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83. Given, xy = ex-y.

Taking log,

log (x + y) = log (ex-y).

=logx + log y = (x-y) log e.

= logx +log y = x -y {Q log e = 1}

Differentiating w r t ‘x’ we get,

$\Rightarrow \frac{1}{x}+\frac{1}{y}\frac{dy}{dx}=1-\frac{dy}{dx}$

$\Rightarrow \frac{1}{y}\frac{dy}{dx}+\frac{dy}{dx}=1-\frac{1}{2}$

$\Rightarrow \frac{dy}{dx}\left(\frac{1}{y}+1\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}\left(\frac{1+y}{y}\right)=\left(\frac{x-1}{x}\right)$

$\Rightarrow \frac{dy}{dx}=\frac{y(x-1)}{x(1+y)}$

The highest order derivation present in the D.E. is $\frac{d^{2}s}{d{t}^2}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

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3. Let the given statement be P(n) i.e.,

P(n): 1+ $\frac{1}{(1+2)}+\frac{1}{(1+2+3)}+?+\frac{1}{(1+2+3+?n)}=\frac{2n}{(n+1)}$

For n=1,

we get,P(1)=1= $\frac{2.1}{1+1}=\frac{2}{2}=1$

which is true.

Let us assume that P(k) is true for some positive integer k.

i.e., $1+\frac{1}{1+2}+\dots +\frac{1}{(1+2+3+?k)}=\frac{2k}{(k+1)}$ ------------------ (1)

Which is true.

Now, let us prove that P(k + 1) is true.

$1+\frac{1}{1+2}+\frac{1}{1+2+3}$ + … $+\frac{1}{(1+2+3+?k)}+\frac{1}{1+2+3+?+k+(k+1)}$

By using eqn (1)

= $\frac{2k}{(k+1)}$ + $\frac{1}{(1+2+3+\dots k+1)}$

?We know that, 1+2+3+ … +n= $\frac{n(n+1)}{2}$

So, we get

= $\frac{2k}{k+1}$ + $\frac{1}{\left(\frac{k+1(k+1+1)}{2}\right)}$

= $\frac{2k}{k+1}$ + $\frac{2}{(k+1)(k+2)}$

= $\frac{2}{k+1}\left\{k+\frac{1}{k+2}\right\}$

= $\frac{2}{\left(k+1\right)}\left\{\frac{k(k+2)+1}{k+2}\right\}$

= $\frac{2}{k+1}\left\{\frac{k^2+2k+1}{k+2}\right\}$

= $\frac{2\left(k^2+2k+1\right)}{k+1(k+2)}$

= $\frac{2{(k+1)}^2}{(k+1)(k+2)}$ = $\frac{2(k+1)}{(k+1+1)}$

? P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all a natural number n.

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82. Given, (cos x)y = (cos y)x

Taking log, y log (cos x) = x log (cos y)

Differentiating w r t ‘x’ we get,

= $y\frac{d}{dx}$ log (cos x) + log (cos x) $\frac{dy}{dx}=x\frac{d}{dx}$ log (cos y) + dog (cos y) $\frac{dx}{dx}$

= y´ $\frac{1}{cosx}\frac{d}{dx}$ cos x + log (cos x) $\frac{dy}{dx}$ = x´ $\frac{1}{cosy}\frac{d}{dx}cosy+log(cosy).$

$\Rightarrow -y\frac{sinx}{cosx}+log(cosx)\frac{dy}{dx}=-x\frac{siny}{cosy}\frac{dy}{dx}+log(cosy)$

= log (cos x) $\frac{dy}{dx}$ + x tan $\frac{dy}{dx}=yctanx+log(cosy)$

$\Rightarrow \frac{dy}{dx}[log(cosx)+xtany]$ = y tan x + log (cos y )

$\Rightarrow \frac{dy}{dx}=\frac{ytanx+log(cosy}{log(cosx)+xtany}.$

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The mode of application is online. Candidates fulfilling the eligibility criteria can fill out the form. Aspirants have to pay a non-refundable application fee to apply for admission. Below are the steps to apply at Bharath College of Science and Management:

Step 1: Visit the official website of Bharath College of Science and Management

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The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4}y}{d{x}^4}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative, its degree is not defined.

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Candidates should have completed a bachelor's degree in any discipline with at least 45% aggregate from a recognised university or college to apply for MA at College of Communication, Culture & Media, MGM University. Students awaiting their examination results can also apply for the MA programme.

The admission process for the Sambalpur University BCom course consists of five broad steps. Students must follow the steps in exact order to get a seat  in the programme. To know the detailed admission process, check out the steps below:

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2. Let the given statement be P(n) i.e.,

P(n)=1³+2³+3³+ … +n³= ${\left(\frac{n(n+1)}{2}\right)}^2$

For, n=1, P(n)=1³=1= ${\left(\frac{1(1+1)}{2}\right)}^2={\left(\frac{1.2}{2}\right)}^2=1^2=1$

which is true.

Consider P(k) be true for some positive integer k

1³+2³+3³+ … +k³= ${\left(\frac{k(k+1)}{2}\right)}^2$ ---------- (1)

Now, let us prove that P(k+1) is true.

Here,  1³+2³+3³+ … +k³+(k+1)³

By using eq (1)

= ${\left(\frac{k(k+1)}{2}\right)}^2+{(k+1)}^3$

= $\frac{k^2(k+1){}^2+4·{(k+1)}^3}{4}$

= $\frac{{(k+1)}^2\left[k^2+4(k+1)\right]}{4}$

= $\frac{{(k+1)}^2\left\{k^2+4k+4\right\}}{4}=\frac{{(k+1)}^2(k+2){}^2}{4}$

= ${\left\{\frac{(k+1)(k+1+1)}{2}\right\}}^2$

? P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, P(n) is true for all natural numbers n.

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