Differentiation (Part-38) | S N De

 

Solve :

$\,20.~~\log \left(\frac{dy}{dx}\right)=3x-5y$

Sol. $~~~~\log \left(\frac{dy}{dx}\right)=3x-5y\\ \Rightarrow \frac{dy}{dx}=e^{3x-5y}\\ \Rightarrow e^{5y}~dy=e^{3x}~dx\\~~~\text{Integrating we get,}\\~~~\int{e^{5y}~dy}=\int{e^{3x}~dx}+c_1,\\ ~~c_1 \rightarrow\text{Constant of Integration}\\ \Rightarrow\frac{e^{5y}}{5}=\frac{e^{3x}}{3}+c_1\\ \Rightarrow 3e^{5y}=5e^{3x}+15c_1\\ \Rightarrow 3e^{5y}-5e^{3x}=c.\text{(ans.)}\\~~[~~c=15c_1]$

$\,21.~~(x-y)(dx+dy)=dx-dy;\\~~~~ \text{given}~~y=-1,\text{when}~~x=0~~\text{(NCERT)}$

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

Now, $\,y(0)=-1~~\text{(given)}\\ \therefore 0-1=\log 1+c\\ \Rightarrow c=-1$

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

$~~x+y=\log|x-y|-1\\ \Rightarrow \log|x-y|=x+y+1~~\text{(ans.)}$

$\,22.~~y^2(x~dy+y~dx)+x~dy-y~dx=0$

Sol. $~~y^2(x~dy+y~dx)+x~dy-y~dx=0\\ \Rightarrow (x~dy+y~dx)+\frac{x~dy-y~dx}{y^2}=0\\ \Rightarrow d(xy)-\frac{y~dx-x~dy}{y^2}=0\\ \Rightarrow d(xy)-d\left(\frac{x}{y}\right)=0\\~~~\text{Integrating we get,}\\~~\int{d(xy)}-\int{d\left(\frac xy\right)}=c\\ \Rightarrow xy-\frac xy=c\\ ~~~c\rightarrow \text{Constant of Integration.}$

$~~23.~~(ax+hy+g)dx+(hx+by+f)dy=0$

Sol. $~(ax+hy+g)dx+(hx+by+f)dy=0\\ \Rightarrow h(y~dx+x~dy)+(ax+g)dx\\~~~~~~+(by+f)dy=0\\ \Rightarrow h~d(xy)+(ax+g)~dx+(by+f)~dy=0\\ ~~\text{Integrating we get,}\\~~~h \int{d(xy)}+\int{(ax+g)dx}\\~~~~~~~+\int{(by+f)~dy}=c_1\\ \Rightarrow hxy+a\cdot\frac{x^2}{2}+gx+b\cdot \frac{y^2}{2}+fy=c_1\\ \Rightarrow 2hxy+ax^2+2gx+by^2+2fy=2c_1\\ \Rightarrow ax^2+2hxy+by^2+2gx+2fy=c\\~~~c ~(=2c_1)\rightarrow \text{Constant of Integration.}$

Long Answer Type Questions :

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

Sol. $~~~(x^2-yx^2)\frac{dy}{dx}+y^2+xy^2=0\\ \Rightarrow x^2(1-y)~\frac{dy}{dx}+y^2(1+x)=0\\ \Rightarrow x^2(1-y)~dy=-y^2(x+1)~dx\\ \Rightarrow \frac{1-y}{y^2}~dy=-\frac{x+1}{x^2}~dx\\ \Rightarrow \int\frac{1}{y^2}~dy-\int{\frac 1y}~dy=-\int{\frac 1x}~dx-\int{\frac{1}{x^2}}~dx+c\\ \Rightarrow \frac{y^{-2+1}}{-2+1}-\log|y|=-\log|x|-\frac{x^{-2+1}}{-2+1}+c\\ \Rightarrow -\frac 1y-\log|y|=-\log|x|+\frac 1x+c\\ \Rightarrow \log|x|-\log|y|=\frac 1x+\frac 1y+c\\ \Rightarrow \log\left|\frac xy\right|=\frac 1x+\frac 1y+c\\~c \rightarrow \text{Constant of Integration.}$

$\,2.~~\frac{dy}{dx}=1+x+y+xy$

Sol. $~~\frac{dy}{dx}=1+x+y+xy\\ \Rightarrow \frac{dy}{dx}=(1+x)+y(1+x)\\ \Rightarrow \frac{dy}{dx}=(1+x)(1+y)\\ \Rightarrow \frac{dy}{1+y}=(1+x)dx\\ \therefore~~\int{\frac{dy}{1+y}}=\int{(1+x)dx}\\ \Rightarrow \log(1+y)=\int{dx}+\int{x~dx}+c_1\\ \Rightarrow \log(1+y)=x+\frac{x^2}{2}+c_1\\ \Rightarrow 2\log(1+y)=2x+x^2+2c_1\\ \Rightarrow \log(1+y)^2=(x^2+2x+1)+(2c_1-1)\\ \Rightarrow \log(1+y)^2=(1+x)^2+c_2~~[*]\\ \Rightarrow (1+y)^2=ce^{(1+x)^2}~~[~~c=e^{c_2}]$

Note[*]: $~~c_2=2c_1-1=\text{Constant of Integration}$

$~~3.~~(1+y^2)(1+\log x)~dx+x~dy=0;\\~~~\text{given}~y=1~~\text{when}~~x=1~~[\text{CBSE}-2000,'03]$

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

For $~~x=1,~~y=1~~\text{(given)}$

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

$~~\frac{(1+\log 1)^2}{2}+\tan^{-1}(1)=c\\ \Rightarrow\frac 12+\frac{\pi}{4}=c.$

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

$~~\frac{(1+\log x)^2}{2}+\tan^{-1}(y)=\frac 12+\frac{\pi}{4}\\ \Rightarrow \frac 12[1+2\log x+(\log x)^2]\\~~~~+\tan^{-1}y=\frac 12+\frac{\pi}{4}\\ \Rightarrow \frac 12+\log x+\frac{(\log x)^2}{2}+\tan^{-1}(y)=\frac 12+\frac{\pi}{4}\\ \Rightarrow \log x+\frac 12 (\log x)^2+\tan^{-1}(y)=\frac{\pi}{4}~~\text{(ans.)}$

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

Sol. $~~~~~\sin x\frac{dy}{dx}+y=y^2\\ \Rightarrow \sin x~dy+y~dx=y^2~dx\\ \Rightarrow \sin x~dy=(y^2-y)~dx\\ \\ \Rightarrow  \frac{dy}{y^2-y}=\frac{dx}{\sin x}\\ \therefore \int{\frac{dy}{y^2-2.y.\frac 12+\frac 14-\frac 14}}=\int{\csc x~dx}\\~~~[\text{Here by csc x, we mean cosec x}]\\ \Rightarrow \int{\frac{dy}{(y-1/2)^2-(1/2)^2}}=\log|\csc x-\cot x|+c_1 \\ \Rightarrow \log\left|\frac{y-1/2-1/2}{y-1/2+1/2}\right|=\log|c(\csc x-\cot x)|\\~~~[~~\text{Here,}~~c_1=\log c]\\  \Rightarrow  \log \left|\frac{y-1}{y}\right|=\log|c\tan(x/2)|~~[*]\\  \Rightarrow (y-1)^2=c^2y^2\tan^2(x/2)~~\text{(ans.)}$

Note[*] : $~~\csc x-\cot x\\=\frac{1}{\sin x}-\frac{\cos x}{\sin x}\\=\frac{1-\cos x}{\sin x}\\=\frac{2\sin^2(x/2)}{2\sin (x/2)\cos(x/2)}\\=\frac{\sin(x/2)}{\cos(x/2)}\\=\tan(x/2)$

$~5.~~y~dy+xe^x\cos^2y~dx=0$

Sol. $~~~y~dy+xe^x\cos^2y~dx=0 \\ \Rightarrow y~dy=-xe^x \cos^2y~dx \\ \Rightarrow \int{y\sec^2y~dy}=-\int{xe^x~dx} \\ \Rightarrow  y\int{\sec^2y~dy}-\int{[\frac{d}{dy}(y)\int{\sec^2y~dy}]dy}\\=-\left[x\int{e^x~dx}-\int{\{\frac{d}{dx}(x)\int{e^x~dx}\}dx}\right] \\ \Rightarrow y \tan y-\int{1.\tan y~dy}\\=-[x.e^x-\int{1.e^x~dx}]\\ \Rightarrow y\tan y-\log|\sec y|\\=-xe^x+e^x+c\\=-e^x(x-1)+c \\ \Rightarrow y\tan y-\log|\sec y|+e^x(x-1)=c$