Class 12th

126. Let $y={(sinx-cosx)}^{sinx-cosx}$

Taking log,

$logy=(sinx-cosx)log(sinx-cosx)$

Differentiating w r t 'x' we get,

$\frac{1}{y}\frac{dy}{dx}=(sinx-cosx)\frac{dlog}{dx}(sinx-cosx)+log(sinx-cosx)\frac{d}{dx}(sinx-cosx).$

$=(sinx-cosx)*\frac{1}{(sinx-cosx)}*(cosx+sinx)+log(sinx-cosx)\left(cosx+sinx\right)$

$=(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}=y(cosx+sinx)[1+log(sinx-cosx)]$

$\frac{dy}{dx}={(sinx-cosx)}^{sinx-cosx}*(cosx+sinx)[1+log(sinx-cosx)]$

125. Let $y=cos(acosx+bsinx)$

So,  $\frac{dy}{dx}=-sin(acosx+bsinx)\frac{d}{dx}(acosx+bsinx)$

$=-sin(acosx+bsinx)[-asinx+bcosx].$

$=sin(acosx+bsinx)(asinx-bcosx)$

124. Let $y={(logx)}^{logx}$

Taking log,

$logy=logxlog(logx)$

Differentiating w r t. x, we get,

$\frac{1}{y}\frac{dy}{dx}=logx\frac{d}{dx}log(logx)+log(logx)\frac{d}{dx}(logx)$

$=logx*\frac{1}{logx}\frac{d}{dx}logx+\frac{log(logx)}{x}$

$=\frac{1}{x}+\frac{log(logx)}{x}$

$\frac{dy}{dx}=\frac{y}{x}[1+log(logx)].$

$={(logx)}^{logx}\left[\frac{1+log(logx)}{x}\right]$

123. Kindly go through the solution

122. Kindly go through the solution

121. Kindly go through the solution

In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.

120. Let $y={\left(5x\right)}^{3cos2x}$

Taking log,

$logy=3cos2x\left[log5x\right]$

Differentiating w r t. x,

$\frac{1}{y}\frac{dy}{dx}=3cos2x\frac{d}{dx}log5x+3log5x\frac{d}{dx}cos2x$

$=3\left[cos2x\cdot \frac{1}{5x}\frac{d}{dx}\left(5x\right)-log5x*sin2x\frac{d}{dx}2x\right]$

$\frac{dy}{dx}=3y\left[cos2x*\frac{5}{5x}-log5x\cdot sin2x\cdot 2\right]$

$=3{\left(5x\right)}^{3cos2x}\left[\frac{cos2x}{x}-2sin2xlog5x\right]$

For a homogenous D.E. of the formula $f\left(\frac{y}{x}\right)$

We put,  $\frac{x}{y}=0=x=vy$

$\therefore$ Option (c) is correct.

The given D.E. is

$\begin{array}{l}2xy+y^2-2x^2\frac{dy}{dx}=0\\ =2x^2\frac{dy}{dx}=2xy+y^2\\ =\frac{dy}{dx}=\frac{2xy+y^2}{2x^2}=\frac{y}{x}+\frac{1}{2}{\left(\frac{y}{x}\right)}^2=f\left(\frac{y}{x}\right)\end{array}$

i.e, the given is homogenous.

Let, $y=vx$ $=\frac{y}{x}=v$ so that $\frac{dy}{dx}=v+x\frac{dv}{dx}$ is the D.E.

Then, $v+x\frac{dv}{dx}=v+\frac{1}{2}{v}^2$

$\begin{array}{l}=x\frac{dv}{dx}=\frac{1}{2}{v}^2\\ =\frac{dv}{v^2}=\frac{dx}{2x}\end{array}$

Now, $={\displaystyle \int \frac{dv}{v^2}={\displaystyle \int \frac{dx}{2x}}}$

$\begin{array}{l}=\frac{v-2+1}{-2+1}=\frac{1}{2}log\left|x\right|+c\\ =\frac{-1}{v}=\frac{1}{2}log\left|x\right|+c\end{array}$

Putting back $\frac{y}{x}=v$ we get,

$=-\frac{x}{y}=\frac{1}{2}log\left|x\right|+c$

$Given\frac{y}{x}=vwhenx=1$ and y= 2

$\begin{array}{l}=-\frac{1}{2}=\frac{1}{2}log\left|1\right|+c\\ =c=-\frac{1}{2}\end{array}$

$\therefore$ The particular solution is,

$\begin{array}{l}=-\frac{x}{y}=\frac{1}{2}log\left|x\right|-\frac{1}{2}\\ =-\frac{2x}{y}=log\left|x\right|-1\\ =y=\frac{-2x}{log\left|x\right|-1}=\frac{2x}{1-log\left|x\right|}\end{array}$