Class 12th

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

$\frac{d^{4}y}{d{x}^4}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

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.

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 )

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

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.

81. Given, yx = xy

Taking log,

x log y .log x

Differentiating w r t 'x' we get,

$\Rightarrow x\frac{d}{dx}logy+logy\frac{dx}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}$

$\Rightarrow \frac{x}{y}·\frac{dy}{dx}+logy=\frac{y}{x}+logx\frac{dy}{dx}$

$\Rightarrow logx\frac{dy}{dx}-\frac{x}{y}\frac{dy}{dx}=logy-\frac{y}{x}$

$\Rightarrow \frac{dy}{dx}\left[\frac{ylogx-x}{y}\right]=\frac{xlogy-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{y(xlogy-y)}{x(ylogx-x)}.$

80. Given, xy + yx = 1

Let 4 = xy and v =., we have,

u + v = 1.

$\Rightarrow \frac{dy}{dx}+\frac{dv}{dy}=0$ ___ (1)

So, u = xy

= log u = y log x(taking log)

Now, differentiating w r t 'x',

$\frac{1}{4}\frac{dy}{dx}=y\frac{d}{dx}logx+logx\frac{dy}{dx}.$

$\frac{dy}{dx}=4\left[\frac{y}{x}+logx\frac{dy}{dx}\right]$

$\Rightarrow x^y·\frac{y}{x}+x^{y}logx·\frac{dy}{dx}$

= xy- 1y + xy log x $\frac{dy}{dx}.$

And v = yx.

log v = x log y.

Differentiating w r t 'x',

$\Rightarrow \frac{1}{v}\frac{dv}{dx}=x\frac{d}{dx}logy+logy\frac{dx}{dx}$

$=\frac{x}{y}\frac{dy}{dx}+logy$

$\Rightarrow \frac{dv}{dx}=v\left[\frac{x}{y}\frac{dy}{dx}+logy\right]$

$=y^x·\frac{x}{y}·\frac{dy}{dx}+y^{x}log·y$

= yx- 1. $x\frac{dy}{dx}$ + yx log y.

So, eqn (1) becomes

xy- 1y + xy log x $\frac{dy}{dx}$ + yx - 1 $\frac{dy}{dx}$ + yx log y = 0

$\Rightarrow \frac{dy}{dx}\left(x^{y}logx+y^{x+1}·x\right)$ = - (xy- 1y + yx log y)

$\Rightarrow \frac{dy}{dx}=\frac{-\left(x^{y-1}·y+y^{x}logy\right)}{\left(x^{y}logx+y^{x-1}·x\right).}$