Summer 2017 – Q.3

Q.3 a) If $latex u=f\left(\frac{y-x}{xy},\;\frac{z-x}{xz}\right),$ show that $latex x^2\frac{\partial u}{\partial z}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial z}=0$ [6M]

$latex u=f\left(\frac{y-x}{xy},\;\frac{z-x}{xz}\right)$

T.P.T. $latex x^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial z}=0$

Let u = u(r, s)

where,

$latex \begin{array}{l}r=\frac{y-x}{xy},\;s=\frac{z-x}{zx}\\\\r=\frac1x-\frac1y,\;s=\frac1x-\frac1z\\\\\begin{array}{cc}\frac{\partial r}{\partial x}=\frac{-1}{x^2},&\frac{\partial s}{\partial x}=\frac{-1}{x^2}\end{array}\\\\\begin{array}{cc}\frac{\partial r}{\partial y}=\frac1{y^2},\;&\frac{\partial s}{\partial z}=\frac1{z^2}\end{array}\\\\\begin{array}{cc}\frac{\partial r}{\partial z}=0,\;&\frac{\partial s}{\partial y}=0\end{array}\end{array}$

We know that,

$latex \begin{array}{l}\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial x}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial x}\\\\\;\;\;\;=\frac{\partial u}{\partial r}\left(\frac{-1}{x^2}\right)+\frac{\partial u}{\partial s}\left(\frac{-1}{x^2}\right)\\\\\;\;\;\;=\frac{-1}{x^2}\frac{\partial u}{\partial r}-\frac1{x^2}\frac{\partial u}{\partial s}\end{array}$

$latex \begin{array}{l}x^2\frac{\partial u}{\partial x}=\frac{-\partial u}{\partial r}-\frac{\partial u}{\partial s}\\\\\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial y}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial y}\\\\\;\;\;\;=\frac{\partial u}{\partial r}\times\frac1{y^2}+\frac{\partial u}{\partial s}\times0\\\\\;\;\;\;=\frac1{y^2}\frac{\partial u}{\partial r}—–\left(i\right)\end{array}$

$latex \begin{array}{l}y^2\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\\\\\frac{\partial u}{\partial z}=\frac{\partial u}{\partial r}\cdot\frac{\partial r}{\partial z}+\frac{\partial u}{\partial s}\cdot\frac{\partial s}{\partial z}\\\\\;\;\;\;=\frac{\partial u}{\partial r}\times0+\frac{\partial u}{\partial s}\times\frac1{z^2}\\\\\;\;\;\;=\frac1{z^2}\frac{\partial u}{\partial s}—–\left(ii\right)\end{array}$

$latex z^2\frac{\partial u}{\partial z}=\frac{\partial u}{\partial s}—–\left(iii\right)$

By adding (i), (ii) and (iii), we get

$latex z^2\frac{\partial u}{\partial x}+y^2\frac{\partial u}{\partial y}+z^2\frac{\partial u}{\partial x}=0$

Hence Proved.

b) Using encoding matrix $latex \begin{bmatrix}1&1\\0&1\end{bmatrix}$, encode and decode the message ‘MUMBAI’. [6M]

Encoding matrix $latex \begin{bmatrix}1&1\\0&1\end{bmatrix}$=A

Message – $latex \begin{array}{c}M\;\;U\;\;M\;\:B\;\;A\;\;I\\13\;\;21\;\;13\;\;2\;\;1\;\;9\end{array}$

Encoded Matrix

$latex \begin{array}{l}\begin{bmatrix}13&21\\13&2\\1&9\end{bmatrix}\;\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}13&34\\13&15\\1&10\end{bmatrix}\\\\A^{-1}=\begin{bmatrix}1&-1\\0&1\end{bmatrix}\end{array}$

Decoded Matrix

$latex \begin{bmatrix}13&34\\13&15\\1&10\end{bmatrix}\;\begin{bmatrix}1&1\\0&1\end{bmatrix}=\begin{bmatrix}13&21\\13&2\\1&9\end{bmatrix}$

c) prove that $latex \log\;\left[\tan\left(\frac\pi4+\frac{ix}2\right)\right]=i\tan^{-1}\left(\sin h\;x\right)$ [8M]