Differentiation (Part-38) | S N De

 

$~7.~~(x^2+xy)~\frac{dy}{dx}=x^2+y^2$

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

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

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

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

$~8.~~\frac{1}{2x}~\frac{dy}{dx}+\frac{x+y}{x^2+y^2}=0$

Sol. $~~~\frac{1}{2x}~\frac{dy}{dx}+\frac{x+y}{x^2+y^2}=0 \\ \Rightarrow \frac{dy}{dx}+\frac{2x^2+2xy}{x^2+y^2}=0\rightarrow(1)$

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

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

$~~~v+x~\frac{dv}{dx}+\frac{2x^2+2x(vx)}{x^2+(vx)^2}=0 \\ \Rightarrow v+x\frac{dv}{dx}+\frac{2x^2(1+v)}{x^2(1+v^2)}=0 \\ \Rightarrow v+x\frac{dv}{dx}=-\frac{2+2v}{1+v^2} \\ \Rightarrow  x\frac{dv}{dx}=-\left(\frac{2+2v}{1+v^2}+v\right) \\ \Rightarrow  x\frac{dv}{dx}=-\left(\frac{2+2v+v+v^3}{1+v^2}\right) \\ \Rightarrow  x\frac{dv}{dx}=-\frac{2+3v+v^3}{1+v^2} \\ \Rightarrow -\frac 13\int{\frac{3+3v^2}{v^3+2v+2}~dv}=\int{\frac{dx}{x}} \\ \Rightarrow  -\frac 13 \log|v^3+3v+2|=\log|x|+\log c_1\\~~~\text{where}~~\log c_1 \rightarrow \text{constant of integration.} \\ \Rightarrow  -\log\left|\frac{y^3}{x^3}+\frac{3y}{x}+2\right|=3\log|x|+3\log c_1 \\ \Rightarrow  -\log\left|\frac{y^3+3x^2y+2x^3}{x^3}\right|=\log |x|^3+\log c_1^3 \\ \Rightarrow  -(\log|y^3+3x^2y+2x^3|-\log|x|^3)\\~~~~=\log|x|^3+\log c_1^3 \\ \Rightarrow  -\log|y^3+3x^2y+2x^3|+\log|x|^3\\~~~~=\log|x|^3+\log c_1^3 \\ \Rightarrow \log|y^3+3x^2y+2x^3|=-\log c_1^3 \\ \Rightarrow \log|y^3+3x^2y+2x^3|=\log (c_1)^{-3} \\ \therefore (y^3+3x^2y+2x^3)^2=c^2~~\text{(ans.)}\\~~~[\text{where}~~(c_1)^{-3}=c^2]$

$~9.~~\frac{dy}{dx}=\frac{y(2y-x)}{x(2y+x)};\\~~~\text{given}~y=1,~\text{when}~x=1.$

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

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

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

$~~~~~v+x\frac{dv}{dx}=\frac{vx(2vx-x)}{x(2vx+x)} \\ \Rightarrow x\frac{dv}{dx}=\frac{x^2(2v^2-v)}{x^2(2v+1)}-v \\ \Rightarrow x\frac{dv}{dx}=\frac{2v^2-v}{2v+1}-v \\ \Rightarrow  x\frac{dv}{dx}=\frac{2v^2-v-v(2v+1)}{2v+1} \\ \Rightarrow  x\frac{dv}{dx}=\frac{2v^2-v-2v^2-v}{2v+1} \\ \Rightarrow  x\frac{dv}{dx}=\frac{-2v}{2v+1}\\ \Rightarrow  -\frac{2v+1}{2v}~dv=\frac 1x~dx \\ \Rightarrow  - \int{\left(1+\frac{1}{2v}\right)~dv}=\int{\frac{dx}{x}} \\ \Rightarrow -\int{dv}-\frac 12\int{\frac{dv}{v}}=\int{\frac{dx}{x}} \\ \Rightarrow -v-\frac 12\log|v|=\log |x|+\log c_1 \\ \Rightarrow v+\frac 12 \log|v|=-2\log|x|-\log c_1 \\ \Rightarrow 2\frac yx+\log\left|\frac yx\right|=-2\log|x|-\log c_1 \\ \Rightarrow 2\frac yx+\log\left|\frac yx\right|+\log|x|^2=-\log c_1 \\ \Rightarrow 2\frac yx+\log\left|\frac yx \times x^2\right|=-\log c_1 \\ \Rightarrow 2\frac yx+\log |xy|=-\log c_1\rightarrow(2)$

Now, $~~\text{given}~y=1,~\text{when}~x=1.$

Hence , by $\,(2)~~$ we get,

$~~~2+\log 1=-\log c_1 \\ \Rightarrow  2+0=-\log c_1 \\ \Rightarrow  \log c_1=-2.$

So, putting the value of $~~\log c_1~$ in $\,(2),\,$ we get,

$~~2\frac yx+\log|xy|=2~~\text{(ans.)}$

$~10.~~x(x-y)~dy+y^2~dx=0.$

Sol. $~~~~x(x-y)~dy+y^2~dx=0 \\ \Rightarrow (x^2-xy)~dy+y^2~dx=0 \\ \Rightarrow  \frac{dy}{dx}=-\frac{y^2}{x^2-xy} \\ \Rightarrow  \frac{dy}{dx}=-\frac{(y/x)^2}{1-(y/x)}$

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

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

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

$~11.~~x^2~dy+(xy+y^2)~dx=0$

Sol. $~~~x^2~dy+(xy+y^2)~dx=0 \\ \Rightarrow  x^2~dy=-(xy+y^2)~dx \\ \Rightarrow  \frac{dy}{dx}=-\frac{xy+y^2}{x^2} \\ \Rightarrow  \frac{dy}{dx}=-\left(\frac yx+\frac{y^2}{x^2}\right)\rightarrow(2)$

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

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

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

$~12.~~x^2y~dx-(x^3+y^3)~dy=0$

Sol. $~~~x^2y~dx-(x^3+y^3)~dy=0 \\ \Rightarrow x^2y~dx=(x^3+y^3)~dy \\ \Rightarrow \frac{dy}{dx}=\frac{x^2y}{x^3+y^3}\rightarrow(1) $

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

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

$~~~~v+x\frac{dv}{dx}=\frac{x^2.vx}{x^3+v^3x^3} \\ \Rightarrow v+x\frac{dv}{dx}=\frac{x^3.v}{x^3(1+v^3)}\\ \Rightarrow v+x\frac{dv}{dx}=\frac{v}{v^3+1} \\ \Rightarrow x\frac{dv}{dx}=\frac{v}{v^3+1}-v \\ \Rightarrow  x\frac{dv}{dx}=\frac{v-v(v^3+1)}{v^3+1} \\ \Rightarrow  x\frac{dv}{dx}=\frac{v-v^4-v}{v^3+1} \\ \Rightarrow x\frac{dv}{dx}=-\frac{v^4}{v^3+1} \\ \Rightarrow \int{\frac{v^3+1}{v^4}~dv}=-\int{\frac{dx}{x}} \\ \Rightarrow \int{\frac{v^3~dv}{v^4}}+\int{\frac{dv}{v^4}}=-\int{\frac{dx}{x}} \\ \Rightarrow  \int{\frac{dv}{v}}+\int{v^{-4}~dv}=-\log|x|+\log c_1 \\ \Rightarrow  \log|v|+\frac{v^{-4+1}}{-4+1}=\log|x|+\log c_1 \\ \Rightarrow  \log|y/x|-\frac 13 \cdot \frac{1}{v^3}=-\log|x|+\log c_1 \\ \Rightarrow  \log|y/x|-\frac 13 \cdot \frac{x^3}{y^3}=-\log|x|+\log c_1 \\ \Rightarrow 3\log|y/x|-\frac{x^3}{y^3}=-3\log|x|+3\log c_1 \\ \Rightarrow \log\left|\frac{y^3}{x^3}\right|-\frac{x^3}{y^3}=-\log|x|^3+3\log c_1 \\ \Rightarrow \log\left|\frac{y^3}{x^3}\right|+\log|x|^3-\frac{x^3}{y^3}=3\log c_1 \\ \Rightarrow \log\left|\frac{y^3}{x^3} \cdot x^3\right|-\frac{x^3}{y^3}=\log c_1^3 \\ \Rightarrow \log|y^3|-\log c_1^3=\frac{x^3}{y^3}\\ \Rightarrow \log\left(\frac{|y|^3}{c_1^3}\right)=\frac{x^3}{y^3} \\ \Rightarrow \left(\frac{y^3}{c_1^3}\right)^2=e^{2x^3/y^3} \\ \Rightarrow y^6=c_1^6~e^{2x^3/y^3} \\ \Rightarrow y^6=c^2~e^{2x^3/y^3}~~~\text{(ans.)}\\~~[\text{where}~~c^2=c_1^6,~\\~~~~~\log c_1\rightarrow \text{constant of integration.}]$