Introduction
The general solution of a linear differential equation with constant coefficients is the sum of the complementary function (CF) and the particular integral (PI):
\[ y(x) = y_c(x) + y_p(x) \]
General Form & Operator Notation
We write the differential equation as:
\[ f(D)y = X(x), \quad D = \frac{d}{dx} \]
Complementary Function (CF)
Solve the homogeneous equation:
\[ f(D)y = 0 \]
The auxiliary equation \( f(m) = 0 \) gives roots that decide the form of CF.
Particular Integral (PI)
A particular solution is found as:
\[ y_p = \frac{1}{f(D)}X(x) \]
Steps to Solve
- Form the auxiliary equation and find CF.
- Apply rules for PI depending on \(X(x)\).
- Write general solution: \( y = y_c + y_p \).
Complementary Function (CF) — Roots of Auxiliary Equation
The Complementary Function (CF) depends on the nature of the roots of the auxiliary (characteristic) equation:
| Case | Roots of Auxiliary Equation | Form of Complementary Function \(y_c\) | Example |
|---|---|---|---|
| 1 | Two distinct real roots \(m_1, m_2\) | \(y_c = C_1 e^{m_1 x} + C_2 e^{m_2 x}\) | \(m^2-5m+6=0 \Rightarrow m=2,3 \Rightarrow y_c=C_1 e^{2x}+C_2 e^{3x}\) |
| 2 | \(n\) distinct real roots \(m_1,\dots,m_n\) | \(y_c = \displaystyle\sum_{k=1}^n C_k e^{m_k x}\) | \(m^3-6m^2+11m-6=0 \Rightarrow m=1,2,3 \Rightarrow y_c=\sum_{k=1}^3 C_k e^{m_k x}\) |
| 3 | One pair of complex conjugate roots \(\alpha \pm i\beta\) | \(y_c = e^{\alpha x}\big(C_1\cos\beta x + C_2\sin\beta x\big)\) | \(m^2+4m+13=0 \Rightarrow m=-2\pm3i \Rightarrow y_c=e^{-2x}(C_1\cos3x + C_2\sin3x)\) |
| 4 | Mixture of distinct real and (one or more) complex roots | Combine the corresponding terms: real roots give \(C e^{mx}\); complex pairs give \(e^{\alpha x}(C_1\cos\beta x+C_2\sin\beta x)\) | \(m=(1,\, -1,\, 2\pm i)\Rightarrow y_c=C_1e^{x}+C_2e^{-x}+e^{2x}(C_3\cos x + C_4\sin x)\) |
| 5 | Repeated real root of multiplicity \(r\) (e.g. \(m\) repeated) | \(y_c = (C_1 + C_2 x + C_3 x^2 + \dots + C_r x^{\,r-1}) e^{m x}\) | \(m^2-4m+4=0 \Rightarrow m=2,2 \Rightarrow y_c=(C_1+C_2x)e^{2x}\) |
| 6 | Repeated complex roots \((\alpha \pm i\beta)\) of multiplicity \(r\) | \(y_c = e^{\alpha x}\Big[(C_1 + C_2 x + \dots + C_r x^{\,r-1})\cos(\beta x) + (D_1 + D_2 x + \dots + D_r x^{\,r-1})\sin(\beta x)\Big]\) | \((m+2)^2+9=0 \Rightarrow m=-2\pm3i\) (repeated) \(y_c = e^{-2x}\big((C_1+C_2x)\cos3x+(D_1+D_2x)\sin3x\big)\) |
Worked Example
Solve \( (D^2 - 3D + 2)y = e^{2x} \):
Auxiliary equation: \( m^2 - 3m + 2 = 0 \implies m=1,2 \)
CF: \( y_c = C_1 e^x + C_2 e^{2x} \)
For PI: try \( y_p = \frac{e^{2x}}{f(2)} \). But \( f(2)=0 \), so multiply by \(x\).
\( y_p = x e^{2x} \)
General solution: \( y = C_1 e^x + C_2 e^{2x} + x e^{2x} \)
Tips & Pitfalls
- Check multiplicity of roots to adjust CF form.
- If denominator vanishes in PI, multiply trial by \(x\).
- Always verify by substitution.
\section{Proof / Derivation of Euler-Lagrange Equation}
Let all possible paths between $ A \And B $ such as $ y_1, y_2, y_3, y_4 $ etc. be represented by a family of curves or paths by the equation :\\
$ y(x,\alpha )=y\left( x \right)+\alpha \,\eta \left( x \right) $\\
&\text{Then}\\
\begin{enumerate}
\item
$.\dfrac{\partial y}{\partial \alpha }=\dfrac{\partial }{\partial \alpha }\left( y\left( x \right)+\alpha \,\eta \left( x \right) \right)=0+\eta \left( x \right)=\eta \left( x \right) $
\item $\dfrac{\partial y'}{\partial \alpha }=\dfrac{\partial }{\partial \alpha }\left( \dfrac{\partial y}{\partial x} \right)=\dfrac{\partial }{\partial \alpha }\left( \dfrac{\partial }{\partial x}\left[ y\left( x \right)+\alpha \,\eta \left( x \right) \right] \right)=\dfrac{\partial }{\partial \alpha }\left( \dfrac{\partial y}{\partial x}+\alpha \,\eta '\left( x \right) \right)=0+\eta '\left( x \right)=\eta '\left( x \right) $
\item $ \dfrac{\partial x}{\partial \alpha }=0 $
\item $ y({{x}_{1}},\alpha )=y\left( {{x}_{1}} \right)+\alpha \eta \left( {{x}_{1}} \right)\Rightarrow {{y}_{1}}={{y}_{1}}+\alpha \eta \left( {{x}_{1}} \right)\Rightarrow \eta \left( {{x}_{1}} \right)=0 $
\item $ y({{x}_{2}},\alpha )=y\left( {{x}_{2}} \right)+\alpha \eta \left( {{x}_{2}} \right)\Rightarrow {{y}_{2}}={{y}_{2}}+\alpha \eta \left( {{x}_{2}} \right)\Rightarrow \eta \left( {{x}_{2}} \right)=0 $
\end{enumerate}
\begin{align*}
&\text{Now $ f=f(y(x,\alpha),y'(x,\alpha),x)$}\\
& I=\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}f(y(x,\alpha),y'(x,\alpha),x)dx\\
&\text{Now $\alpha$ is a parameter on which I depends and for maximum or minimum values of I we will have }\\
& \Rightarrow \frac{\partial I}{\partial \alpha }=\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\frac{\partial y}{\partial \alpha }+\frac{\partial f}{\partial y'}\frac{\partial y'}{\partial \alpha }+\frac{\partial f}{\partial x}\frac{\partial x}{\partial \alpha } \right]\,}dx=0 \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+\frac{\partial f}{\partial y}\eta '\left( x \right)+\frac{\partial f}{\partial x}\times 0 \right]\,}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+\frac{\partial f}{\partial y'}\left( \frac{d}{dx}\eta \left( x \right) \right) \right]\,}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+ \right]\,}dx+\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\frac{\partial f}{\partial y'}\left( \frac{d}{dx}\eta \left( x \right) \right)}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+ \right]\,}dx+\frac{\partial f}{\partial y'}\left[ \eta \left( x \right) \right]_{x={{x}_{1}}}^{x={{x}_{2}}}-\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)\eta \left( x \right)}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+ \right]\,}dx+\frac{\partial f}{\partial y'}\left[ \eta \left( {{x}_{2}} \right)-\eta \left( {{x}_{1}} \right) \right]-\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)\eta \left( x \right)}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+ \right]\,}dx+\frac{\partial f}{\partial y'}\left[ 0-0 \right]-\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)\eta \left( x \right)}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}\eta \left( x \right)+ \right]\,}dx-\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)\eta \left( x \right)}dx \\
& =\int\limits_{x={{x}_{1}}}^{x={{x}_{2}}}{\left[ \frac{\partial f}{\partial y}-\frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right) \right]\eta \left( x \right)\,}dx=0 \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)-\frac{\partial f}{\partial y}=0 \\
\intertext{as $ \eta(x) dx $ is arbitrary}
\end{align*}
\section{Applications}
\subsection{Prove that Straight line is the shortest path between two points A and B lying on a plane.}
Let $ ds $ be the distance between Two points P and Q lying close to each other on the path between two points A and B on a plane. Then :
\begin{figure}
\begin{center}
\begin{tikzpicture}
\fill[blue!10] (-2,-1) rectangle (6,6);
\draw[thick, black, ->] (0,0)--(4,0);
\draw[thick, black, ->] (0,0)--(0,4);
\node[below] at (2,0) {$x-axis$}
\node[left=0.3cm,rotate=90] at (0,2) {$y-axis$}
\draw (1,1)..controls (2,3) and (2,3)..(4,4);
\node [] at (1,1) {$A$}
\node [] at (4,4) {$B$}
\node [] at (1.5,1.7) {$P$}
\node [] at (2,2.5) {$Q$}
\end{tikzpicture}
\caption{ The Curve between A and B on X-Y Plane .}
\label{fig:example-diagram2}
\end{center}
\end{figure}
\begin{align*}
& ds=\sqrt{d{{x}^{2}}+d{{y}^{2}}} \\
& =\sqrt{1+{{\left( \dfrac{dy}{dx} \right)}^{2}}}dx \\
& =\sqrt{1+y{{'}^{2}}}dx \\
\intertext{So the path distance between A and B is :}
S&=\int{ds=}\int{\sqrt{1+y{{'}^{2}}}dx} \\
& =\int{f\left( y' \right)dx} \\
& \Rightarrow f\left( y' \right)=\sqrt{1+y{{'}^{2}}} \\
& \Rightarrow \dfrac{d}{dx}\left( \dfrac{\partial f}{\partial y'} \right)-\dfrac{\partial f}{\partial y}=0 \\
& \Rightarrow \dfrac{d}{dx}\left( \dfrac{\partial }{\partial y'}\sqrt{1+y{{'}^{2}}} \right)-\dfrac{\partial }{\partial y}\sqrt{1+y{{'}^{2}}}=0 \\
& \Rightarrow \dfrac{d}{dx}\left( \dfrac{y'}{\sqrt{1+y{{'}^{2}}}} \right)-0=0 \\
& \Rightarrow \dfrac{y'}{\sqrt{1+y{{'}^{2}}}}=k \\
\intertext{Where $k$ is a constant}
& \Rightarrow y{{'}^{2}}={{k}^{2}}+{{k}^{2}}y{{'}^{2}} \\
& \Rightarrow {{y}^{'}}^{2}\left( 1-{{k}^{2}} \right)={{k}^{2}} \\
& \Rightarrow y'=\sqrt{\dfrac{{{k}^{2}}}{1-{{k}^{2}}}}=m \\
& \Rightarrow \dfrac{dy}{dx}=m \\
& \Rightarrow y=mx+c \\
\end{align*}
Which is a straight line .\\
\hrule
\subsection{\textbf{Brachistochrone Problem: Prove that the curve in a vertical space between two points A and B such that B is not exactly below A is a cycloid when a particle takes minimum time to descend from A to B under the action of gravity with all other forces absent. or the curve of quickest descent is a cycloid or Brachistochrone ( A greek word) }}\\
Let P and Q be two points on the curve between A and B in a vertical space as shown in the figure . The point A is at the origin and B is at B($x,y$) .A particle when released from A comes to the point P after descending through a height $x$ and attains a speed $v=\sqrt{2gx}$ that remains uniform between P and Q assuming the points P and Q to be very close such that the path between them can be straight,T is the time that the particle takes to travel from A to B by integrating the time taken to travel from P to Q which is given by $dt$ :
\begin{figure}[h]
\begin{center}
\resizebox{0.3\textwidth}{!}
{
\begin{tikzpicture}
\fill[blue!10] (-5,1) rectangle (6,-7);
\coordinate (A) at (2, 2);
\coordinate (B) at (3, 4);
\coordinate (C) at (1, 3);
\node[] at (3,-4) {$B$};
\node[above] at (0,0) {$A$};
\node[above] at (1,-1.5) {$P$};
\node[left] at (2,-2.5) {$Q$};
\draw[thick, balck,->] (0,0)--(5,0)
\node[below] at (2.5,0) {$x-axis$};
\node[left] at (0,-2.5) {$-y-axis$};
\draw[thick, red] (0,0) .. controls (1.5, -3) .. (3,-4);
\draw[thick, balck,->] (0,0)--(0,-5)
\end{tikzpicture}
}
\caption{Brachistochrone }
\label{Fig:brachistochrone}
\begin{align*}
& dt=\frac{ds}{v} \\
& =\frac{ds}{\sqrt{2gx}} \\
& =\frac{\sqrt{d{{x}^{2}}+d{{y}^{2}}}}{\sqrt{2gx}} \\
& =\sqrt{\frac{1+{{\left( \frac{dy}{dx} \right)}^{2}}}{2gx}}dx \\
& =\sqrt{\frac{1+{{y}^{'}}^{2}}{2gx}}dx \\
& \Rightarrow T=\int{dt} \\
& =\int{\sqrt{\frac{1+{{y}^{'}}^{2}}{2gx}}dx} \\
& =\int{f\left( x,y' \right)dx} \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)-\frac{\partial f}{\partial y}=0 \\
\end{align*}
\end{center}
\end{figure}
\[\begin{align}
& \Rightarrow \frac{d}{dx}\left( \frac{\partial }{\partial y'}\left( \sqrt{\frac{1+{{y}^{'}}^{2}}{2gx}} \right) \right)-\frac{\partial }{\partial y}\left( \frac{1+{{y}^{'}}^{2}}{2gx} \right) \\
& \Rightarrow \frac{d}{dx}\left( \frac{y'}{\sqrt{2gx\left( 1+y{{'}^{2}} \right)}} \right)-0=0 \\
\end{align}\]
\[\begin{align}
& \Rightarrow \frac{y'}{\sqrt{2gx\left( 1+y{{'}^{2}} \right)}}=k \\
& \Rightarrow y{{'}^{2}}=2gx{{k}^{2}}+2{{k}^{2}}gxy{{'}^{2}} \\
& \Rightarrow y'=\sqrt{\frac{2g{{k}^{2}}x}{\left( 1-2{{k}^{2}}gx \right)}} \\
& \Rightarrow y'=\sqrt{\frac{x}{1/2g{{k}^{2}}-x}} \\
& =\sqrt{\frac{x}{a-x}} \\
& \Rightarrow \frac{dy}{dx}=\sqrt{\frac{x}{a-x}} \\
& \Rightarrow dy=\sqrt{\frac{x}{a-x}}dx \\
& =\sqrt{\frac{a{{\sin }^{2}}\alpha }{a-a{{\sin }^{2}}\alpha }}2a\sin \alpha \cos \alpha d\alpha \\
& =2a{{\sin }^{2}}\alpha d\alpha \\
& =a\left( 1-\cos 2\alpha \right)d\alpha \\
& \Rightarrow y=a\left( \alpha -\frac{\sin 2\alpha }{2} \right)=\frac{a}{2}\left( 2\alpha -\sin 2\alpha \right) \\
& x=a{{\sin }^{2}}\alpha =\frac{a}{2}\left( 1-\cos 2\alpha \right) \\
& \Rightarrow 2\alpha =\theta \\
& x=\frac{a}{2}\left( 1-\cos \theta \right)\And y=\frac{a}{2}\left( \theta -\sin \theta \right)
\end{align}\]
the expression on $x$ and $y $represents a cycloid.
\vfill
\subsection{\textbf{Prove that the curve which when revolves about a fixed axis generates minimum surface area when it is a catenary. }}\\
Let the curve between A and B which when revolves about the vertical axis generates area A and dA is the area generated when an elementary thickness ds of the curve at distance x from the vertical axis revolves .Then dA is :
\[\begin{align}
& dA=2\pi xds \\
& =2\pi x\sqrt{d{{x}^{2}}+d{{y}^{2}}} \\
& =2\pi x\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}dx \\
& =2\pi x\sqrt{1+y{{'}^{2}}}dx \\
& \Rightarrow A=\int{2\pi x\sqrt{1+y{{'}^{2}}}dx} \\
& =2\pi \int{x\sqrt{1+y{{'}^{2}}}dx} \\
& =2\pi \int{f(x,y')dx} \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)-\frac{\partial f}{\partial y}=0 \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial }{\partial y'}\left( x\sqrt{1+y{{'}^{2}}} \right) \right)-\frac{\partial }{\partial y}\left( x\sqrt{1+y{{'}^{2}}} \right)=0 \\
& \Rightarrow \frac{d}{dx}\left( \frac{xy'}{\sqrt{1+y{{'}^{2}}}} \right)-0=0 \\
& \Rightarrow \frac{xy'}{\sqrt{1+y{{'}^{2}}}}=a \\
& \Rightarrow {{x}^{2}}y{{'}^{2}}={{a}^{2}}+{{a}^{2}}y{{'}^{2}} \\
& \Rightarrow y{{'}^{2}}\left( {{x}^{2}}-{{a}^{2}} \right)={{a}^{2}} \\
& \Rightarrow y'=\frac{a}{\sqrt{{{x}^{2}}-{{a}^{2}}}} \\
& \Rightarrow dy=\int{\frac{a}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}dx \\
\end{align}\]
\begin{figure}
\begin{center}
\begin{tikzpicture}
\fill[blue!20] (-5,-1) rectangle (5,5);
\draw[thick] (0,3) ellipse (1 and 0.2);
\draw[thick] (0,1) ellipse (1 and 0.2);
\draw[black,thick,<->] (0,0)--(0,4);
\draw (-1,1) .. controls (-0.5,2) and (-0.5,2) .. (-1,3);
\draw (1,1) .. controls (0.5,2) and (0.5,2) .. (1,3);
\draw [thick,<->](0,2)--(0.8,2);
\node[below] at (0.3,2) {$ x$};
\draw[thick,black](0.6,2)--(0.7,2.3);
\node[below,rotate=60]at(0.6,2.3) {$ ds $};
\node[right]at(1,1) {$A$}
\node[right]at(1,3) {$B$}
\end{tikzpicture}
\end{center}
\caption {Minimum surface are of revolution by a catenary about the vertical axis}
\label{Fig:Diagram5}
\end{figure}
\[\begin{align}
& =a{{\cosh }^{-1}}\left( \frac{x}{a} \right)+b \\
& \Rightarrow {{\cosh }^{-1}}\left( \frac{x}{a} \right)=\frac{y-b}{a} \\
& \Rightarrow x=a\cosh \left( \frac{y-b}{a} \right) \\
\end{align}\]
\subsection{\textbf{Geodesic on the surface of a Spherical surface is a part of its great circle.}}\\
Consider the Sphere $ {x}^{2}}+{{y}^{2}}+{{z}^{2}}={{r}^{2} $\\
We have to find the curve with shortest length $S$ between two points A and B such that the curve connecting the points will have all points on the surface of the sphere. Let P and Q are two points in between A and B which are close to each other such that the distance between them is :\\
\begin{align*}
& ds=\sqrt{d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}}} \\
\intertext{Where in spherical coordinates :}
& x=r\sin \theta \cos \phi \Rightarrow dx=r\cos \theta \cos \phi d\theta -r\sin \theta \sin \phi d\phi \\
& y=r\sin \theta \sin \phi \Rightarrow dy=r\cos \theta \sin \phi d\theta +r\sin \theta \cos \phi d\phi \\
& z=r\cos \theta \Rightarrow dz=-r\sin \theta d\theta \\
& \Rightarrow d{{x}^{2}}+d{{y}^{2}}= ={{r}^{2}}\left( {{\cos }^{2}}\theta \left( {{\cos }^{2}}\phi +{{\sin }^{2}}\phi \right){{\left( d\theta \right)}^{2}}+{{\sin }^{2}}\theta \left( {{\cos }^{2}}\phi +{{\sin }^{2}}\phi \right){{\left( d\phi \right)}^{2}} \right) \\
& ={{r}^{2}}\left( {{\cos }^{2}}\theta {{\left( d\theta \right)}^{2}}+{{\sin }^{2}}\theta {{\left( d\phi \right)}^{2}} \right) \\
& \Rightarrow d{{x}^{2}}+d{{y}^{2}}+d{{z}^{2}}={{r}^{2}}\left( {{\cos }^{2}}\theta {{\left( d\theta \right)}^{2}}+{{\sin }^{2}}\theta {{\left( d\phi \right)}^{2}}+{{\sin }^{2}}\theta {{\left( d\theta \right)}^{2}} \right)={{r}^{2}}\left( {{\left( d\theta \right)}^{2}}+{{\sin }^{2}}\theta {{\left( d\phi \right)}^{2}} \right) \\
& \Rightarrow ds=\sqrt{{{\left( dx \right)}^{2}}+d{{y}^{2}}+d{{z}^{2}}} \\
& =\sqrt{{{\left( r\cos \theta \cos \phi d\theta -r\sin \theta \sin \phi d\phi \right)}^{2}}+{{\left( r\cos \theta \sin \phi d\theta +r\sin \theta \cos \phi d\phi \right)}^{2}}+{{\left( -r\sin \theta d\theta \right)}^{2}}} \\
& =r\sqrt{{{\left( d\theta \right)}^{2}}+{{\sin }^{2}}\theta {{\left( d\phi \right)}^{2}}} \\
& =r\sqrt{1+{{\sin }^{2}}\theta {{\left( \frac{d\phi }{d\theta } \right)}^{2}}}d\theta \\
& =r\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}}d\theta \\
& \Rightarrow S=\int{r\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}}d\theta } \\
& =\int{f\left( \theta ,\phi ' \right)d\theta } \\
& \Rightarrow \frac{d}{d\theta }\left( \frac{\partial f}{\partial \phi '} \right)-\frac{\partial f}{\partial \phi }=0 \\
& \Rightarrow \frac{d}{d\theta }\left( \frac{\partial }{\partial \phi '}\left[ r\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}} \right] \right)-\frac{\partial }{\partial \phi }\left[ r\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}} \right]=0 \\
& \Rightarrow \frac{d}{d\theta }\left( \frac{r\left( {{\sin }^{2}}\theta \right)\phi '}{\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}}} \right)-0=0 \\
& \Rightarrow \frac{r\left( {{\sin }^{2}}\theta \right)\phi '}{\sqrt{1+\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}}}}=a \\
& \Rightarrow \left( {{\sin }^{4}}\theta \right)\phi {{'}^{2}}={{a}^{2}}+{{a}^{2}}\left( {{\sin }^{2}}\theta \right)\phi {{'}^{2}} \\
& \Rightarrow \phi {{'}^{2}}{{\sin }^{2}}\theta \left( {{\sin }^{2}}\theta -{{a}^{2}} \right)={{a}^{2}} \\
& \Rightarrow \phi '=\frac{a}{\sin \theta \sqrt{{{\sin }^{2}}\theta -{{a}^{2}}}}=\frac{a}{\sin \theta \sin \theta \sqrt{1-{{a}^{2}}\cos e{{c}^{2}}\theta }} \\
& =\frac{a\cos e{{c}^{2}}\theta }{\sqrt{1-{{a}^{2}}\left( 1+{{\cot }^{2}}\theta \right)}}=\frac{a\cos e{{c}^{2}}\theta }{\sqrt{1-{{a}^{2}}-{{a}^{2}}{{\cot }^{2}}\theta }} \\
& =\frac{a\cos e{{c}^{2}}\theta d\theta }{a\sqrt{\left( 1-{{a}^{2}} \right)/{{a}^{2}}-{{\cot }^{2}}\theta }}=\frac{\cos e{{c}^{2}}\theta }{\sqrt{{{k}^{2}}-{{\cot }^{2}}\theta }} \\
& \Rightarrow \frac{d\phi }{d\theta }=\frac{\cos e{{c}^{2}}\theta }{\sqrt{{{k}^{2}}-{{\cot }^{2}}\theta }} \\
& \Rightarrow d\phi =\frac{\cos e{{c}^{2}}\theta \,d\theta }{\sqrt{{{k}^{2}}-{{\cot }^{2}}\theta }} \\
& \Rightarrow d\phi =\frac{-dx}{\sqrt{{{k}^{2}}-{{x}^{2}}}}(\text{Taking }\cot \theta =x) \\
& \Rightarrow \phi =-{{\sin }^{-1}}\left( x/k \right)+c \\
& \Rightarrow {{\sin }^{-1}}\left( x/k \right)=c-\phi \\
& \Rightarrow x=k\sin \left( c-\phi \right)=k\sin c\cos \phi -k\cos c\sin \phi \\
& \Rightarrow \cot \theta =k\sin c\cos \phi -k\cos c\sin \phi \\
& \Rightarrow \cos \theta =k\sin c\left( \sin \theta \cos \phi \right)-\cos c\left( \sin \theta \sin \phi \right) \\
& \text{Multiplying both sides by r:} \\
& r\cos \theta =k\sin c\left( r\sin \theta \cos \phi \right)-k\cos c\left( r\sin \theta \sin \phi \right) \\
& \Rightarrow z=\left( k\sin c \right)x-\left( k\cos c \right)y \\
& \Rightarrow z=Ax+By
\end{align*}
Where A and B are two constants.The equation $ z=Ax+By $ represents the plane that passes through the origin or center of the sphere and the curve between A and B is a part of the intersection of this plane and the sphere which a part of the great circle as the plane cuts the sphere into a circular region or the great circle of the sphere. \\
\subsection{\textbf{Proof of Snell's Law:}}
Let $ ds $ is the distance between two points P and Q situated very close to each other and $\mu$ is the refractive index of the medium .T is the time time taken to travel from A to B whcich is otained by integrating the time $ dt $ between P and Q given by :
\begin{align*}
& dt=\frac{ds}{v} \\
& =\frac{ds}{c/\mu } \\
& =\frac{\mu ds}{c} \\
& =\frac{\mu }{c}\sqrt{d{{x}^{2}}+d{{y}^{2}}} \\
& =\frac{\mu }{c}\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}dx \\
& =\frac{\mu }{c}\sqrt{1+y{{'}^{2}}}dx \\
& \Rightarrow T=\int{\frac{\mu }{c}\sqrt{1+y{{'}^{2}}}dx} \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial f}{\partial y'} \right)-\frac{\partial f}{\partial y}=0 \\
& \Rightarrow \frac{d}{dx}\left( \frac{\partial }{\partial y'}\left[ \frac{\mu }{c}\sqrt{1+y{{'}^{2}}} \right] \right)-\frac{\partial }{\partial y}\left[ \frac{\mu }{c}\sqrt{1+y{{'}^{2}}} \right]=0 \\
& \Rightarrow \frac{d}{dx}\left[ \frac{\mu y'}{\sqrt{1+y{{'}^{2}}}} \right]-0=0 \\
& \Rightarrow \frac{\mu y'}{\sqrt{1+y{{'}^{2}}}}=k \\
& \Rightarrow \frac{\mu \tan \theta }{\sqrt{1+{{\tan }^{2}}\theta }}=k \\
& \Rightarrow \mu \sin \theta =k \\
& \Rightarrow {{\mu }_{1}}\sin {{\theta }_{1}}={{\mu }_{2}}\sin {{\theta }_{2}} \\
& \Rightarrow \frac{{{\mu }_{2}}}{{{\mu }_{1}}}=\frac{\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}} \\
\end{align*}