Differentiation (Part-38) | S N De

 

$\,11.\,ydx+(1+x^2)\tan^{-1}x \,dy=0$

Sol. $~~~~~ydx+(1+x^2)\tan^{-1}x \,dy=0\\ \Rightarrow \frac{dx}{(1+x^2)\tan^{-1}x}+\frac{dy}{y}=0 \\~~~\text{Integrating we get,}\\ ~~~\int{\frac{dx}{(1+x^2)\tan^{-1}x}}+\int{\frac{dy}{y}}=\log c\\ \Rightarrow \int{\frac{\sec^2z dz}{\sec^2z \cdot z}}+\log|y|=\log c \\~~~\text{let,}\,\, x=\tan z \Rightarrow dx=\sec^2z dz \\ \Rightarrow \int{\frac{dz}{z}}+\log|y|=\log c\\ \Rightarrow \log|z|+\log|y|=\log c \\ \Rightarrow \log|zy|=\log c\\ \Rightarrow yz=c \\ \Rightarrow y\tan^{-1}x=c~~\text{(ans.)}$

$c \rightarrow \,\,\text{constant of Integration.}$

$\,12.\,$ Show that, the general solution of  $\,(1+x^2)dy+(1+y^2)dx=0\,\,$ is $\,x+y=c(1-xy),\,$ where $\,c\,$ is an arbitrary constant.

Sol. $~~~~(1+x^2)dy+(1+y^2)dx=0\\ \Rightarrow \frac{dy}{1+y^2}+\frac{dx}{1+x^2}=0\\~~~\text{Integrating we get,}\\ ~~~\int{\frac{dy}{1+y^2}}+\int{\frac{dx}{1+x^2}}=c\\ \Rightarrow \tan^{-1}\left(\frac{x+y}{1-xy}\right)=\tan^{-1}c\\ \Rightarrow \frac{x+y}{1-xy}=c\\ \Rightarrow x+y=c(1-xy)~~\text{(ans.)}$

$c \rightarrow \,\,\text{constant of Integration.}$

$\,13.\,$ Show that, the general solution of the equation $\,dy=y\log xdx\,\,$ is $\,y=c(\frac xe)^x,\,$ where $\,c\,$ is an arbitrary constant.

Sol. $~~~~~dy=y\log xdx\\ \Rightarrow \frac{dy}{y}=\log x \,dx \\ \text{Integrating we get,}\\ ~~\int{\frac{dy}{y}}=\int{\log x\,dx}\\ \Rightarrow \log y=\log x \cdot x-\int{\frac 1x \cdot x\,dx}+\log c\\ \Rightarrow \log y=x\log x-x+\log c\\ \Rightarrow \log y=\log(x^x)-\log(e^x)+\log c\\ \Rightarrow \log y=\log\left[(\frac xe)^x \cdot c\right]\\ \Rightarrow y=c\left(\frac xe\right)^x$

$c \rightarrow \,\,\text{constant of Integration.}$

Find the particular solution of each of the following equations: 

$\,14.~~~a^2y\frac{dy}{dx}+b^2x=0;\,\text{given}~~y=\frac{b}{\sqrt 2},\\~~~~~~~~\text{when}\,\,x=\frac{a}{\sqrt 2}.$

Sol. $~~~~~~a^2y\frac{dy}{dx}+b^2x=0\\ \Rightarrow a^2y~dy+b^2x~dx=0\\ \text{Integrating we get,}\\~~~a^2 \int{ydy}+b^2\int{xdx}=c\\ \Rightarrow a^2 \cdot \frac{y^2}{2}+b^2 \cdot \frac{x^2}{2}=c \rightarrow(1)$

Now, $\text{given}~~y=\frac{b}{\sqrt 2},\,\text{when}\,\,x=\frac{a}{\sqrt 2}.$

Hence, from $\,(1),\,$ we get,

$~~~~~a^2 \cdot \frac 12\left(\frac{b}{\sqrt 2}\right)^2+b^2 \cdot \frac 12\left(\frac{a}{\sqrt 2}\right)^2=c\\ \Rightarrow \frac{a^2b^2}{4}+\frac{a^2b^2}{4}=c\\ \Rightarrow c=\frac{a^2b^2}{2}.$

Now, putting the value of $\,c\,$ in $\,(1),\,$ we get,

$~~~\frac{a^2y^2}{2}+\frac{b^2x^2}{2}=\frac{a^2b^2}{2}\\ \Rightarrow b^2x^2+a^2y^2=a^2b^2\,\,\text{(ans.)}$

$15.~~~\frac{dy}{dx}=\frac{\cos^2y}{1+x^2};\\~~~\text{given}~~y=0,\text{when}~~x=0.$

Sol. $~~\frac{dy}{dx}=\frac{\cos^2y}{1+x^2} \\ \Rightarrow \sec^2y ~dy=\frac{dx}{1+x^2}\\~~~\text{Integrating we get,}\\ \int{\sec^2y dy}=\int{\frac{dx}{1+x^2}}+c\\ \Rightarrow \tan y=\tan^{-1}x+c\rightarrow(1)$

Now, $\,y(0)=0$

Hence, by $\,(1),$ we get,

$\,0=0+c\Rightarrow c=0.$

Putting the value of $\,c\,$ in $\,(1),~~$ we get,

$\,\tan y=\tan^{-1}x~~\text{(ans.)}$

$\,16.\,v\frac{dv}{dx}=n^2x;\,\text{given}~~v=u,~~\text{when}~~x=a.$

Sol. $~~v\frac{dv}{dx}=n^2x\\ \Rightarrow vdv=n^2xdx\\ ~\text{Integrating we get,}\\~~~\int{vdv}=n^2\int{xdx}+c\\ \Rightarrow \frac{v^2}{2}=n^2\cdot\frac{x^2}{2}+c\rightarrow(1)$

Now, $~~v(a)=u~~\text{(given)}$

Hence, from $\,(1),\,$ we get,

$~~\frac{u^2}{2}=n^2\cdot \frac{a^2}{2}+c \\ \Rightarrow c=\frac{u^2-n^2a^2}{2}.$

Now, putting the value of $\,c\,$ in $\,(1),$ we get,

$~~~\frac{v^2}{2}=\frac{n^2x^2}{2}+\frac{u^2-n^2a^2}{2}\\ \Rightarrow v^2-n^2x^2=u^2-n^2a^2~~~\text{(ans.)}$

$\,17.~~\frac{dy}{dx}=y\sec x;~~\text{given}~y=1,\text{when}~~x=\frac{\pi}{6}$

Sol. $~~~\frac{dy}{dx}=y\sec x\\ \Rightarrow \frac{dy}{y}=\sec xdx\\~~\text{Integrating we get,}\\~~\int{\frac{dy}{y}}=\int{\sec x~dx}+c\\ \Rightarrow \log|y|=\log|\sec x+\tan x|+c\rightarrow(1)$

Now, $~~~y(\frac{\pi}{6})=1.\\ \therefore ~~\log|1|=\log|\sec(\pi/6)+\tan(\pi/6)|+c\\ \Rightarrow 0=\log|(2/\sqrt 3)+(1/\sqrt 3)|+c\\ \Rightarrow 0=\log\left|\frac{3}{\sqrt 3}\right|+c\\ \Rightarrow c+\log(\sqrt 3)=0 \\ \Rightarrow c=-\frac 12 \log 3$

Now, putting the value of $\,c,\,$ in $\,(1),$ we get,

$~~\log|y|=\log|\sec x+\tan x|-\frac 12 \log 3\\ \Rightarrow 2\log|y|+\log 3=2\log|\sec x+\tan x|\\ \Rightarrow \log|3y^2|=\log|(\sec x+\tan x)^2|\\ \Rightarrow ~~3y^2=(\sec x+\tan x)^2~~\text{(ans.)}$

$\,18.~~y~dx+x~dy=xy(dy-dx);~~\\~~\text{given}~~y=1,~~\text{when}~x=1.$

Sol. $~~~~~y~dx+x~dy=xy(dy-dx)\\ \Rightarrow d(xy)=xy~d(y-x)\\ \Rightarrow \frac{d(xy)}{xy}=d(y-x)\\~~~\text{Integrating we get,}\\~~ \int{\frac{d(xy)}{xy}}=\int{d(y-x)}+c\\ \Rightarrow \log|xy|=y-x+c\rightarrow(1)$

Now, $~~y(1)=1\\ \Rightarrow \log 1=1-1+c\\ \Rightarrow c=0.$

So, putting the value of $\,c,\,$ in $\,(1),\,$ we get,

$~~\log|xy|=y-x~~\text{(ans.)}$

$\,19.~~\frac{dy}{dx}=e^{4x-3y}; \\~~~\text{given}~~y=0,~~\text{when}~~x=0.\,\,\text{(NCERT)}$

Sol. $~~\frac{dy}{dx}=e^{4x-3y} \\ \Rightarrow e^{3y}~dy=e^{4x}~dx \\~~~\text{Integrating we get,}\\~~~\int{e^{3y}~dy}=\int{e^{4x}~dx}+c\\ \Rightarrow \frac{e^{3y}}{3}=\frac{e^{4x}}{4}+c\rightarrow(1)\\~~c\rightarrow\text{constant of integration}$

Now, $~~y(0)=0\\ ~~\therefore \frac 13=\frac 14+c\\ \Rightarrow c=\frac 13-\frac 14=\frac{1}{12}.$

So, putting the value of $\,c,\,$ in $\,(1),\,$ we get,

$~~~~~\frac{e^{3y}}{3}=\frac{e^{4x}}{4}+\frac{1}{12}\\ \Rightarrow 4e^{3y}-3e^{4x}=1~~\text{(ans.)}$