Separable

Solve: $\frac{d^2y}{dx^2} = 2 + \frac{1}{x}$

As the equation is of first order, integrate the function twice, i.e. $\frac{dy}{dx} = 2x + ln\,x + C$ and $y = x^2 + x\,ln\,x + Cx + K$

$y = x^2 + x\,ln\,x + Cx + K$

$\mathbf{\frac{dx}{dy} = 1 - y}$

$dx = (1 - y)\,dy$

$\therefore\;\;\;\;\;\;x = y - \frac{y^2}{2} + C$

$\mathbf{\frac{dy}{dx}\;=\;a^2\;+\;y^2}$

$dx\;=\;\frac{dy}{a^2\;+\;y^2}$

$x\;=\;\frac{1}{a}\;arctan\,(\frac{y}{a})\;+\;c$

$\mathbf{\frac{dy}{dx}\;=\;\frac{x^2\;-\;1}{y}}$

$y\,dy\;=\;(x^2\;-\;1)\,dx$

$\therefore\;\;\;\;\;\frac{y^2}{2}\;=\;\frac{x^3}{3}\;-\;x\;+\;C$

$\mathbf{\frac{dy}{dx}\;=\;\frac{y}{x^2\;-\;1}}$

$\frac{dy}{y}\;=\;\frac{dx}{x^2\;-\;1}$

$\therefore\;\;\;\;\;ln\,y\;=\;\frac{1}{2}\;ln\;\;\left(\frac{x\;+\;1}{x\;-\;1} \right)}$

$\therefore\;\;\;\;\;y\;=\;C\;\sqrt{\;\left(\frac{x\;+\;1}{x\;-\;1} \right)}$

Separate the Variables

Integrating

$\frac{dy}{dx}=\frac{1}{cos^2 \tfrac{1}{2}x}$

This is the same as $y=\int \frac{1}{\cos^2 \tfrac{1}{2}x}dx$

which we integrate in the normal way to yield $y=2\tan \tfrac{1}{2} x + C$

$y=2\tan \tfrac{1}{2} x + C$

Solve the following equation given that y=0 at x=0 $\frac{dy}{dx}=\sec y$

$\sec y \equiv \frac{1}{\cos y}$ so $\frac{dy}{dx}=\sec y$ becomes $\int \cos y\;dy=\int dx$ $\therefore\;\;\;\;\;\;\;\sin y=x+A$

But when x=0, y=0 and so A=0

$y=\sin^{-1}x$

Solve the following equation: $\frac{r \tan \theta }{a^2-r^2} = d \theta$ when r=0 at $theta=\frac{\pi }{4}$

Rearrange and integrate $\int {\frac{r}{a^2 - r^2}\dr = \int \frac{1}{\tan \theta} d \theta$

$-\tfrac{1}{2}\ln(a^2-r^2)=\ln \sin \theta +\ln C$ or $\ln \left (\frac{1}{\sqrt{a^2-r^2}} \right )=\ln \left | \sin \theta \cdot C \right |$

$\therefore\;\;\;\;\;\;\frac{1}{\sqrt{a^2 - r^2}} = C \sin \theta$ but r=0 when $theta=\frac{\pi }{4}$ $\therefore\;\;\;\;\;\frac{1}{a}=\frac{C}{\sqrt{2}}$ thus $C=\frac{\sqrt{2}}{a}$

Therefore $\frac{1}{(a^2 - r^2)} = \frac{2}{a^2} \sin^2\theta$

$(a^2-r^2)=\frac{a^2}{2} \textup{cosec}^2 \theta$

If a and b are the radii of concentric spherical conductors at potentials of $V_1$ respectively, then V is the potential at a distance r from the centre. Find the value of V if: $\frac{d}{dr}\left(r^2 \frac{dV}{dr} \right)=0}$ and $V=V_1$ at r=a and $V=0$ at r=b

$\frac{d}{dr}\left(r^2\frac{dV}{dr} \right)=0$

$\therefore\;\;\;\;\;\;r^2\frac{dV}{dr} \right)=A$

$\frac{dV}{A}=\frac{dr}{r^2}$

$\therefore\;\;\;\;\;\;\frac{V}{A}=-\frac{1}{r}\;+\;B$

Substituting in the given values for V and r $\frac{V_1}{A}=-\frac{1}{a}+B$ and $\frac{0}{A}=-\frac{1}{b}+B$ $\therefore\;\;\;\;\;\;B=\frac{1}{b}$ $\therefore\;\;\;\;\;\;\frac{V_1}{A}=-\frac{1}{a}+\frac{1}{b}=-\frac{b-a}{ab}$ Thus $A=-\frac{abV_1}{b-a}$

$V=\left(\frac{1}{r}-\frac{1}{b} \right)\left(\frac{abV_1}{b-a} \right)$