Differentiation (Part-38) | S N De

 

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

Sol. $~~~\frac{dy}{dx}=\frac{y-x}{x+y}\rightarrow(1)$

Let $~~~~y=vx \\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~v+x\frac{dv}{dx}=\frac{vx-x}{x+vx} \\ \Rightarrow x\frac{dv}{dx}=\frac{x(v-1)}{x(1+v)}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{v-1-v(v+1)}{v+1} \\ \Rightarrow x\frac{dv}{dx}=\frac{v-1-v^2-v}{v+1} \\ \Rightarrow x\frac{dv}{dx}=\frac{-1-v^2}{v+1} \\ \therefore-\int{\frac{v+1}{v^2+1}~dv}=\int{\frac 1x~dx} \\ \Rightarrow -\frac 12 \int{\left[\frac{2(v+1)}{v^2+1}\right]~dv}=\int{\frac{dx}{x}} \\ \Rightarrow -\frac 12 \left[\int{\frac{2v}{v^2+1}~dv}+2\int{\frac{dv}{v^2+1}}\right]\\~~~~=\log x+\log c_1\\~~~~~~\log c_1 \rightarrow \text{constant of integration.} \\ \Rightarrow -\left[\int{\frac{d(v^2+1)}{v^2+1}}+2\int{\frac{dv}{v^2+1}}\right]\\~~~=2\log x+2\log c_1\\~~~~~~~[\because d(v^2+1)=2v~dv] \\ \Rightarrow -\left[\log|v^2+1|+2\tan^{-1}(v)\right]\\~~=\log x^2+2\log c_1 \\ \Rightarrow -\log \left(\frac{y^2}{x^2}+1\right)-2\tan^{-1}\left(\frac yx\right)\\~~=\log x^2+2\log c_1 \\ \Rightarrow \log\left(\frac{y^2}{x^2}+1\right)+2\tan^{-1}(y/x)\\~~=-\log x^2-2\log c_1 \\ \Rightarrow \log \left(\frac{x^2+y^2}{x^2}\right)+2\tan^{-1}(y/x)\\~~=-\log x^2-2\log c_1 \\ \Rightarrow \log(x^2+y^2)-\log x^2+2\tan^{-1}(y/x)\\~~=-\log x^2+c~~[*] \\ \Rightarrow \log(x^2+y^2)+2\tan^{-1}(y/x)=c$

Note[*] : $~~-2\log c_1=c ;\\~~~~\text{Here}~\log c_1\rightarrow \text{constant of integration.}$

$~2.~~\frac{dy}{dx}=\frac{3x+2y}{2x-3y}$

Sol. $~~~\frac{dy}{dx}=\frac{3x+2y}{2x-3y}\rightarrow(1)$

Let $~~~~~y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~v+x\frac{dv}{dx}=\frac{3x+2(vx)}{2x-3(vx)} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{x(3+2v)}{x(2-3v)} \\ \Rightarrow x\frac{dv}{dx}=\frac{3+2v}{2-3v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{3+2v-v(2-3v)}{2-3v} \\ \Rightarrow x\frac{dv}{dx}=\frac{3+2v-2v+3v^2}{2-3v} \\ \Rightarrow x\frac{dv}{dx}=\frac{3(1+v^2)}{2-3v} \\ \Rightarrow \frac{2-3v}{3(1+v^2)}~dv=\frac 1x~dx \\ \therefore\frac 13 \int{\frac{3\left(\frac 23-v\right)}{v^2+1}~dv}=\int{\frac{dx}{x}}\\ \Rightarrow \frac 23 \int{\frac{1}{v^2+1}~dv}-\frac 12\int{\frac{2v}{v^2+1}~dv}=\int{\frac{dx}{x}} \\ \Rightarrow \frac 23 \tan^{-1}v-\frac 12\log|v^2+1|=\log x+\log c_1 \\ \Rightarrow \frac 23 \tan^{-1}\frac yx-\frac 12\log|(y/x)^2+1|\\~~~~=\log x+\log c_1 \\ \Rightarrow 4 \tan^{-1}\frac yx-3\log\left|\frac{x^2+y^2}{x^2}\right|\\~~~~=6\log x+6\log c_1\\~~~[\text{Multiplying both sides by}~~6.] \\ \Rightarrow 4\tan^{-1}\frac yx-\log\left|\frac{(x^2+y^2)^3}{x^6}\right|=\log x^6+c\\~~~~~[~\text{where}~~c=\log c_1\rightarrow \text{constant of integration}] \\ \Rightarrow 4\tan^{-1}\frac yx-[\log(x^2+y^2)^3-\log x^6]\\~~~~=\log x^6+c \\ \Rightarrow 4\tan^{-1}\frac yx-\log(x^2+y^2)^3+\log x^6\\~~~~=\log x^6+c \\ \Rightarrow 4\tan^{-1}\frac yx-3\log(x^2+y^2)=c~~\text{(ans.)}$

$~3.~~(2x+y)~dy=(x-2y)~dx$

Sol. $~~~(2x+y)~dy=(x-2y)~dx \\ \Rightarrow  \frac{dy}{dx}=\frac{x-2y}{2x+y}\rightarrow(1)$

Let $~~y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~v+x\frac{dv}{dx}=\frac{x-2vx}{2x+vx} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{x(1-2v)}{x(2+v)} \\ \Rightarrow x\frac{dv}{dx}=\frac{1-2v}{2+v}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{1-2v-v(2+v)}{2+v} \\ \Rightarrow x\frac{dv}{dx}=\frac{1-2v-2v-v^2}{2+v} \\ \Rightarrow x\frac{dv}{dx}=\frac{-(v^2+4v-1)}{2+v} \\ \Rightarrow \frac{2+v}{v^2+4v-1}~dv=-\frac 1x~dx \\ \Rightarrow  \frac 12\int{\frac{2v+4}{v^2+4v-1}~dv}=-\int{\frac{dx}{x}} \\ \Rightarrow \frac 12\log|v^2+4v-1|\\~~~~=-\log x+\log c_1 \\ \Rightarrow \log\left|\frac{y^2}{x^2}+\frac{4y}{x}-1\right|\\~~~~=-2\log x+2\log c_1 \\ \Rightarrow  \log\left|\frac{y^2}{x^2}-+\frac{4y}{x}-1\right|+\log x^2=\log c_1^2 \\ \Rightarrow \log \left|\left(\frac{y^2}{x^2}-+\frac{4y}{x}-1\right)x^2\right|=\log c_1^2 \\ \Rightarrow \log|y^2+4xy-x^2|=\log c_1^2 \\ \Rightarrow y^2+4xy-x^2=c_1^2 \\ \Rightarrow  x^2-4xy-y^2=-c_1^2 \\ \Rightarrow  x^2-4xy-y^2=c~~\text{(ans.)}\\~~~[~~\text{where}~~c=-c_1^2~\\~~~\log c_1 \rightarrow \text{constant of integration.}]$

$~4.~~\frac{dy}{dx}=\frac{2x+3y}{3x+2y}$

Sol. $~~~\frac{dy}{dx}=\frac{2x+3y}{3x+2y}\rightarrow(1)$

Let $~~~~y=vx \\ \therefore \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~v+x\frac{dv}{dx}=\frac{2x+3(vx)}{3x+2(vx)} \\ \Rightarrow x\frac{dv}{dx}=\frac{x(2+3v)}{x(3+2v)}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{2+3v}{3+2v}-v \\ \Rightarrow  x\frac{dv}{dx}=\frac{2+3v-v(3+2v)}{3+2v} \\ \Rightarrow  x\frac{dv}{dx}=\frac{2+3v-3v-2v^2}{3+2v} \\ \Rightarrow  x\frac{dv}{dx}=\frac{2-2v^2}{3+2v}\\ \Rightarrow  \frac{3+2v}{-2v^2+2}~dv=\frac 1x~dx \\ \Rightarrow  -\frac 12 \int{\left(\frac{3+2v}{v^2-1}\right)~dv}=\int{\frac{dx}{x}} \\ \Rightarrow  \frac 32\int{\frac{dv}{v^2-1}}+\frac 12\int{\frac{2v~dv}{v^2-1}}=-\int{\frac{dx}{x}} \\ \Rightarrow  \frac 32.\frac 12 \log\left|\frac{v-1}{v+1}\right|+\frac 12 \log|v^2-1|\\~~~=-\log x+\log c_1\\~~~~~[\text{where}~~\log c_1 \rightarrow \text{constant of integration}] \\ \Rightarrow \frac 32 \log\left|\frac{(y/x)-1}{(y/x)-1}\right|+\log\left|\frac{y^2}{x^2}-1\right|\\~~~~=-2\log x+2\log c_1 \\ \Rightarrow  \frac 32 \log\left|\frac{y-x}{y+x}\right|+\log\left|\frac{y^2-x^2}{x^2}\right|\\~~+\log x^2=\log c_1^2 \\ \Rightarrow 3\log\left|\frac{y-x}{y+x}\right|\\~~~+2\left(\log\left|\frac{y^2-x^2}{x^2} \times x^2\right|\right)=2\log c_1^2 \\ \Rightarrow 3(\log|y-x|-\log|y+x|)\\~~+2\log\left|(y^2-x^2)\right|=2\log c_1^2 \\ \Rightarrow  3\log|y-x|-3\log|y+x|\\~~+2\log|(y+x)(y-x)|=2\log c_1^2 \\ \Rightarrow  3\log|y-x|-3\log|y+x|+2\log|y+x|\\~~~+2\log|y-x|=2\log c_1^2 \\ \Rightarrow  5\log|y-x|-\log|y+x|=\log c_1^4 \\ \Rightarrow  \log|y-x|^5=\log|y+x|+\log c\\~~~~[\text{where}~~c_1^4=c] \\ \Rightarrow  \log|y-x|^5=\log|c(y+x)| \\ \therefore [(y-x)^5]^2=[c(y+x)]^2 \\ \Rightarrow  (y-x)^{10}=c^2(x+y)^2~~\text{(ans.)}$

$~5.~~x\frac{dy}{dx}=y(\log y-\log x+1)$

Sol. $~~x\frac{dy}{dx}=y(\log y-\log x+1)\\ \Rightarrow \frac{dy}{dx}=\frac yx \left(\log \frac yx+1\right)\rightarrow(1)$

Let $~~y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~v+x\frac{dv}{dx}=v(\log v+1) \\ \Rightarrow  v+x\frac{dv}{dx}=v\log v+v  \\ \Rightarrow  x\frac{dv}{dx}=v\log v  \\ \Rightarrow \int{\frac 1v \cdot \frac{1}{\log v}~dv}=\int{\frac 1x~dx}\\ \Rightarrow \int{\frac{d(\log v)}{\log v}}=\int{\frac{dx}{x}} \\ \Rightarrow  \log|\log v|=\log|x|+\log c\\~~~\log c \rightarrow \text{constant of integration.} \\ \Rightarrow \log |\log v|=\log|xc| \\ \Rightarrow  |\log v|=|xc| \\ \Rightarrow |v|=e^{|xc|} \\ \Rightarrow v^2=e^{2xc} \\ \Rightarrow \frac{y^2}{x^2}=e^{2xc} \\ \Rightarrow  y^2=x^2e^{2xc}~~~\text{(ans.)}$

$~6.~~\frac{dy}{dx}+\frac yx=\frac{y^2}{x^2}$

Sol. $~~\frac{dy}{dx}+\frac yx=\frac{y^2}{x^2}\rightarrow(1)$

Let $~~~y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}\rightarrow(2)$

So, from $~(1),(2)~$ we get,

$~~~~~~v+x\frac{dv}{dx}+v=v^2 \\ \Rightarrow x\frac{dv}{dx}=v^2-2v \\ \Rightarrow \frac{1}{v^2-2v}~dv=\frac 1x~dx \\ \Rightarrow \int{\frac{dv}{(v^2-2v+1)-1}}=\int{\frac{dx}{x}} \\ \Rightarrow \int{\frac{dv}{(v-1)^2-1^2}}=\int{\frac{dx}{x}} \\ \Rightarrow \frac 12\log\left|\frac{v-1-1}{v-1+1}\right|=\log|x|+\log c_1 \\ \Rightarrow \log\left|\frac{v-2}{v}\right|=2\log |x|+2\log c_1 \\ \Rightarrow  \log \left|\frac{\frac yx-2}{\frac yx}\right|=\log |x|^2+\log c_1^2 \\ \Rightarrow \log \left|\frac{y-2x}{y}\right|=\log(x^2c)\\~~~[\text{where}~~c_1^2=c,\\c_1 \rightarrow \text{constant of integration.}] \\ \Rightarrow \left|\frac{y-2x}{y}\right|=x^2c \\ \Rightarrow \left(\frac{y-2x}{y}\right)^2=x^4c^2 \\ \Rightarrow (y-2x)^2=x^4y^2c^2~~\text{(ans.)}$