Differentiation (Part-38) | S N De

29. $x\left(\frac{dy}{dx}+y\right)=1-y$

Sol.

$~~~x\left(\frac{dy}{dx}+y\right)=1-y$

$\Rightarrow \frac{dy}{dx}+y=\frac 1x-\frac yx$

$\Rightarrow \frac{dy}{dx}+\left(1+\frac 1x\right)y=\frac 1x \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\left(1+\frac 1x\right)~dx}}$

$=e^{x+\log_e x}$

$=e^x \cdot e^{\log_e x}$

$=xe^{x}$

Now, multiplying both sides of $\,(1)\,$ by $~xe^x~$ we get,

$~~xe^x~\frac{dy}{dx}+(xe^x+e^x)y=e^x$

$\Rightarrow \frac{d}{dx}(yxe^x)=e^x$

$\therefore yxe^x=\displaystyle\int{e^x~dx}$

$\Rightarrow yxe^x=e^x+c$

$\Rightarrow xy=1+ce^{-x}~~\text{(ans.)}$

30(i). $~(x+2y^3)~\frac{dy}{dx}=y$

Sol.

$~~~(x+2y^3)~\frac{dy}{dx}=y$

$\Rightarrow \frac{dx}{dy}=\frac{x+2y^3}{y}$

$\Rightarrow \frac{dx}{dy}-\frac 1y \cdot x=2y^2 \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\left(-\frac 1y\right)~dy}}$

$=e^{-\log_e y}$

$=e^{\log (y)^{-1}}$

$=y^{-1} = \frac 1y$

Now, multiplying both sides of $\,(1)\,$ by $~\frac 1y~$ we get,

$~~\frac 1y \cdot \frac{dx}{dy}-\frac{1}{y^2} \cdot x=2y$

$\text{or,}~~ \frac{d}{dy} \left(\frac 1y \cdot x\right)=2y$

$\therefore \frac xy=\displaystyle\int{2y~dy}$

$\Rightarrow \frac xy=2 \cdot \frac{y^2}{2}+c$

$\Rightarrow \frac xy=y^2+c$

$\therefore x=y^3+cy~~\text{(ans.)}$

30(ii). $~y~dx-(x+2y^2)~dy=0$

Sol.

$~~~y~dx-(x+2y^2)~dy=0$

$\Rightarrow \frac{dx}{dy}-\frac 1y \cdot x=2y \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\left(-\frac 1y\right)~dy}} = e^{-\log_e y} = y^{-1} = \frac 1y$

Now, multiplying both sides of $\,(1)\,$ by $~\frac 1y~$ we get,

$~~\frac 1y \cdot \frac{dx}{dy}-\frac{1}{y^2}\cdot x=2$

$\therefore \frac{d}{dy} \left(\frac 1y \cdot x\right)=2$

$\Rightarrow \frac xy=\int{2~dy}$

$\Rightarrow \frac xy=2y+c$

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

31. $~~1+y^2+\left(x-e^{\tan^{-1}y}\right)\frac{dy}{dx}=0$

Sol.

$~~~1+y^2+\left(x-e^{\tan^{-1}y}\right)\frac{dy}{dx}=0$

$\Rightarrow \frac{dx}{dy}(1+y^2)+ x=e^{\tan^{-1}y}$

$\Rightarrow \frac{dx}{dy}+\frac{1}{1+y^2} \cdot x=\frac{e^{\tan^{-1}y}}{1+y^2} \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\frac{dy}{1+y^2}}}=e^{\tan^{-1}y}$

Now, multiplying both sides of $\,(1)\,$ by $~e^{\tan^{-1}y}~$ we get,

$~~e^{\tan^{-1}y}\cdot \frac{dx}{dy}+\frac{e^{\tan^{-1}y}}{1+y^2} \cdot x=\frac{(e^{\tan^{-1}y})^2}{1+y^2}$

$\Rightarrow \frac{d}{dy} \left(xe^{\tan^{-1}y}\right)=\frac{(e^{\tan^{-1}y})^2}{1+y^2}$

$\therefore xe^{\tan^{-1}y}=\displaystyle\int{\frac{(e^{\tan^{-1}y})^2}{1+y^2}~dy}$

$\Rightarrow xe^{\tan^{-1}y}=\displaystyle \int{e^{2\tan^{-1}y}~d(\tan^{-1}y)}$

$\Rightarrow xe^{\tan^{-1}y}=\frac 12 e^{2\tan^{-1}y}+k$

$\Rightarrow 2 xe^{\tan^{-1}y}=e^{2\tan^{-1}y}+c~~\text{(ans.)}\\~~~[~\text{where}~~c=2k]$

32. $~~1+y^2+\left(x-e^{-\tan^{-1}y}\right)\frac{dy}{dx}=0$

Sol.

$~~1+y^2+\left(x-e^{-\tan^{-1}y}\right)\frac{dy}{dx}=0$

$\Rightarrow (1+y^2)\frac{dx}{dy}+x= e^{-\tan^{-1}y}$

$\Rightarrow \frac{dx}{dy}+\frac{1}{1+y^2} \cdot x=\frac{e^{-\tan^{-1}y}}{1+y^2}\rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\frac{1}{1+y^2}~dy}}=e^{\tan^{-1}y}$

Now, multiplying both sides of $\,(1)\,$ by $~e^{\tan^{-1}y}~$ we get,

$~~e^{\tan^{-1}y}~\frac{dx}{dy}+\frac{e^{\tan^{-1}y}}{1+y^2} \cdot x=\frac{1}{1+y^2}$

$\Rightarrow \frac{d}{dy}\left(xe^{\tan^{-1}y}\right)=\frac{1}{1+y^2}$

$\therefore xe^{\tan^{-1}y}=\displaystyle\int{\frac{dy}{1+y^2}}$

$\Rightarrow xe^{\tan^{-1}y}=e^{\tan^{-1}y}+c~~\text{(ans.)}$

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

Sol.

$~~~y^2+\left(x-\frac 1y\right)\frac{dy}{dx}=0$

$\Rightarrow y^2~\frac{dx}{dy}+x=\frac 1y$

$\Rightarrow \frac{dx}{dy}+\frac{1}{y^2} \cdot x=\frac{1}{y^3} \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\displaystyle\int{\frac{dy}{y^2}}}=e^{-1/y}$

Now, multiplying both sides of $\,(1)\,$ by $~e^{-1/y}~$ we get,

$~~e^{-1/y} \cdot \frac{dx}{dy} +\frac{e^{-1/y}}{y^2} \cdot x= \frac{e^{-1/y}}{y^3}$

$\Rightarrow \frac{d}{dy} \left(x e^{-1/y}\right)=\frac{e^{-1/y}}{y^3}$

$\therefore x e^{-1/y}=\displaystyle\int{\frac{1}{y^3}~e^{-1/y}~dy}\rightarrow(2)$

Now, let $~~\frac 1y=z \Rightarrow -\frac{1}{y^2}~dy=dz$

Hence from $\,(2)\,$ we get,

$~~~x e^{-1/y}=-\displaystyle\int{ze^{-z}~dz}$

$\Rightarrow xe^{-1/y}=-\left(-ze^{-z}+\displaystyle\int{e^{-z}~dz}\right)$

$\Rightarrow xe^{-1/y}=-\left(-ze^{-z}-e^{-z}\right)$

$\Rightarrow xe^{-1/y}=ze^{-z}+e^{-z}+c$

$\Rightarrow xe^{-1/y}=\frac 1ye^{-1/y}+e^{-1/y}+c$

$\Rightarrow x=\frac 1y+1+ce^{1/y}$

$\therefore ~x-\frac 1y-1=ce^{1/y}~~~\text{(ans.)}$

34. $~~dx+x~dy=e^{-y}~dy$

Sol.

$~~~dx+x~dy=e^{-y}~dy \implies \frac{dx}{dy}+x=e^{-y}\sec^2y \rightarrow(1)$

The integrating factor (I.F.) of $\,(1)$

$=e^{\int{dy}}=e^y$

Now, multiplying both sides of $\,(1)\,$ by $~e^y~$ we get,

$~~e^y~\frac{dx}{dy}+xe^y=\sec^2y \Rightarrow \frac{d}{dy}(xe^y)=\sec^2y$

$\therefore xe^y=\displaystyle\int{\sec^2y ~dy}$

$\text{or,}~~xe^y=\tan y+c~~\text{(ans.)}$

35. $~~(x^2y^3+2xy)~dy=dx$

Sol.

$~~~(x^2y^3+2xy)~dy=dx$

$\Rightarrow \frac{dx}{dy}=x^2y^3+2xy$

$\Rightarrow \frac{1}{x^2}\cdot \frac{dx}{dy}-\frac 2x~y=y^3 \rightarrow(1)$

Now, let $~-\frac 1x=z \Rightarrow \frac{1}{x^2}~\frac{dx}{dy}=\frac{dz}{dy}$

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

$~~\frac{dz}{dy}+2yz=y^3 \rightarrow(2)$

The integrating factor (I.F.) of $\,(2)$

$=e^{\int{2y~dy}}=e^{2 \cdot \frac{y^2}{2}}=e^{y^2}$

Now, multiplying both sides of $\,(2)\,$ by $~e^{y^2}~$ we get,

$~~~e^{y^2}~\frac{dz}{dy}+2ye^{y^2}z=y^3e^{y^2} \Rightarrow \frac{d}{dz}\left(ze^{y^2}\right)=y^3e^{y^2}$

$\therefore ze^{y^2}=\displaystyle\int{y^3e^{y^2}~dy}\rightarrow(3)$

Let $~~y^2=v \Rightarrow y~dy=\frac 12~dv\rightarrow(4)$

So, from $\,(3)\,$ and $\,(4)\,$ we get,

$~~ -\frac 1x~e^{y^2}=\frac 12\displaystyle \int{ve^v~dv}$

$\Rightarrow -\frac 1x e^{y^2}=\frac 12(ve^v-e^v)+k$

$\Rightarrow -\frac 1xe^{y^2}=\frac 12(y^2-1) e^{y^2}+k$

$\therefore \frac 1x=\frac 12(1-y^2)+ce^{-y^2}~~\text{(ans.)}\\~~~~[~\text{where}~~c=-k~]$