# UD-TRG-Math-Nagpur-Summer-17

$latex \frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}=\begin{vmatrix}\frac{1+y^2}{\displaystyle\left(1-xy\right)^2}&\frac{1+x^2}{\displaystyle\left(1-xy\right)^2}\\frac1{1+x^2}&\frac1{1+y^2}\end{vmatrix}$

$latex =\left[\left(\frac{1+y^2}{\left(1-xy\right)^2}\times\frac1{\left(1+y^2\right)}\right)\right]-\left[\left(\frac{1+x^2}{\left(1-xy\right)^2}\times\frac1{\left(1+x^2\right)}\right)\right]$

$latex \begin{array}{l}=\frac1{\left(1-xy\right)^2}-\frac1{\left(1-xy\right)^2}\\=0\end{array}$

$latex \therefore\frac{\partial\left(u,v\right)}{\partial\left(x,y\right)}=0$

As the jacobian of u & v with respect to x,y vanishes,so u & v are functionally related

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$latex =\cos\theta\pm\sqrt{-\sin^2\theta}$

$latex x=\cos\theta+i\sin\theta$

similarly,$latex y=\cos\phi\pm i\sin\phi$

$latex \therefore$ one of the value 7 x & y is

$latex x=\cos\theta+i\sin\theta\;\&\;y=\cos\phi+i\sin\phi$

$latex \begin{array}{l}x^m=\cos m\theta+i\sin m\theta\;\;\&\\y^m=\cos n\phi+i\sin n\phi\end{array}$

$latex \left(\because\;by\;De-Moivre’s\;tn^m\right)$

$latex x^my^n=\left(\cos m\theta+i\sin m\theta\right)\left(\cos n\phi+i\sin n\phi\right)$

$latex \therefore x^my^n=\cos\left(m\theta+n\phi\right)+i\sin\left(m\theta+n\phi\right)……\left(i\right)$

(By continued product formula)

$latex \frac1{x^my^n}=\left[\cos\left(m\theta\right)+n\phi+i\sin\left(m\theta+n\phi\right)\right]$

$latex \frac1{x^my^n}=\left[\cos\left(m\theta\right)+n\phi+i\sin\left(m\theta+n\phi\right)\right]$

(By De-Moivre’s theorem)

$latex x^my^n+\frac1{x^my^n}=2\cos\left(m\theta+n\phi\right)$

(By (i) & (ii) )

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$latex \frac{\partial z}{\partial v}=\frac u{2\sqrt{uv}}$

$latex \frac{\partial z}{\partial w}=0$

Then by using chain rule,we have

$latex \frac{\partial\phi}{\partial u}=\frac{\partial\phi}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial\phi}{\partial y}\frac{\partial u}{\partial u}+\frac{\partial\phi}{\partial z}\frac{\partial z}{\partial u}$

$latex \Rightarrow\frac{\partial\phi}{\partial u}=\frac w{2\sqrt{un}}\frac{\partial\phi}{\partial y}+\frac v{2\sqrt{uv}}\frac{\partial\phi}{\partial z}…..\left(i\right)$

$latex \frac{\partial\phi}{\partial v}=\frac{\partial\phi}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial\phi}{\partial y}\frac{\partial u}{\partial v}+\frac{\partial\phi}{\partial z}\frac{\partial z}{\partial v}$

$latex \Rightarrow\frac{\partial\phi}{\partial v}=\frac w{2\sqrt{vw}}\frac{\partial\phi}{\partial x}+\frac u{2\sqrt{uv}}\frac{\partial\phi}{\partial z}……\left(ii\right)$

$latex \frac{\partial\phi}{\partial w}=\frac{\partial\phi}{\partial x}\frac{\partial x}{\partial w}+\frac{\partial\phi}{\partial y}\frac{\partial y}{\partial w}+\frac{\partial\phi}{\partial z}\frac{\partial z}{\partial w}$

$latex \frac{\partial\phi}{\partial w}=\frac v{2\sqrt{vw}}\frac{\partial\phi}{\partial x}+\frac u{2\sqrt{uw}}\frac{\partial\phi}{\partial y}….\left(iii\right)$

Multiplying eqn (i) eqn (ii) & eqn (iii) by u,v & w repectively
we get,

$latex u\frac{\partial\phi}{\partial u}=\frac{wu}{2\sqrt{uw}}\frac{\partial\phi}{\partial y}+\frac{uv}{2\sqrt{uv}}\frac{\partial\phi}{\partial z}$

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As $latex v=\tan^{-1}x+\tan^{-1}y$

$latex \begin{array}{l}\tan v=\tan\left(\tan^{-1}x+\tan^{-1}y\right)\\\;\;\;\;\;\;\;\;=\frac{\tan\left(\tan^{-1}x+\tan\left(\tan^{-1}y\right)\right)}{1-\tan\left(\tan^{-1}x\right)\tan\left(\tan^{-1}y\right)}\end{array}$

$latex \because\tan\left(A+B\right)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$latex \begin{array}{l}=\frac{x+y}{1-xy}\\=u\end{array}$

The required relationship is

u=tanv

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use De-Moivre’s theorem to solve

$latex x+x^5+x^3+1=0$

soln-Given eqn is

$latex x^8+x^5+x^3+1=0$

$latex \begin{array}{l}\therefore x^5\left(x^3+1\right)+\left(x^3+1\right)=0\\\left(x^5+1\right)\left(x^3+1\right)=0\end{array}$

$latex \therefore x^5+1=0\;\;\&\;x^3+1=0$

$latex \therefore x=\left(-1\right)^{1/5}\;\;and\;\;x=\left(-1\right)^{1/3}$

Now,
$latex x=\left(-1\right)^{1/5}=\left[\cos\Pi+i\sin\Pi\right]^{1/5}$

$latex =\left[\cos\left(2n\Pi+\Pi\right)+i\sin\left(2n\Pi+\Pi\right)\right]^{1/5}$

$latex =\cos\left[\frac{2n\Pi+\Pi}5\right]+\sin\left[\frac{2n\Pi+\Pi}5\right]$

(By De-Moivre’s theorem)

Putting n=0,1,2,3,4,
we get the roots

Therefore Roots are

$latex \cos\frac\pi5+i\sin\frac\pi5,\cos\frac{3\pi}5+i\sin\frac{3\pi}5,\cos\pi+i\sin\pi$

$latex \cos\frac{7\pi}5+i\sin\frac{7\pi}5,\cos\frac{5\pi}5+i\sin\frac{9\pi}5$

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But $latex \cos\frac{7\pi}5+i\sin\frac{7\pi}5$

$latex =\cos\left(2\pi-\frac{3\pi}5\right)+i\sin\left(2\pi-\frac{3\pi}5\right)$

$latex =\cos\frac{3\pi}5-i\sin\frac{3\pi}5$

& $latex \cos\frac{9\pi}5+i\sin\frac{9\pi}5=\cos\left(2\pi-\frac\pi5\right)+i\sin\left(2\pi-\frac\pi5\right)$

$latex =\cos\frac\pi5-i\sin\frac\pi5$

Therefore roots of equation $latex x^5+1=0$ are

$latex \cos\frac\pi5+i\sin\frac\pi5,\;\cos\frac{3\pi}5+i\sin\frac{3\pi}5,\;\cos\pi+i\sin\pi$

Now, $latex x=\left(-1\right)^{1/3}$

$latex \begin{array}{l}=\left[\cos\pi+i\sin\pi\right]^{1/3}\\=\left[\cos\left(2n\pi+\pi\right)+i\sin\left(2n\pi+\pi\right)^{1/3}\right]\end{array}$

$latex x=\cos\left(\frac{2n\pi+\pi}3\right)+i\sin\left(\frac{2n\pi+\pi}3\right)$

(Because by De-Moivre’s theorem)

Putting n=0,1,2

we get the roots

therefore roots are

$latex \cos\frac\pi3+i\sin\frac\pi3,\;\cos\pi+i\sin\pi,\;\cos\frac{5\pi}3+i\sin\frac{5\pi}3$

But, $latex \cos\frac{5\pi}3+i\sin\frac{5\pi}3=\cos\left(\frac{2\pi-\pi}3\right)+i\sin\left(\frac{2\pi-\pi}3\right)$

$latex =\frac{\cos\pi}3-\frac{i\sin\pi}3$

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Q41a) If $latex y=\left(\sin^{-1}x\right)^2\;$ , prove that

$latex \left(1-x^2\right)y_n+2-\left(2n+1\right)xy_{n+1}-n^2yn=0$

Given $latex y=\left(\sin^{-1}x\right)^2…..\left(i\right)$

Differentiating equation (i) with respect to x
we get,

$latex y_1=2\left(\sin^{-1}x\right)\times\frac1{\sqrt{1-x^2}}$

$latex \left(\sqrt{1-x^2}\right)y_1=2\left(\sin^{-1}x\right)$

squaring on both sides
we get,

$latex \left(1-x^2\right)y_1^2=4\left(\sin^{-1}x\right)2$

$latex \left(1-x^2\right)y_1^2=4y…….\left(ii\right)$

Again differentiating equation (ii) with respect to x
we get

$latex \begin{array}{l}\left(1-x^2\right)2y_1.y_2+\left(-2x\right)y_1^2=4y_1\\\left(1-x^2\right)2y_1.y_2-2xy_1^2=4y_1\end{array}$

Dividing both side by 2y1
we get

$latex \left(1-x^2\right)y_2-xy_1-2=0$

Differentiating equation (iii) n times
with respect to x by using

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Q10 The differential equation for a ckt in which selfinductance & capacitance neutualite each other is $latex \frac{Ld^2i}{dt^2}+\frac ic=0$
Find the current i.

soln- Given D.E is

$latex L\frac{d^2i}{dt^2}+\frac iC=0$

$latex \left(Ln^2+\frac1C\right)i=0$

It’s A.F. is

$latex Lm^2+\frac1C=0$

$latex \therefore m=\pm\frac1{\sqrt{LC}}$

$latex i=C_1\cos\frac1{\sqrt{LC}}+C_2\sin\frac1{\sqrt{LC}}…..\left(i\right)$
(RHS=0)

At t=0, i=0

equation (i) $latex L=C_2\sin\frac1{\sqrt{LC}}……….\left(ii\right)$

Substituting $latex \sin\frac1{\sqrt{LC}}=1$ in equation (ii)

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$latex \therefore x=\frac9{\sqrt3}$

Similarly,

$latex y=\frac b{\sqrt3}\;,\;z=\frac c{\sqrt3}$

Therefore Maximum volume of a rectangular parallelopiped

$latex =8\frac a{\sqrt3}.\frac b{\sqrt3}.\frac c{\sqrt3}$

The required volume is

$latex V_{max}=8\frac{abc}{3\sqrt3}$

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therefore Augmented matrix [A:B] is given by

$latex \left[A:B\right]=\begin{bmatrix}1&1&1&.&6\1&2&5&.&10\2&3&\lambda&.&\mu\end{bmatrix}$

$latex \begin{array}{l}R_2\rightarrow R_2-R_1\\R_3\rightarrow R_3-2R_1\end{array}$

$latex \left[A:B\right]=\begin{bmatrix}1&1&1&.&6\0&1&4&.&4\0&1&\lambda-2&.&\mu-12\end{bmatrix}$

$latex \begin{array}{l}R_1\rightarrow R_1-R_2\\R_3\rightarrow R_3-R_2\end{array}$

$latex \left[A:B\right]=\begin{bmatrix}1&0&-3&.&2\0&1&4&.&4\0&0&\lambda-6&.&\mu-16\end{bmatrix}$

If $latex \lambda\neq6\;,\;\mu\;$ may have any value

P(A)=P(A:B)=3=No. of lenknocons
(here P denotes rank)

Therefore system has unique solution.

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Heibnitz rule i.e.

$latex D^n\left(u.v.\right)=unV+nu_{u-1}V_1+\frac{n\left(n-1\right)}{2!}$

$latex un-2V2+…..+Vnu=0$

We get,

$latex D^n\left[\left(1-x^2\right)y_2\right]-D^n\left[xy_1\right]=0$

$latex \left[\left(1-x^2\right)\left(y_2\right)n+\left(y_2\right)n-1{\left(1-x^2\right)}_1+\frac{n\left(n-1\right)}{2!}\left(y_2\right)n-2{\left(1-x^2\right)}_2\right]-\left[x\left(y_1\right)n+n\left(y_1\right)n-1{\left(x\right)}_1\right]=0$

$latex \left[\left(1-x^2\right)y_{n+2}+ny_{n+1}\left(-2x\right)+\frac{n\left(n-1\right)}{2!}y_n\left(-2\right)\right]-\left[xy_{n+1}+ny_n\right]=0$

$latex \left(1-x^2\right)y_{n+2}-\left(2n+1\right)xy_{n+1}-n^2y_n+ny_n-ny_n=0$

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