Differentiation (Part-38) | S N De

 $~16.~xy\frac{dy}{dx}=\frac{1+y^2}{1+x^2}(1+x+x^2)$

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

$~17.~~(e^y+1)\cos x~dx+e^y\sin x~dy=0$

Sol. $~~~(e^y+1)\cos x~dx+e^y\sin x~dy=0 \\ \Rightarrow \int{\frac{\cos x}{\sin x}~dx}=-\int{\frac{e^y}{e^y+1}~dy}\\ \Rightarrow \int{\cot x~dx}=-\int{\frac{d(e^y+1)}{e^y+1}}\\ \Rightarrow \log|\sin x|=-\log|e^y+1|+\log c\\ \Rightarrow \log|\sin x|+\log|e^y+1|=\log c\\ \Rightarrow  \log\left(|\sin x|(e^y+1)\right)=\log c\\ \Rightarrow  |\sin x|(e^y+1)=c~~\text{(ans.)}$

$~18.~~(x+2)~\frac{dy}{dx}=4x^2y$

Sol. $~~~(x+2)~\frac{dy}{dx}=4x^2y \\ \Rightarrow \frac 1y~dy=4 \cdot \frac{x^2}{x+2}~dx\\ \therefore\int{\frac{dy}{y}}=4\int{\frac{x^2-4+4}{x+2}}\\ \Rightarrow  \log|y|=4\int{\frac{x^2-2^2}{x+2}~dx}+16 \int{\frac{dx}{x+2}}\\ \Rightarrow \log|y|=4\int{\frac{(x+2)(x-2)}{(x+2)}~dx}\\~~~~~~~~~~~+16 \log|x+2|+c \\ \Rightarrow \log|y|=4\int{(x-2)~dx}+16 \log|x+2|+c\\ \Rightarrow  \log|y|=4\left[\frac{x^2}{2}-2x\right]+16\log|x+2|+c\\ \Rightarrow \log|y|=2x^2-8x+16\log|x+2|+c\\~~~c \rightarrow\text{constant of  integration.}$

$~19.~~\frac{dy}{dx}+\frac{y(y-1)}{x(x-1)}=0$

Sol. $~~~\frac{dy}{dx}+\frac{y(y-1)}{x(x-1)}=0\\ \Rightarrow \frac{dy}{dx}=-\frac{y(y-1)}{x(x-1)}\\ \Rightarrow \frac{1}{y(y-1)}~dy=-\frac{1}{x(x-1)}~dx\\ \therefore \int{\frac{dy}{y(y-1)}}=-\int{\frac{dx}{x(x-1)}}\\ \Rightarrow \int{\frac{dy}{y^2-y}}=-\int{\frac{dx}{x^2-x}}\\ \Rightarrow \int{\frac{dy}{\left(y^2-2.y.\frac 12+\frac 14\right)-\frac 14}}=-\int{\frac{dx}{\left(x^2-2.x.\frac 12+\frac 14\right)-\frac 14}}\\ \Rightarrow \int{\frac{dy}{\left(y-\frac 12\right)^2-\left(\frac 12\right)^2}}=-\int{\frac{d}{\left(x-\frac 12\right)^2-\left(\frac 12\right)^2}}\\ \Rightarrow \frac{1}{2 \times \frac 12}\cdot\log\left|\frac{y-\frac 12-\frac 12}{y-\frac 12+\frac 12}\right|=\frac{-1}{2 \times \frac 12}\cdot\log\left|\frac{x-\frac 12-\frac 12}{x-\frac 12+\frac 12}\right|\\~~~~~~~~+\log c\\ \Rightarrow \log\left|\frac{y-1}{y}\right|=-\log\left|\frac{x-1}{x}\right|+\log c \\ \Rightarrow \log\left|\frac{y-1}{y}\right|+\log\left|\frac{x-1}{x}\right|=\log c\\ \Rightarrow \log\left|\frac{(y-1)(x-1)}{yx}\right|=\log c \\ \Rightarrow \left|\frac{(x-1)(y-1)}{xy}\right|=c \\ \Rightarrow \{(x-1)(y-1)\}^2=(cxy)^2~~\text{(ans.)}$

$\,20.~~e^x\tan y~dx+(1-e^x)\sec^2y~dy=0$

Sol. $~~e^x\tan y~dx+(1-e^x)\sec^2y~dy=0\\ \Rightarrow e^x\tan y~dx=-(1-e^x)\sec^2y~dy \\ \Rightarrow e^x\tan y~dx=(e^x-1)\sec^2y~dy \\ \Rightarrow \frac{e^x}{e^x-1}~dx=\frac{\sec^2y}{\tan y}~dy\\ \therefore \int{\frac{e^x}{e^x-1}~dx}=\int{\frac{d(\tan y)}{\tan y}}\\ \Rightarrow \int{\frac{d(e^x-1)}{e^x-1}}=\int{\frac{d(\tan y)}{\tan y}}\\~~~\because~d(e^x-1)=e^x~dx,\\~~~~~~~~~d(\tan y)=\sec^2y~dy \\ \Rightarrow \log|e^x-1|=\log|\tan y|-\log c\\~~~\text{where}~~(-\log c) \rightarrow \text{constant of integration.}\\ \Rightarrow \log|-(1-e^x)|+\log c=\log|\tan y|\\ \Rightarrow \log|c(1-e^x)|=\log|\tan y|\\ \therefore ~ \tan y=c(1-e^x)~\text{(ans.)}$

$\,21.~y~dx+\sqrt{1-x^2}\sin^{-1}x~dy=0$

Sol. $~~~~~y~dx+\sqrt{1-x^2}\sin^{-1}x~dy=0 \\ \Rightarrow y~dx=-\sqrt{1-x^2}\sin^{-1}x~dy \\ \Rightarrow  \int{\frac{1}{\sqrt{1-x^2}}.\frac{1}{\sin^{-1}x}}~dx=-\int{\frac 1y~dy}\\ \Rightarrow  \int{\frac 1z~dz}=-\int{\frac 1y~dy}\\~~~\left[\because~\sin^{-1}x=z \\~~~~~~\Rightarrow \frac{1}{\sqrt{1-x^2}}dx=dz\right]\\ \Rightarrow \log|z|=-\log|y|+\log c\\~~~~\text{where}~~\log c\rightarrow \text{constant of integration.}\\ \Rightarrow  \log|z|+\log|y|=\log c \\ \Rightarrow \log|yz|=\log c \\ \Rightarrow \log|y\sin^{-1}x|=\log c \\ \Rightarrow |y\sin^{-1}x|=c~~\text{(ans.)}$

$\,22.~~(1+e^{2x})~dy+e^x(1+y^2)~dx=0\\~~\text{it being given that}~~y=1~~\text{when}~~x=0.$

Sol. $~~(1+e^{2x})~dy=-e^x(1+y^2)~dx\\ \Rightarrow \int{\frac{1}{1+y^2}~dy}=-\int{\frac{e^x}{e^{2x}+1}~dx}\\ \Rightarrow \tan^{-1}y=-\int{\frac{dz}{z^2+1}}\\~~~\left[\text{where}~~z=e^x ~~ \Rightarrow dz=e^x~dx\right] \\ \Rightarrow \tan^{-1}y=-\tan^{-1}z+c \\ \Rightarrow\tan^{-1}y=-\tan^{-1}(e^x)+c\rightarrow(1)$

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

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

$\,\, \tan^{-1}(1)=-\tan^{-1}(e^0)+c\\ \Rightarrow \frac{\pi}{4}=-\tan^{-1}(1)+c \\ \Rightarrow \frac{\pi}{4}=-\frac{\pi}{4}+c\\ \Rightarrow \frac{\pi}{4}+\frac{\pi}{4}=c\\ \Rightarrow c=\frac{\pi}{2}$

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

$\tan^{-1}(e^x)+\tan^{-1}y=\frac{\pi}{2}~~\text{(ans.)}$

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