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

Marks : 4

Year : 2015

Comparing it with $Mdx+Ndy=0$ we get,

$M=1+e^{\dfrac xy}$

$N=e^{\dfrac xy}(1-\dfrac xy)$

$\dfrac {\partial M}{\partial y}=e^{\dfrac xy}\Bigg(\dfrac {-x}{y^2}\Bigg)$

$\dfrac {\partial N}{\partial x}=e^{\dfrac xy}\times \dfrac {\partial}{\partial x}\Bigg(1-\dfrac xy\Bigg) + \Bigg(1-\dfrac xy\Bigg)\dfrac {\partial}{\partial x}e^{\dfrac xy} $

$\therefore\dfrac {\partial N}{\partial x}=e^{\dfrac xy}\times \Bigg(\dfrac {-1}y\Bigg) + \Bigg(1-\dfrac xy\Bigg)e^{\dfrac xy} \times\dfrac 1y$

$=e^{\dfrac xy}\Bigg(1-\dfrac xy\Bigg)\dfrac 1y-\dfrac 1ye^{\dfrac xy}$

$=e^{\dfrac xy}\Bigg[\dfrac 1y-\dfrac x{y^2}-\dfrac 1y\Bigg]$

$=\dfrac {-x}{y^2}e^{x/y}$

Now $\dfrac {\partial M}{\partial y}=\dfrac {\partial N}{\partial x},$ the equation is exact

$\therefore \int Mdx=\int 1+e^{x/y}dx=x+ ye^{x/y} + c$

$\int$ N free from x=0

$\therefore $ The solution is $x+ye^{x/y} + c$