Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2015

$$\dfrac {dy}{dx}=\dfrac {y+1}{(y+2)e^y-x}$$ $$ \therefore [(y+2)e^y-x]dy=(y+1)dx$$

Comparing it with

$Mdx + Ndy = 0$ we get

$M=(y+1)\space and \space N= -[(y+2)e^y-x]\\ \dfrac {\partial M}{\partial y}=1 \hspace {1cm} \dfrac {\partial N}{\partial x}=-[0+0-1]\\ \dfrac {\partial M}{\partial y}=1 \hspace {1cm} \dfrac {\partial N}{\partial x}=1\\ \therefore \dfrac {\partial M}{\partial y}=\dfrac {\partial N}{\partial x}$

Which is exact

$\therefore \text{Solution is } \\ \int Mdx=\int (y+1)dx\\ =xy+x........ (1)\\ \int (N free from x) dy=\int -(y+2)e^ydy$

Integration by parts:

$=-\Bigg[(y+2)\int e^ydy-\int \Bigg(\dfrac d{dy}(y+2)\int e^ydy\Bigg)dy\Bigg]\\ =-\Big[(y+2)e^y-\int e^ydy\Big]\\ =-\Big[(y+2)e^y- e^y\Big]\\ =e^y[1-y-2]\\ =-e^y(-y+1).... (2)\\ \therefore \text{Solution is }\\ \text{Adding 1 & 2} \\ x(y+1)-e^y(y+1)=c\\ \therefore (y+1)(x-e^y)=c$