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

Marks : 6

Year : 2014

Comparing it with $Mdx + Ndy = 0$

We get $M = xy^2 – e^{1/x^3}$ and $N = -x^2y$

$\dfrac {\partial M}{\partial y}=2xy\space \space \space \dfrac {\partial N}{\partial x}=-2xy$

$\therefore\dfrac {\partial M}{\partial y}-\dfrac {\partial N}{\partial x}/N=\dfrac {2xy-(-2xy)}{-x^2y}$

$=\dfrac {4xy}{-x^2y}=\dfrac {-4}x=f(x)$

$I.F=e^{\int f(x)dx}$

$=e^{\int -4/x \space dx}=e^{-4\log x}=e^{\log \dfrac 1{x^4}}=\dfrac 1{x^4}$

Multiplying the given equation by $\dfrac 1{x^4}$ we get

$\dfrac {xy^2-e^{1/x^3}}{x^4}dx-\dfrac {x^2y}{x^4}dy=0$ which is exact

$\therefore \int M \space \space dx=\int \Bigg(\dfrac {y^2}{x^3}dx-\dfrac {e^{1/x^2}}{x^4}\Bigg) dx$

Let $I_1=\int \dfrac{y^2}{x^3} -dx=\dfrac {-y^2}{2x^2} $

$I_2=\int\dfrac {e^{1/x^3}}{x^4}dx$

Put $\dfrac 1{x^3}=t$

$\therefore \dfrac {-3}{x^4}dx=dt$

$I_2=\int \dfrac {-3e^{1/x^3}}{-3x^4}dx$

$=\dfrac 1{-3}\int e^t.dt$

$=\dfrac {-1}3e^t$

Resubstituting value of t we get

$I_2=\dfrac {-1}3e^{1/x^3}$

$\therefore \int Mdx=\dfrac {-y^2}{2x^2}-\Bigg[\dfrac {-1}3e^{1/x^3}\Bigg]$

$=\dfrac {1}3e^{1/x^3}-\dfrac {y^2}{2x^2}$

$\int $ (N free from x)dy =0

The solution is $\dfrac 13e^{1/x^3}-\dfrac {y^2}{2x^2}$