derived as a consequence of LR-1, the Multivariable Chain Rule, and the Fundamental Theorem of Calculus (FTC):

Clearly, $\int\limits_{a(t)}^{b(t)}f(x, t)\;dx\quad\quad\quad(2)$

on the left side of (1) is a function of $t$.

Let $u = a(t),\quad\quad\quad(3)$ $v = b(t),\quad\quad\quad(4)$ $w = t,\quad\quad\quad(5)$

(2) can be expressed as $G(u,v,w) = \int\limits_{u}^{v}f(x,w)\;dx.$

Hence, by the chain rule, $\frac{d}{d t}\int\limits_{u}^{v}f(x, w)\;dx = \frac{\partial}{\partial u}G(u,v,w)\cdot \frac{du}{dt} + \frac{\partial}{\partial v} G(u,v,w)\cdot \frac{dv}{dt} + \frac{\partial}{\partial w}G(u,v,w)\cdot \frac{dw}{dt}$

where $\frac{\partial}{\partial u}G(u,v,w)=\frac{\partial}{\partial u}\int\limits_{u}^{v}f(x, w)\;dx$ $= \frac{\partial}{\partial u}\left(-\int\limits_{v}^{u}f(x, w)\;dx\right)$ $= -\frac{\partial}{\partial u}\int\limits_{v}^{u}f(x, w) \;dx$ $\overset{\textbf{FTC}}{=} -f(u, w)$ $\overset{(3), (5)}{=} -f(a(t), t),$ $\frac{\partial}{\partial v}G(u,v,w)=\frac{\partial}{\partial v}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{FTC}}{=}f(v, w) \overset{(4), (5)}{=} f(b(t), t)$

and, $\frac{\partial}{\partial w}G(u,v,w)=\frac{\partial}{\partial w}\int\limits_{u}^{v}f(x, w)\;dx \overset{\textbf{LR-1}}{=} \int\limits_{u}^{v}\frac{\partial}{\partial w}f(x, w)\;dx\overset{(3), (4), (5)}{=} \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$

It follows that $\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx =-f(a(t),t)\cdot a'(t) + f(b(t),t)\cdot b'(t) + \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx,$

i.e., $\frac{d}{d t}\int\limits_{a(t)}^{b(t)}f(x, t)\;dx = f(b(t),t)\cdot b'(t) -f(a(t),t)\cdot a'(t)+ \int\limits_{a(t)}^{b(t)}\frac{\partial}{\partial t}f(x, t)\;dx.$

# A Semi-Rigorous Derivation of Leibniz’s Rule

Leibniz’s rule (LR-1) states:

Let $f(x, \beta)$ be continuous and have a continuous derivative $\frac{\partial}{\partial \beta}$ in a domain of $x\beta-$plane that includes the rectangle $a \le x \le b, \beta_1 \le \beta \le \beta_2,$ $\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx =\int\limits_{a}^{b}\frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

I will derive LR-1 semi-rigorously as follows:

Let $g(t) = \int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx.\quad\quad\quad(1-1)$

Integrate (1-1) with respect to $t$ from a constant $\alpha$ to a variable $\beta$, we have $\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{\alpha}^{\beta}\left(\int\limits_{a}^{b} \frac{\partial}{\partial t}f(x, t)\;dx\right)\;dt$ $\overset{(\star)}{=}\int\limits_{a}^{b}\left(\int\limits_{\alpha}^{\beta} \frac{\partial}{\partial t}f(x, t)\;dt\right)\;dx$ $=\int\limits_{a}^{b}f(x, \beta) - f(x, \alpha)\; dx$ $=\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.$

That is, $\int\limits_{\alpha}^{\beta} g(t)\;dt = \int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx.\quad\quad\quad(1-2)$

While $\int\limits_{\alpha}^{\beta}g(t)\;dt$ and $\int\limits_{a}^{b}f(x,\beta)\;dx$ are functions of $\beta$, $\int\limits_{a}^{b}f(x,\alpha)\;dx$ is a constant.

Since $\frac{d}{d\beta}\int\limits_{\alpha}^{\beta} g(t)\;dt \overset{\textbf{FTC}}{= }g(\beta), \quad\frac{d}{d\beta}\left(\int\limits_{a}^{b}f(x, \beta)\;dx - \int\limits_{a}^{b}f(x,\alpha)\; dx\right)=\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx,$

differentiate (1-2) with respect to $\beta$ gives $g(\beta) = \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx\overset{(1-1)}{\implies} \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx= \frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx;$

i.e., $\frac{d}{d\beta}\int\limits_{a}^{b}f(x, \beta)\;dx = \int\limits_{a}^{b} \frac{\partial}{\partial \beta}f(x, \beta)\;dx.$

In the three-dimensional $x, y, z$-space, the double integral of a continuous function with two independent variables, $V=\iint_{R} f(x, y) dx dy$, may be interpreted as a volume between the surface $z=f(x, y)$ and the $x, y$-plane:

Fig. 1 $V = \iint_R f(x,y) dx dy$

We see from Fig. 1 that on one hand, $V=\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy,\quad\quad\quad(2-1)$

but on the other hand, $V=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(2-2)$

Since (2-1) and (2-2) amounts to the same thing, it must be true that $\int\limits_{c}^{d}\int\limits_{a}^{b}f(x,y)\;dx dy=\int\limits_{a}^{b}\int\limits_{c}^{d}f(x,y)\;dy dx.\quad\quad\quad(\star)$

In other words, the order of integration can be interchanged.