Maths

121. Kindly go through the solution

6. Let (r + 1)th be the general term of ( $x^2- yx){}^{12}$

So, T_r-1 = ¹²C_r (x²)^12–r (–yx)^r

= (–1)^r12C_rx^24–2ry^rx^r

= (–1)^r12C_r $x^{24-2r+r}{y}^r$

= (-1)^r12C­_r $x^{24-r}{y}^r$

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

Hence, (D) is homogenous

$\therefore$ Option (D) is correct.

5. Let (r + 1)^th term be the general term of (x²–y)⁶.

So, T_r-1 = ⁶C_r (x²)^6-r (-y)^r

= (–1)^r .⁶C_r . $x^{12-r}$ . $y^r$

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,

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

119. Let $y=si{n}^{3}x+co{s}^{6}x$

So,  $\frac{dy}{dx}=\frac{d}{dx}\left(si{n}^{3}x+co{s}^{6}x\right)$

$=3si{n}^{2}x\frac{d}{dx}sinx+6co{s}^{5}x\frac{d}{dx}cosx$

$=3si{n}^{2}xcosx-6co{s}^{5}xsinx$

$=3sinxcosx\left(sinx-2co{s}^4\right)$

4. Let a⁵b⁷ occurs in (r + 1)^th term of [removed]a – 2b)¹².

Now, T_r+1 = ¹²C_ra^12-r (–2b)^r

= (–1)^r12C_ra^12–r . 2^r. b^r

Comparing indices of a and b in T_r-1 with a⁵ and b⁷ we get,  r = 7

So, co-efficient of a⁵b⁷ is (–1)⁷¹²C₇ 2⁷

The given D.E.is

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

i.e, the given D.E. is homogenous.

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

Then, $v+x\frac{dv}{dx}=v-cosecv$

$\begin{array}{l}=x\frac{dv}{dx}=-cosecv\\ =\frac{dv}{cosecv}=-\frac{dx}{x}\\ =sinvdv=-\frac{dx}{x}\end{array}$

Integrating both sides we get,

$\begin{array}{l}{\displaystyle \int sinvdv=-{\displaystyle \int \frac{dx}{x}}}\\ =-cosv=-log\left|x\right|+c\\ =cosv=log\left|x\right|-c\end{array}$

Putting back $v=\frac{y}{x}$ we get,

$=cos\frac{y}{x}=log\left|x\right|-c$

Given, $y=0,when,x=1$

$\begin{array}{l}=cos0=log1-c\\ =c=-1\end{array}$

$\therefore$ The required particular solution is

$\begin{array}{l}cos\left(\frac{y}{x}\right)=log\left|x\right|+1=log\left|x\right|+log\left|c\right|\\ =cos\left(\frac{y}{x}\right)=log\left|cx\right|\end{array}$