Differentiation (Part-38) | S N De

 

Solve [$\,1-11$]: 

$\,1.~~(1+x^2)dx+(1+y^2)dy=0$  

Sol. $~~(1+x^2)dx+(1+y^2)dy=0\rightarrow(1)\\ \text{Integrating (1) we get,}\\ ~~~\int(1+x^2)dx+\int(1+y^2)dy=\frac c3\\ \Rightarrow x+\frac{x^3}{3}+y+\frac{y^3}{3}=\frac c3 \\ \Rightarrow x^3+y^3+3(x+y)=c\\~~~~~~~~~~~[\text{c =constant of integration.}]$ 

$\,2.~~e^{x-y}dx+e^{y-x}dy=0$

Sol. $~~~~~~e^{x-y}dx+e^{y-x}dy=0\\ \Rightarrow e^{2x}dx+e^{2y}dy=0 \\~~~~~ \text{Integrating we get,} \\~~~\int{e^{2x}dx}+\int{e^{2y}dy}=\frac c2 \\ \Rightarrow \frac{e^{2x}}{2}+\frac{e^{2y}}{2}=\frac c2\\ \Rightarrow e^{2x}+e^{2y}=c~~\text{(ans.)}$

$~~c \rightarrow \text{constant of integration}$

$\,3.~~~y(1+x^2)dy=x(1+y^2)dx$

Sol. $~~~~~~~y(1+x^2)dy=x(1+y^2)dx\\ \Rightarrow \frac{y}{1+y^2}dy=\frac{x}{1+x^2}dx \\ ~~~\text{Integrating we get,}\\ ~~\int{\frac{y}{1+y^2}dy}=\int{\frac{x}{1+x^2}dx }+\frac 12\log c\\ \Rightarrow \frac 12\log(1+y^2)=\frac 12\log(1+x^2)+\frac 12\log c \\ \Rightarrow \log(1+y^2)=\log(1+x^2)+\log c \\ \Rightarrow \log(1+y^2)=\log[c(1+x^2)]\\ \Rightarrow 1+y^2=c(1+x^2)$

$~~c \rightarrow \text{constant of integration}$

$\,4.~~\sec^2x\tan^2ydx+\sec^2y\tan^2xdy=0$

Sol. $~~\sec^2x\tan^2ydx+\sec^2y\tan^2xdy=0\\ \Rightarrow \frac{\sec^2x}{\tan^2x}dx+\frac{\sec^2y}{\tan^2y}dy=0\\ \Rightarrow \frac{\sec^2x \cos^2x}{\sin^2x}dx+\frac{\sec^2y\cos^2y}{\sin^2y}dy=0\\ \Rightarrow \csc^2xdx+\csc^2ydy=0\\~~~ \text{Integrating we get,}\\ ~~~\int{\csc^2x}dx+\int{\csc^2y}dy=0\\ \Rightarrow -\cot x-\cot y=-c \\ \Rightarrow \frac{1}{\tan x}+\frac{1}{\tan y}=c\\ \Rightarrow \tan x+\tan y=c\tan x\tan y~~\text{(ans.)}$

$~~c \rightarrow \text{constant of integration}$

$\,5.~~\sqrt{1+x^2}~~dy+\sqrt{1+y^2}~~dx=0$

Sol. $~~~\sqrt{1+x^2}~~dy+\sqrt{1+y^2}~~dx=0\\ \Rightarrow \frac{dx}{\sqrt{1+x^2}}+\frac{dy}{\sqrt{1+y^2}}=0\\ ~~~\text{Integrating we get,}\\~~~~\int{\frac{dx}{\sqrt{1+x^2}}}+\int{\frac{dy}{\sqrt{1+y^2}}}=\log c\\ \Rightarrow \log(x+\sqrt{x^2+1})+\log(y+\sqrt{y^2+1})\\~~~~~~~~~~~=\log c\\ \Rightarrow \log[{(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})}]=\log c\\ \Rightarrow (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=c$

$~~c \rightarrow \text{constant of integration}$

$\,6.~~~\frac{dy}{dx}=\sqrt{4-y^2}$

Sol. $~~~~~\frac{dy}{dx}=\sqrt{4-y^2}\\ \Rightarrow \frac{dy}{\sqrt{4-y^2}}=dx\\ \text{Integrating we get,}\\~~~\int{\frac{dy}{\sqrt{4-y^2}}}=\int{dx}+c\\ \Rightarrow \sin^{-1}\left(\frac{y}{2}\right)=x+c.$

$~~c \rightarrow \text{constant of integration}$

$\,7.~~y(1+x)~dx+x(1+y)~dy=0$

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

$~~c \rightarrow \text{constant of integration}$

$\,8(i).~~\frac{dy}{dx}=e^{x+y}+x^2e^y$

Sol. $~~~~\frac{dy}{dx}=e^{x+y}+x^2e^y\\ \Rightarrow e^{-y}~dy=(e^x+x^2)dx\\~~~~\text{Integrating we get,}\\~~~\int{ e^{-y}}~~dy=\int{(e^x+x^2)}~~dx+\frac c3\\ \Rightarrow -e^{-y}=e^x+\frac{x^3}{3}+\frac c3\\ \Rightarrow 3(e^x+e^{-y})+x^3+c=0.$

$~~c \rightarrow \text{constant of integration}$

$\,8(ii).y\frac{dy}{dx}=xe^{x^2+y^2}$

Sol. $~~~y\frac{dy}{dx}=xe^{x^2+y^2}\\ \Rightarrow y\frac{dy}{dx}=xe^{x^2}.e^{y^2}\\ \Rightarrow \frac{y}{e^{y^2}}~dy=(xe^{x^2})~dx\\~~~\text{Integrating we get,}\\~~~~\int{\frac{y}{e^{y^2}}~dy}=\int{(xe^{x^2})~dx}-\frac c2\\ \Rightarrow \frac 12\int{e^{-y^2}}d(y^2)=\frac{1}{2}\int{e^{x^2}d(x^2)}-\frac c2\\ \Rightarrow \frac 12(-e^{-y^2})=\frac 12e^{x^2}-\frac c2\\ \Rightarrow e^{x^2}+e^{-y^2}=c\\ \Rightarrow e^{-(x^2+y^2)}+1=ce^{-x^2}\,\,\text{(ans.)}$

$~~c \rightarrow \text{constant of integration}$

$\,9.~~\frac{dy}{dx}=y\sin2x~~~\text{given,}~~y(0)=1.$

Sol. $~~\frac{dy}{dx}=y\sin2x\\ \Rightarrow \frac{dy}{y}=\sin2x ~dx\\ ~~\text{Integrating we get,}\\~~~\int{\frac{dy}{y}}=\int{\sin2x}+c\\ \Rightarrow \log|y|=-\frac 12 \cos2x+c\rightarrow(1)\\~~~\text{Now,}~~y(0)=1\\ \Rightarrow \log 1=-\frac 12 \cdot 1+c\\ \Rightarrow 0=-\frac 12+c\\ \Rightarrow c=\frac 12.$

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

$\,\log|y|=-\frac 12\cos2x+\frac 12\\ \Rightarrow 2\log|y|+\cos2x=1~~\text{(ans.)}$

$\,10.~~ 3^{y-x}dy-3^{x-y}~dx=0$

Sol. $~~ 3^{y-x}dy-3^{x-y}~dx=0\\ \Rightarrow 3^{2y}dy-3^{2x}dx=0\\~~~\text{Integrating we get,}\\~~~\int{3^{2y}}~~dy-\int{3^{2x}}~~dx=c_1\\ \Rightarrow \frac{3^{2y}}{2\log 3}-\frac{3^{2x}}{2\log 3}=\frac{c}{2\log 3}\\ \Rightarrow 3^{2y}=3^{2x}+c$

$~~c \rightarrow \text{constant of integration}$