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

Marks : 6

Year : 2015

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

$M=xy\cos x-y\sin x-2$

$\dfrac {\partial M}{\partial y}=x\cos x-\sin x$

$N=x\sin x$

$\dfrac {\partial N}{\partial x}=x\cos x+\sin x$

$\dfrac {\partial M}{\partial y}-\dfrac {\partial N}{\partial x}=x\cos x-\sin x -x\cos x-\sin x$

$=-2\sin x$

$\therefore \dfrac {\dfrac {\partial M}{\partial y}-\dfrac {\partial N}{\partial x}}N=\dfrac {-2\sin x}{x\sin x}=\dfrac {-2}x=f(x) $

$\therefore I.F.=e^{\int -2/x}dx$

$=e^{-2\log x}$

$=e^{\log 1/x^2}$

$=\dfrac 1{x^2}$

Multiplying By I.F. we get

$\dfrac {\sin x}x dy+\Bigg(\dfrac yx\cos x-\dfrac y{x^2}\sin x-\dfrac 2{x^2}\Bigg)dx=0$

Which is exact.

$\int Mdx=\int\Bigg(\dfrac yx\cos x-\dfrac y{x^2}\sin x-\dfrac 2{x^2}\Bigg)dx$

$=\int\Bigg[\dfrac yx\Bigg[1-\dfrac {x^2}{2!}+\dfrac {x^4}{4!}-----\Bigg]-\dfrac y{r^2}\Bigg[x-\dfrac {x^3}{3!}+\dfrac {x^5}{5!}---\Bigg]-\dfrac 2{x^2}\Bigg]dx$

$\Bigg[y\Bigg\{\Bigg[\dfrac 1x-\dfrac 1x\Bigg]-\Bigg[\dfrac x{2!}-\dfrac x{3!}\Bigg]----\Bigg\}-\dfrac 2{x^2}\Bigg]dx$

$\therefore \int Mdx=y\dfrac {x^2}2\times \dfrac {(3!.2!)}{2!\times3!}+j\dfrac {x^4(5!-4!)}{5!\times4!}+\dfrac 2x$

$\int$ N free space x=0

$\therefore$ The solution Is $y\dfrac {x^2}2\times \dfrac {(3!-2!)}{2!\times3!}+y\dfrac {x^4}4\times \dfrac {(5!-4!)}{5!\times4!}+\dfrac 2x + c$