Differentiation (Part-38) | S N De

 

$~23.~~\frac{dy}{dx}=\frac{\cos(\log x)}{\log y}$

Sol. $~~~~\frac{dy}{dx}=\frac{\cos(\log x)}{\log y}\\ \Rightarrow \log y ~dy=\cos(\log x)~dx\\ \therefore \int{\log y~dy}=\int{\cos(\log x)~dx} \\ \Rightarrow \log y\int{dy}-\int{\left[\left\{\frac{d}{dy}(\log y)\right\}\int{dy}\right]dy}\\=\frac 12[x\cos(\log x)+x\sin(\log x)]\\ \Rightarrow y\log y-\int{\frac 1y \cdot y~dy }\\~~~=\frac 12[x\cos(\log x)+x\sin(\log x)]+c_1\\ \Rightarrow 2(y\log y-y)\\~~~=x[\cos(\log x)+\sin(\log x)]+c\\ \Rightarrow 2y(\log y-1)=x[\cos(\log x)+\sin(\log x)]+c\\~~~[\text{where}~~c=2c_1(c_1 \rightarrow \text{const. of integration.})]$

Note : 

$~24.~~\left(y-x~\frac{dy}{dx}\right)=a\left(y^2+\frac{dy}{dx}\right)$

Sol. $~~~\left(y-x~\frac{dy}{dx}\right)=a\left(y^2+\frac{dy}{dx}\right)\\ \Rightarrow (x+a)~\frac{dy}{dx}=y-ay^2 \\ \Rightarrow  \frac{dy}{y(1-ay)}=\frac{dx}{x+a}\\ \therefore \int{\frac{dy}{y(1-ay)}}=\int{\frac{dx}{x+a}}\\ \Rightarrow \int{\frac{1-ay+ay}{y(1-ay)}~dy}=\int{\frac{dx}{x+a}}\\ \Rightarrow \int{\frac{dy}{y}}+\int{\frac{a~dy}{1-ay}}=\int{\frac{dx}{x+a}}\\ \Rightarrow \log|y|-\log|1-ay|\\~~~~~~~~=\log|x+a|-\log c\\~~~[~~-\log c\rightarrow\text{const. of integration.}]\\ \Rightarrow \log|y|+\log c=\log|(x+a)(1-ay)|\\ \Rightarrow \log|cy|=\log|(x+a)(1-ay)|\\ \Rightarrow  c^2y^2=[(x+a)(1-ay)]^2~~~\text{(ans.)}$

$~25.~~\frac{dy}{dx}+\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}=0$

Sol. $~~~\frac{dy}{dx}+\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}=0 \\ \Rightarrow \frac{dy}{dx}=-\frac{\sqrt{(x^2-1)(y^2-1)}}{xy}\\ \Rightarrow \frac 12 \int{\frac{2y}{\sqrt{y^2-1}}~dy}=-\int{\frac{\sqrt{x^2-1}}{x}~dx}\rightarrow(1)$

Let $\,I_1=\int{\frac{2y}{\sqrt{y^2-1}}~dy}\\~~~~=\int{\frac{d(y^2-1)}{\sqrt{y^2-1}}}\\~~~~=\int{\frac{dz}{\sqrt z}}~~[\text{where}~~z=y^2-1]\\~~~~=\int{z^{-1/2}~dz}\\~~~~=\frac{z^{-1/2+1}}{-1/2+1}\\~~~~=2\sqrt z\\~~~~=2\sqrt{y^2-1}\rightarrow(2)$

Let $~~I_2=\int{\frac{\sqrt{x^2-1}}{x}~dx}\\~~~~=\int{\frac{\sqrt{x^2-1}}{x^2}~x~dx}\rightarrow(3)$

Suppose, $~~~x^2-1=z^2\\ \Rightarrow 2x~dx=2z~dz\\ \Rightarrow x~dx=z~dz$

Then from $\,(3),~$ we get,

$~~I_2=\int{\frac{z. z~dz}{z^2+1}}\\~~~~=\int{\frac{z^2}{z^2+1}~dz}\\~~~~=\int{\left(1-\frac{1}{z^2+1}\right)~dz}\\~~~~=\int{dz}-\int{\frac{dz}{z^2+1}}\\~~~~=z-\tan^{-1}z\\~~~~=\sqrt{x^2-1}-\tan^{-1}(\sqrt{x^2-1})\rightarrow(4)$

Hence, by $\,(1),(2),(4)~~$ we get,

$~~\frac 12.2\sqrt{y^2-1}\\~~~~=-[\sqrt{x^2-1}-\tan^{-1}(\sqrt{x^2-1})]+c\\ \Rightarrow \sqrt{x^2-1}+\sqrt{y^2-1}-\sec^{-1}(x)=c\\~~~[\because ~~\tan^{-1}\sqrt{x^2-1}=\sec^{-1}(x)]$

$\,26.~~$ Show that, the general solution of the differential equation $~~\sqrt{1-x^2}~dy+\sqrt{1-y^2}~dx=0~$ is $~~x\sqrt{1-y^2}+y\sqrt{1-x^2}=c,~$ where $\,c\,$ is an arbitrary constant.

Sol. $~~\sqrt{1-x^2}~dy+\sqrt{1-y^2}~dx=0 \\ \Rightarrow \sqrt{1-x^2}~dy=-\sqrt{1-y^2}~dx\\ \Rightarrow \int{\frac{dy}{\sqrt{1-y^2}}}=-\int{\frac{dx}{\sqrt{1-x^2}}}\\ \Rightarrow \sin^{-1}y=-\sin^{-1}x+c_1\\ \Rightarrow \sin^{-1}x+\sin^{-1}y=c_1\\ \Rightarrow\sin^{-1}[x\sqrt{1-y^2}+y\sqrt{1-x^2}]=c_1\\ \Rightarrow x\sqrt{1-y^2}+y\sqrt{1-x^2}=\sin(c_1)\\ \Rightarrow x\sqrt{1-y^2}+y\sqrt{1-x^2}=c~~\text{(showed)}$

Find the particular solution of each of the following equations: 

$~27.~\tan x~\frac{dy}{dx}=1+y^2;\\~~\text{given}~y=1,~~\text{when}~x=\frac{\pi}{2}.$

Sol. $~~\tan x~\frac{dy}{dx}=1+y^2 \\ \Rightarrow \frac{dy}{1+y^2}=\frac{1}{\tan x}~ dx\\ \therefore \int{\frac{dy}{1+y^2}}=\int{\cot x~dx}\\ \Rightarrow \tan^{-1}y=\log|\sin x|+c\rightarrow(1)\\ \therefore \tan^{-1}(1)=\log|\sin(\pi/2)|+c\\ \Rightarrow \frac{\pi}{4}=\log 1+c \\ \Rightarrow c= \frac{\pi}{4}\rightarrow(2)$

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

$~~\tan^{-1}y-\log|\sin x|=\frac{\pi}{4}.$

$~28.~~(1-x^2)~\frac{dy}{dx}=2y,\\~~~\text{given}~~~y=1,\text{when}~~x=2.$

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

Now, $~~\text{given}~y=1,\text{when}~x=2.$

Hence, from $\,(1),~$ we get, 

$~~~~~1^2=\left(\frac{1+2}{1-2}\right)^2c\\ \Rightarrow 1=9c \\ \Rightarrow c=\frac 19.$

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

$~~y^2=\frac 19 \left(\frac{1+x}{1-x}\right)^2~~\text{(ans.)}$

$~29.~~\frac{dy}{dx}=\frac{1+y^2}{xy};\\~~~~\text{given}~y=0,~~\text{when}~x=1.$

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

Now, $~~~\text{given}~~y=0,~~\text{when}~x=1.$

So, from $\,(1)~$ we get, 

$~~~~1+0^2=1^2c \\ \Rightarrow c=1.$

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

$~~1+y^2=x^2~~\text{(ans.)}$

$~30.~~y-x~\frac{dy}{dx}=2\left(1+x^2~\frac{dy}{dx}\right); \\~~~\text{given}~~y=1,~\text{when}~~~x=1.$

Sol. $~~~~y-x~\frac{dy}{dx}=2\left(1+x^2~\frac{dy}{dx}\right)\\ \Rightarrow y-x~\frac{dy}{dx}=2+2x^2~\frac{dy}{dx} \\ \Rightarrow y~dx-x~dy=2~dx+2x^2~dy \\ \Rightarrow (y-2)~dx=(2x^2+x)~dy \\ \Rightarrow \frac{1}{2x^2+x}~dx=\frac{1}{y-2}~dy \\ \Rightarrow \int{\frac{(2x+1)-2x}{x(2x+1)}~dx}=\int{\frac{1}{y-2}~dy} \\ \Rightarrow \int{\frac{dx}{x}}-\int{\frac{2}{2x+1}~dx}=\int{\frac{dy}{y-2}}\\ \Rightarrow\log|x|-\log|2x+1|\\~~~~=\log|y-2|+\log c~~~[*]\\ \Rightarrow\log|x|-\log c=\log|2x+1|+\log|y-2|\\ \Rightarrow \log \left|\frac xc\right|=\log|(2x+1)(y-2)| \\ \therefore (2x+1)^2(y-2)^2=\frac{x^2}{c^2}\rightarrow(1)$

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

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

$~~~(2\times 1+1)^2 (1-2)^2=\frac{1^2}{c^2}\\ \Rightarrow ~9=\frac{1}{c^2} \\ \Rightarrow c^2=\frac 19.$

Hence, putting the value of $\,c,\,$ we get,

$~~(2x+1)^2(y-2)^2=9x^2~~\text{(ans.)}$

Note [*] : $~~\log c \rightarrow \text{constant of integration.}$