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

Marks : 3

Year : 2014

The given diff. equation is of type $Mdx+Ndy=0$

$$\therefore M=x\sqrt{x^2+y^2}-y \space \&\space N=y\sqrt{x^2+y^2}-x$$

$\dfrac {\partial M}{\partial y}=\dfrac x{2\sqrt{x^2+y^2}}.2y-1=\dfrac {xy}{\sqrt{x^2+y^2}}-1\\ \dfrac {\partial N}{\partial x}=\dfrac y{2\sqrt{x^2+y^2}}.2x-1=\dfrac {xy}{\sqrt{x^2+y^2}}-1\\ \dfrac {\partial M}{\partial y}=\dfrac {\partial N}{\partial x} \text{ which is exact }\\ \therefore \int Mdx=\int(x\sqrt{x^2+y^2}-y)dx\\ Let \space U=x^2+y^2\space diff. \space w.r.t.x\\ \therefore dU=2xdx\\ \therefore \dfrac {dU}2=xdx\\ \text{ Substituting value of U }\\ \therefore \dfrac {(x^2+y^2)^{3/2}}3-yx=c\\ \int \text{ N free from x dy}=\int 0dw=0\\ \therefore \text{ Solution is }\\ \dfrac {(x^2+y^2)^{3/2}}3-yx=c\\ (x^2+y^2)^{3/2} -3yx=3c$