Differentiation (Part-38) | S N De

 

$~13.~~\frac{dy}{dx}=\frac yx+\cot\frac yx$

Sol. $~~~\frac{dy}{dx}=\frac yx+\cot\frac yx \rightarrow(1)$

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

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

$~~~~v+x\frac{dv}{dx}=v+\cot v \\ \Rightarrow x\frac{dv}{dx}=\cot v \\ \Rightarrow  \int{\frac{dv}{\cot v}}=\int{\frac{dx}{x}}\\ \Rightarrow \int{\tan v}~dv=\int{\frac{dx}{x}} \\ \Rightarrow \log|\sec v|=\log |x|+\log c_1 \\ \Rightarrow \log\left|\sec\frac yx\right|-\log|x|=\log c_1 \\ \Rightarrow \log \left|\frac{\sec\frac yx}{x}\right|=\log c_1 \\ \Rightarrow \frac{1}{x\cos \frac yx}=c_1 \\ \Rightarrow   x\cos \frac yx=\frac{1}{c_1} \\ \Rightarrow  x\cos\frac yx=c~~\text{(ans.)}\\~~~~[\text{where}~~\log c_1 \rightarrow \text{constant of integration},~~c=1/c_1.]$

$~14.~~\frac{dy}{dx}=\frac yx-\tan\frac yx$

Sol. $~~~\frac{dy}{dx}=\frac yx-\tan\frac yx\rightarrow(1)$

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

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

$~~~~v+x\frac{dv}{dx}=v-\tan v \\ \Rightarrow x\frac{dv}{dx}=-\tan v \\ \Rightarrow \int{\frac{ dv}{\tan v}}=-\int{\frac{dx}{x}} \\ \Rightarrow \int{\cot v~dv}=-\log|x|+\log c \\ \Rightarrow \log|\sin v|+\log|x|=\log c \\ \Rightarrow  \log\left|\sin\frac yx\right|+\log|x|=\log c \\ \Rightarrow \log|x\sin\frac yx|=\log c \\ \Rightarrow \left|x\sin\frac yx\right|=c~~\text{(ans.)}\\~~~~[~\text{where}~\log c\rightarrow \text{constant of integration.}]$

$~15.~~x^2~dy+(x^2-xy+y^2)~dy=0$

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

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

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

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

$~16.~~(x^2-2xy)~dy\\~~~+(x^2-3xy+2y^2)~dx=0$

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

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

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

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

$~17.~~y^2~dx+(x^2-xy)~dy=0$

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

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

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

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

$~18.~~\frac{dy}{dx}=\frac{y(y+x)}{x(y-x)}$

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

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

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

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

$~19.~~x~dy-y~dx=(x^2+y^2)~dx$

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

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

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

$~~~dv=(1+v^2)dx \\ \therefore \int{\frac{dv}{1+v^2}}=\int{dx} \\ \text{or,}~~ \tan^{-1} v=x+c \\ \text{or,}~~ \tan^{-1}(y/x)=x+c \\ \text{or,}~~\frac yx=\tan(x+c) \\ \text{or,}~~ y=x\tan(x+c)~~\text{(ans.)}\\~~~c \rightarrow \text{constant of integration.}$

Note : $~~d(y/x)\\=[\frac{d}{dx}(y/x) ]dx\\=\left(\frac 1x \frac{dy}{dx}-\frac {y}{x^2}\right)~dx\\=\frac 1x~dy-\frac{y}{x^2}~dx\\=\frac{x~dy-y~dx}{x^2}$