Differentiation (Part-38) | S N De

 

$~6.~e^y\frac{dy}{dx}+x^3\cos x^2=0$

Sol. $~~e^y\frac{dy}{dx}+x^3\cos x^2=0 \\ \Rightarrow  e^y~dy+x^3\cos x^2~dx=0 \\ \therefore\int{e^y~dy}+\int{x^3\cos x^2~dx}=c \\ \Rightarrow  e^y+I_1=c_1 \rightarrow(1)\\ ~~~c_1 \rightarrow \text{constant of Interest.}$

Now, $~~I_1=\int{x^3\cos x^2~dx}\\ ~~\text{Let}~~x^2=z \\ \Rightarrow  2x~dx=dz\\ \therefore I_1=\frac 12\int{x^2\cos x^2 \cdot 2x~dx}\\~~~~~~~=\frac 12\int{z\cos z~dz}\\~~~~~~~=\frac 12\left[z\int{\cos z~dz}-\int{\{\frac{d}{dz}(z)\int{\cos z ~dz}}\}dz\right]\\~~~~~~~=\frac 12\left[z\sin z-\int{1.\sin z~dz}\right]\\~~~~~~~=\frac 12[z\sin z+\cos z]\\~~~~~~~=\frac 12[x^2\sin x^2+\cos x^2]\rightarrow(2)$

Hence, from $\,(1)~$ and $\,(2),\,$ we get,

$\,e^y+\frac 12\left(x^2\sin x^2+\cos x^2\right)=c_1\\ \Rightarrow 2e^y+x^2\sin x^2+\cos x^2=c\\~~~~\text{where}~~c=2c_1.$

$~7.~\frac{dy}{dx}=\log(x+1)$

Sol. $~~~~~\frac{dy}{dx}=\log(x+1)\\ \Rightarrow dy=\log(x+1)dx\\ \therefore \int{dy}=\int{\log(x+1)~dx}+c \\ \Rightarrow  y=x\log(x+1)-\int{\{\frac{d}{dx}~\log(x+1)\int{dx}\}~dx}+c\\ \Rightarrow y=x\log(x+1)-\int{\frac{x}{x+1}~dx}+c \\ \Rightarrow y=x\log(x+1)-\int{[1-\frac{1}{x+1}]dx}+c\\ \Rightarrow y=x\log(x+1)-\int{dx}+\int{\frac{dx}{x+1}}+c\\ \Rightarrow y=x\log(x+1)-x+\log(x+1)+c\\ \Rightarrow y=(x+1)\log(x+1)-x+c$

$~~~~~~~c\rightarrow \text{Constant of Integration.}$

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

Sol. $~~\frac{dy}{dx}=\frac{y^2-y-2}{x^2-x+1}\\ \Rightarrow \frac{1}{y^2-y+1}~dy=\frac{1}{x^2-x+1}~dx\\ \therefore~\int{\frac{dy}{\left(y^2-2y.\frac 12+\frac 14\right)+1-\frac 14}}=\int{\frac{dx}{\left(x-\frac 12\right)^2+(\sqrt{3}/2)^2}}\\ \Rightarrow  \int{\frac{dy}{\left(y-\frac 12\right)^2+(\sqrt 3/2)^2}}=\int{\frac{dx}{\left(x-\frac 12\right)^2+(\sqrt{3}/2)^2}}\\ \Rightarrow \frac{2}{\sqrt 3}~\tan^{-1}\left(\frac{y-\frac 12}{\sqrt 3/2}\right)=\frac{2}{\sqrt 3}~\tan^{-1}\left(\frac{x-\frac 12}{\sqrt 3/2}\right)+c_1 \\ \Rightarrow \frac{2}{\sqrt 3}\left[\tan^{-1}\left(\frac{y-\frac 12}{\sqrt 3/2}\right)-\tan^{-1}\left(\frac{x-\frac 12}{\sqrt 3/2}\right)\right]=c_1\\ \Rightarrow\tan^{-1}\left(\frac{2y-1}{\sqrt 3}\right)-\tan^{-1}\left(\frac{2x-1}{\sqrt 3}\right)=\frac{\sqrt 3c_1}{2} \\ \Rightarrow \tan^{-1}\left[\frac{\frac{2y-1}{\sqrt 3}-\frac{2x-1}{\sqrt 3}}{1+\frac{2y-1}{\sqrt 3} \cdot \frac{2x-1}{\sqrt 3}}\right]=c_2~[*]\\ \Rightarrow \frac{(2y-1)-(2x-1)}{\sqrt 3+\frac{1}{\sqrt 3}(2x-1)(2y-1)}=\tan c_2\\ \Rightarrow \frac{2(y-x)}{3+(2x-1)(2y-1)}=\frac{1}{\sqrt 3} \tan c_2\\ \Rightarrow \frac{2(y-x)}{3+4xy-2x-2y+1}=c_3~[**]\\ \Rightarrow \frac{2(y-x)}{2(2+2xy-x-y)}=c_3 \\ \Rightarrow \frac{y-x}{2+2xy-x-y}=c_3=-\frac 1c,~~\text{(say)}\\ \Rightarrow  c(x-y)=2(xy+1)-x-y~~\text{(ans.)}$

Note[*] : $~~c_2=\frac{\sqrt 3 c_1}{2}$

Note[**] : $~~c_3=\frac{1}{\sqrt 3}\tan c_2$

$~9.~\frac{dy}{dx}=\frac{y^2-y-2}{x^2+2x-3}$

Sol. $~~\frac{dy}{dx}=\frac{y^2-y-2}{x^2+2x-3}\\ \Rightarrow \frac{1}{y^2-y-2}~dy=\frac{1}{x^2+2x-3}~dx\\ \Rightarrow \frac{1}{\left(y^2-2y.\frac 12+\frac 14\right)-2-\frac 14}~dy=\frac{1}{(x^2+2.x.1+1)-4}~dx\\ \Rightarrow \int{\frac{dy}{\left(y-\frac 12\right)^2-\frac 94}}=\int{\frac{dx}{(x+1)^2-2^2}}\\ \Rightarrow \frac{1}{2\times \frac 32}\log\left|\frac{y-\frac 12-\frac 32}{y-\frac 12+\frac 32}\right|=\frac{1}{2 \times 2}\log \left|\frac{x+1-2}{x+1+2}\right|+\log c_1\\ \Rightarrow \frac 13\log \left|\frac{y-2}{y+1}\right|=\frac 14\log \left|\frac{x-1}{x+3}\right|+\log c_1\\ \Rightarrow 12 \times \frac 13\log \left|\frac{y-2}{y+1}\right|=12 \times \frac 14\log \left|\frac{x-1}{x+3}\right|+12\log c_1 \\ \Rightarrow 4\log \left|\frac{y-2}{y+1}\right|=3\log \left|\frac{x-1}{x+3}\right|+\log c~~[*] \\ \Rightarrow \log\left|\left(\frac{y-2}{y+1}\right)^4\right|=\log\left|c\left(\frac{x-1}{x+3}\right)^3\right|\\ \Rightarrow \left(\frac{y-2}{y+1}\right)^4=c\left(\frac{x-1}{x+3}\right)^3$

Note[*]: $~~12 \log c_1=\log c$

$~10.~\cos y\log(\sec x+\tan x)~dx\\~~~~=\cos x\log(\sec y+\tan y)~dy$

Sol. $~~\cos y\log(\sec x+\tan x)~dx\\~~~~=\cos x\log(\sec y+\tan y)~dy \\ \Rightarrow \frac{\log(\sec x+\tan x)}{\cos x}~dx=\frac{\log(\sec y+\tan y)}{\cos y}~dy\rightarrow(1)$

Let $~~I_1=\int{\log(\sec x+\tan x)\sec x}~dx\\=\log(\sec x+\tan x)\int{\sec x}~dx \\-\int{[\frac{d}{dx}\{\log(\sec x+\tan x)\}\int{\sec x}~dx}]dx\\=[\log(\sec x+\tan x)]^2 \\-\int{\frac{\sec x(\sec x+\tan x)}{(\sec x+\tan x)}\{\log(\sec x+\tan x)\}}~dx\\=[\log(\sec x+\tan x)]^2\\-\int{\sec x\log(\sec x+\tan x)}~dx\\=[\log(\sec x+\tan x)]^2-I_1\\ \Rightarrow 2I_1=[\log(\sec x+\tan x)]^2\\ \Rightarrow I_1=\frac 12[\log(\sec x+\tan x)]^2\rightarrow(2)$

Similarly, $~I_2=\int{\log(\sec y+\tan y)\sec y}~dy\\ \Rightarrow I_2=\frac 12[\log(\sec y+\tan y)]^2\rightarrow(3)$

From $~(1),(2),(3)~$ we get,

$~\frac 12[\log(\sec x+\tan x)]^2\\~~~~=\frac 12[\log(\sec y+\tan y)]^2+\frac 12 c\\ \Rightarrow [\log(\sec x+\tan x)]^2\\~~~~~=[\log(\sec y+\tan y)]^2+c~~\text{(ans.)}$

$~11.~x^2\frac{dy}{dx}=y^2-5y+6$

Sol. $~~x^2\frac{dy}{dx}=y^2-5y+6\\ \Rightarrow \frac{1}{y^2-5y+6}~dy=\frac{1}{x^2}~dx\\ \Rightarrow \frac{dy}{\left(y^2-2.y.\frac 52+\frac{25}{4}\right)+6-\frac{25}{4}}=\frac{dx}{x^2}\\ \Rightarrow \int{\frac{dy}{\left(y-\frac 52\right)^2-(1/2)^2}}=\int{\frac{dx}{x^2}}\\ \Rightarrow \frac{1}{2.\frac 12}\log\left|\frac{y-\frac 52-\frac 12}{y-\frac 52+\frac 12}\right|=\frac{x^{-2+1}}{-2+1}+c_1\\ \Rightarrow \log \left|\frac{2y-5-1}{2y-5+1}\right|=-\frac 1x+c_1\\ \Rightarrow \log\left|\frac{2(y-3)}{2(y-2)}\right|=-\frac 1x+c_1\\ \Rightarrow \log\left|\frac{y-3}{y-2}\right|=-\frac 1x+c_1\\ \Rightarrow \frac{y-3}{y-2}=e^{-1/x}.e^{c_1}\\ \Rightarrow (y-3)=c(y-2)e^{-1/x}\\~~\text{where}~~c=e^{c_1}$

$~12.~~x\sqrt{y^2-1}~dx-y\sqrt{x^2-1}~dy=0$

Sol. $~~x\sqrt{y^2-1}~dx-y\sqrt{x^2-1}~dy=0\\ \Rightarrow x\sqrt{y^2-1}~dx=y\sqrt{x^2-1}~dy\\ \Rightarrow \frac 12\int{\frac{2x}{\sqrt{x^2-1}}~dx}=\frac 12\int{\frac{2y}{\sqrt{y^2-1}}~dy}\\ \Rightarrow \frac 12\int{\frac{d(x^2-1)}{\sqrt{x^2-1}}}=\frac 12\int{\frac{d(y^2-1)}{\sqrt{y^2-1}}}\\ \Rightarrow \frac 12. 2\sqrt{x^2-1}=\frac 12.2\sqrt{y^2-1}+c\\ \Rightarrow \sqrt{x^2-1}-\sqrt{y^2-1}=c~~\text{(ans.)}$

$~13.~~y(1-x^2)~dy=x(1+y^2)~dx$

Sol. $~~y(1-x^2)~dy=x(1+y^2)~dx\\ \Rightarrow\frac{y}{1+y^2}~dy=\frac{x}{1-x^2}~dx \\ \Rightarrow \frac 12\int{\frac{2y}{1+y^2}~dy}=-\frac 12\int{\frac{-2x}{1-x^2}~dx}\\ \Rightarrow \frac 12 \int{\frac{d(y^2+1)}{y^2+1}}=-\frac 12 \int{\frac{d(1-x^2)}{1-x^2}}+c_1 \\ \Rightarrow \frac 12 \log|y^2+1|+\frac 12 \log|1-x^2|=c_1\\ \Rightarrow \log|(y^2+1)(1-x^2)|=2c_1\\ \Rightarrow (y^2+1)(1-x^2)=e^{c_1} \\ \Rightarrow (y^2+1)(1-x^2)=c\\~~~\text{where}~~c=e^{c_1}$

$~14.~\cos x(1+\cos y)~dx\\~~~~~-\sin y(1+\sin x)~dy=0$

Sol. $~~\cos x(1+\cos y)~dx\\~~~~~-\sin y(1+\sin x)~dy=0 \\ \Rightarrow  \cos x(1+\cos y)~dx=\sin y(1+\sin x)~dy\\ \Rightarrow \int{\frac{\cos x}{1+\sin x}~dx}=\int{\frac{\sin y}{1+\cos y}~dy}\\ \Rightarrow  \int{\frac{d(1+\sin x)}{(1+\sin x)}}=\int{\frac{-d(1+\cos y)}{(1+\cos y)}}\\ \Rightarrow  \log|1+\sin x|=-\log|1+\cos y|+\log c ~[*]\\ \Rightarrow  \log|(1+\sin x)|+\log|(1+\cos y)|=\log c \\ \Rightarrow \log|(1+\sin x)(1+\cos y)|=\log c \\ \Rightarrow |(1+\sin x)(1+\cos y)|=c$

Note[*] : $~~\log c\,$ being the constant of Integration.

$~15.~~(e^x+1)y~dy-(y^2+1)e^x~dx=0;~~\text{given}~y=0,~\text{when}~x=0.$

Sol. $~~(e^x+1)y~dy-(y^2+1)e^x~dx=0 \\ \Rightarrow  (e^x+1)y~dy=(y^2+1)e^x~dx\\ \Rightarrow  \frac{y}{y^2+1}~dy=\frac{e^x}{e^x+1}~dx \\ \Rightarrow  \frac 12\int{\frac{2y}{y^2+1}~dy}=\int{\frac{e^x}{e^x+1}~dx} \\ \Rightarrow \frac 12 \int{\frac{d(y^2+1)}{y^2+1}}=\int{\frac{d(e^x+1)}{e^x+1}}\\ \Rightarrow  \frac 12\log|(y^2+1)|=\log|(e^x+1)|+\log(c_1)\\ \Rightarrow  \log|(y^2+1)|=2\log|(e^x+1)|+2\log(c_1)\\ \Rightarrow  \log|y^2+1|=\log|e^x+1|^2+\log(c_1)^2\\ \Rightarrow \log|y^2+1|=\log|(e^x+1)c_1^2|\\ \Rightarrow  (y^2+1)=c_1^2(e^x+1)^2\rightarrow(1)$

For $~~y=0,~x=0,~$ we get by $\,(1),$

$~~0+1=c_1^2(e^0+1)^2 \\ \Rightarrow 1=4c_1^2 \\ \Rightarrow c_1^2=\frac 14$

Hence, putting the value of $\,c_1^2,$ we get by $\,(1),$

$~~(y^2+1)=\frac 14(e^x+1)^2 \\ \Rightarrow  4(y^2+1)=(e^x+1)^2~~\text{(ans.)}$