Differentiation (Part-38) | S N De

 

$~20.~~x~dy-y~dx=\sqrt{x^2+y^2}~dx$

Sol. $~~~x~dy-y~dx=\sqrt{x^2+y^2}~dx \\ \text{or,}~~x~dy=(\sqrt{x^2+y^2}+y)~dx \\ \text{or,}~~ \frac{dy}{dx}=\frac{\sqrt{x^2+y^2}+y}{x} \\ \text{or,}~~ \frac{dy}{dx}=\sqrt{1+(y/x)^2}+\frac yx\rightarrow(1)$

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

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

$~~~~v+x\frac{dv}{dx}=\sqrt{1+v^2}+v \\ \text{or,}~~x\frac{dv}{dx}=\sqrt{1+v^2} \\ \text{or,}~~\int{\frac{dv}{\sqrt{1+v^2}}}=\int{\frac{dx}{x}} \\ \text{or,}~~\log|v+\sqrt{v^2+1}|=\log|x|+\log c \\~~~\text{where}~~\log c \rightarrow \text{constant of integration.} \\ \text{or,}~~ \log\left|\frac yx+\sqrt{(y/x)^2+1}\right|-\log|x|=\log c \\ \text{or,}~~ \log\left|\frac yx+\frac{\sqrt{y^2+x^2}}{x}\right|-\log|x|=\log c \\ \text{or,}~~ \log\left|\frac{y+\sqrt{x^2+y^2}}{x}\right|-\log|x|=\log c \\ \text{or,}~~\log\left|\frac{y+\sqrt{x^2+y^2}}{x^2}\right|=\log c \\ \text{or,}~~ |y+\sqrt{x^2+y^2}|=cx^2~~~\text{(ans.)}$

$~21.~~(3xy+y^2)~dx+(x^2+xy)~dy=0$

Sol. $~~~(3xy+y^2)~dx+(x^2+xy)~dy=0 \\ \text{or,}~~ \frac{dy}{dx}=-\frac{3xy+y^2}{x^2+xy} \\ \text{or,}~~\frac{dy}{dx}=-\frac{3(y/x)+(y/x)^2}{1+(y/x)}\rightarrow(1)$

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

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

$~~~~v+x\frac{dv}{dx}=-\frac{3v+v^2}{1+v} \\ \text{or,}~~ x\frac{dv}{dx}=-\left(\frac{3v+v^2}{1+v}+v\right) \\ \text{or,}~~ x\frac{dv}{dx}=-\frac{3v+v^2+v(1+v)}{1+v} \\ \text{or,}~~ x\frac{dv}{dx}=-\frac{2v^2+4v}{1+v} \\ \text{or,}~~ \frac{v+1}{2v^2+4v}~dv=-\frac 1x~dx \\ \text{or,}~~ \frac 14\displaystyle\int{\frac{4v+4}{2v^2+4v}~dv}=-\int{\frac{dx}{x}} \\ \text{or,}~~ \frac 14\displaystyle \int{\frac{d(2v^2+4v)}{2v^2+4v}}=-\log|x|+\log c_1\\ \text{or,}~~ \frac 14\log|2v^2+4v|=-\log|x|+\log c_1 \\ \text{or,}~~ \log|2v^2+4v|^{1/4}+\log|x|=\log c_1 \\ \text{or,}~~\log\left |\left(2\frac{y^2}{x^2}+4\frac yx\right)^{1/4} \cdot x\right|=\log c_1 \\ \text{or,}~~ \left(2\frac{y^2}{x^2}+4\frac yx\right)^{1/4} \cdot x=c_1 \\ \text{or,}~~ \left(2\frac{y^2}{x^2}+4\frac yx\right)\cdot x^4=c_1^4 \\ \text{or,}~~  (2x^2y^2+4x^3y)=2c^2\\~~~~~[~\text{where}~~c_1^4=2c^2] \\ \text{or,}~~2x^2(y^2+2xy)=2c^2 \\ \text{or,}~~ x^2(y^2+2xy)=c^2~~~\text{(ans.)}$

Note : $~~\log c_1 \rightarrow \text{constant of integration.}$

$~22.~~xy~\frac{dy}{dx}-y^2=(x+y)^2~e^{-y/x}$

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

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

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

$~~~x(vx)\left(v+x\frac{dv}{dx}\right)-(vx)^2=(x+vx)^2e^{-v} \\ \text{or,}~~vx^2\left(v+x\frac{dv}{dx}\right)-v^2x^2=x^2(1+v)^2e^{-v} \\ \text{or,}~~v\left(v+x\frac{dv}{dx}\right)-v^2=(1+v)^2e^{-v} \\ \text{or,}~~ v^2+vx\frac{dv}{dx}-v^2=(1+v)^2e^{-v} \\ \text{or,}~~ vx\frac{dv}{dx}=(1+v)^2e^{-v} \\ \text{or,}~~ \frac{ve^v}{(1+v)^2}~dv=\frac{1}{x}~dx \\ \text{or,}~~ \displaystyle\int{\frac{(1+v-1)e^v}{(1+v)^2}~dv}=\int{\frac{dx}{x}} \\ \text{or,}~~ \displaystyle\int{e^v\left(\frac{1}{1+v}-\frac{1}{(1+v)^2}\right)~dv}=\int{\frac{dx}{x}} \\ \text{or,}~~\frac{e^v}{1+v}=\log|x|+\log c~~[*] \\ \text{or,}~~\frac{e^{y/x}}{1+(y/x)}=\log|cx| \\ \text{or,}~~ \frac{xe^{y/x}}{x+y}=\log|cx| \\ \text{or,}~~ xe^{y/x}=(x+y)\log|cx|~~\text{(ans.)}$

Note[*] : $~~\int{e^v(f(v)+f'(v))~dv}=e^v~f(v)+c\\~~~\log c \rightarrow \text{constant of integration.}$

$~23.~~\left(x+y\cos \frac yx\right)~dx=x\cos\frac yx~dy$

Sol. $~~~\left(x+y\cos \frac yx\right)~dx=x\cos\frac yx~dy \\ \text{or,}~~ \frac{dy}{dx}=\frac{x+y\cos \frac yx}{x\cos \frac yx} \\ \text{or,}~~ \frac{dy}{dx}=\frac{1+\frac yx \cos\frac yx}{\cos \frac yx}\rightarrow(1)$

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

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

$~~~v+x\frac{dv}{dx}=\frac{1+v \cos v}{\cos v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{1+v\cos v}{\cos v}-v \\ \text{or,}~~x\frac{dv}{dx}=\frac{1+v\cos v-v\cos v}{\cos v} \\ \text{or,}~~ x\frac{dv}{dx}=\frac{1}{\cos v} \\ \text{or,}~~\displaystyle\int{\cos v~dv}=\int{\frac{dx}{x}} \\ \text{or,}~~\sin v=\log|x|+c \\ \text{or,}~~ \sin\frac yx=\log|x|+c~~\text{(ans.)}\\~~[~c~\rightarrow \text{constant of integration.}]$

$~24.~~\left(1+e^{x/y}\right)~dx+e^{x/y}\left(1-\frac xy\right)~dy=0$

Sol. $~~~\left(1+e^{x/y}\right)~dx+e^{x/y}\left(1-\frac xy\right)~dy=0 \\ \text{or,}~~ \frac{dx}{dy}=-\frac{e^{x/y}\left(1-\frac xy\right)}{1+e^{x/y}}\rightarrow(1)$

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

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

$~~v+y\frac{dv}{dy}=-\frac{e^v(1-v)}{1+e^v} \\ \text{or,}~~ y\frac{dv}{dy}=\frac{-e^v+ve^v}{1+e^v}-v \\ \text{or,}~~ y~\frac{dv}{dy}=\frac{-e^v+ve^v-v(1+e^v)}{1+e^v} \\ \text{or,}~~y~\frac{dv}{dy}=\frac{-e^v+ve^v-v-ve^v}{1+e^v} \\ \text{or,}~~ y~\frac{dv}{dy}=\frac{-(e^v+v)}{1+e^v} \\ \text{or,}~~\displaystyle \int{\frac{1+e^v}{v+e^v}~dv}=-\int{\frac{dy}{y}} \\ \text{or,}~~ \displaystyle\int{\frac{d(v+e^v)}{v+e^v}}=-\log|y|+\log c \\ \text{or,}~~\log|v+e^v|+\log|y|=\log c \\ \text{or,}~~\log\left|\left(\frac xy+e^{x/y}\right).y\right|=\log c \\ \text{or,}~~\log|x+ye^{x/y}|=\log c \\ \text{or,}~~ x+ye^{x/y}=c~~\text{(ans.)}$

Note : $~~\log c \rightarrow \text{constant of integration.}$

$~25(i)~\left(y\sin\frac yx-x\cos\frac yx\right)x~dy\\~~~~=\left(x\cos\frac yx+y\sin\frac yx\right)y~dx$

Sol. $~~\left(y\sin\frac yx-x\cos\frac yx\right)x~dy\\~~~=\left(x\cos\frac yx+y\sin\frac yx\right)y~dx \\ \text{or,}~~ \frac{dy}{dx}=\frac{y\left(x\cos\frac yx+y\sin\frac yx\right)}{x\left(y\sin\frac yx-x\cos\frac yx\right)} \rightarrow(1)$

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

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

$~~~~~v+x\frac{dv}{dx}=\frac yx \cdot\frac{\cos (y/x)+(y/x)\sin (y/x)}{(y/x)\sin (y/x)-\cos (y/x)} \\ \text{or,}~~ v+x\frac{dv}{dx}=v \cdot \left(\frac{\cos v+v\sin v}{v\sin v-\cos v}\right) \\ \text{or,}~~x\frac{dv}{dx}=\frac{v\cos v+v^2\sin v}{v\sin v-\cos v}-v \\ \text{or,}~~ x~\frac{dv}{dx}=\frac{v\cos v+v^2\sin v-v(v\sin v-\cos v)}{v\sin v-\cos v} \\ \text{or,}~~x~\frac{dv}{dx}=\frac{v\cos v+v^2\sin v-v^2\sin v+v\cos v}{v\sin v-\cos v} \\ \text{or,}~~ x~\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v} \\ \text{or,}~~\frac{v\sin v-\cos v}{2v\cos v}~dv=\frac 1x~dx \\ \text{or,}~~ \frac 12\displaystyle\int{\tan v}~dv-\frac 12\int{\frac{dv}{v}}=\int{\frac{dx}{x}} \\ \text{or,}~~\frac 12\left( \log|\sec v|-\log|v|\right)\\~~~~=\log|x|+\log c_1 \\ \text{or,}~~ -\log|\cos v|-\log|v|\\~~~~=2\log|x|+2\log c_1 \\ \text{or,}~~ -(\log|\cos v|+\log|v|)\\~~~~=\log|x|^2+\log c_1^2 \\ \text{or,}~~ \log\left|\cos(y/x)\cdot (y/x)\right|+\log|x|^2=c \\ \text{or,}~~\log\left|\left(\cos\frac yx\right) \cdot \frac yx\cdot x^2\right|=c \\ \text{or,}~~ \log|xy\cos(x/y)|=c~~\text{(ans.)}\\~~~~[\text{where}~c=-\log c_1^2]$

Note : $~~\log c_1 \rightarrow \text{constant of integration.}$

$~25(ii)~~x(y~dx+x~dy)\cos(xy)\\~~~~+\sin(xy)~dx=0$

Sol. $~~x(y~dx+x~dy)\cos(xy)+\sin(xy)~dx=0 \\ \text{or,}~~ x~d(xy)\cos(xy)+\sin(xy)~dx=0 \\ \text{or,}~~x~d(xy)\cos(xy)=-\sin(xy)~dx \\ \text{or,}~~ \frac{\cos(xy)}{\sin(xy)}~d(xy)=-\frac{dx}{x} \\ \text{or,}~~\displaystyle\int{\cot(xy)~d(xy)}=-\int{\frac{dx}{x}} \\ \text{or,}~~ \log|\sin(xy)|=-\log|x|+\log c \\ \text{or,}~~\log|\sin(xy)|+\log|x|=\log c \\ \text{or,}~~ \log|x\sin(xy)|=\log c \\ \text{or,}~~ |x\sin(xy)|=c~~\text{(ans.)} $

Note : $~~\log c \rightarrow \text{constant of integration.}$