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

Marks : 6

Year : 2013

$$(x^3y^4+ x^2y^3+xy^2+y)dx + (x^4y^3-x^3y^2-x^2y + x)dy = 0$$

$\therefore [x^2y^3(xy+1)+y(xy+1)]dx+[x^4y^3-x^3y^2-x^2y + x]dy = 0$

$\therefore (xy+1)(x^2y^3+y)dx + [x(x^3y^3+1)-x^2y(xy+1)]dy=0$

Expanding $x^3y^3+1$ as $(a^3+b^3)$

We get

$\therefore (xy+1)(x^2y^3+y)dx+[x(xy+1)(x^2y^2-xy+1)-x^2y(xy+1)]dy=0$

Dividing by $(xy+1)$

we get

$y(x^2y^2+1)dx + x (x^2y^2-2xy+1)dy = 0 -----(1)$

$\therefore M=y(x^2y^2+1) \space \& \space N=(x^2y^2-2xy+1)x$

Comparing it with

$Mdx + Ndy =0$

$Mx-Ny=xy(x^2y^2+1)-xy(x^2y^2-2xy+1)$

$=xy(x^2y^2+1-x^2y^2+2xy-1)$

$=xy(2xy)$

$=2x^2y^2 \neq 0$

$\therefore I.f =\dfrac 1{Mx-Ny}=\dfrac 1{2x^2y^2}$

Multiplying eq(1) by I.f we get

$\dfrac {(x^2y^2+1)y}{2x^2y^2}dx + x\Bigg(\dfrac {x^2y^2-2xy+1}{2x^2y^2}\Bigg)dy=0$

$\therefore \Bigg(\dfrac y2+\dfrac 1{2x^2y}\Bigg)dx+\Bigg(\dfrac x2-\dfrac 1y+\dfrac 1{2xy^2}\Bigg)dy=0$

Which is exact

$\int M \space dx=\int\dfrac y2+\dfrac 1{2y}x^{-2}$

$=\dfrac {xy}2+\dfrac 1{2y}\dfrac {x^{-1}}{-1}$

$\int $ (N neglecting x)dy=$\int \dfrac {-1}y dy=-\log y$

$\therefore$ complete solution is

$\dfrac {xy}2-\dfrac 1{2xy}-\log y +c =0$