written 7.8 years ago by | modified 2.1 years ago by |
Mumbai University > First Year Engineering > sem 2 > Applied Maths 2
Marks : 6
Year : 2013
written 7.8 years ago by | modified 2.1 years ago by |
Mumbai University > First Year Engineering > sem 2 > Applied Maths 2
Marks : 6
Year : 2013
written 7.8 years ago by | • modified 2.5 years ago |
We have equation as $Mdx + Ndy = 0$
$M = 2xy^4e^y + 2xy^3 + ys$
$N = x^2y^4e^y – x^2y^2 – 3x$
$\dfrac {\partial M}{\partial y}=2x(y^4e^y+4y^3e^y)+6xy^2+1$
$\dfrac {\partial N}{\partial y}=2xy^4e^y -2xy^2-3$
$\dfrac {\dfrac {\partial M}{\partial y}-\dfrac {\partial N}{\partial y}}M=\dfrac {2xy^4e^y-2xy^2-3-2xy^4e^y-8xy^3e^y-6xy^2-1}{2xy^4e^y+2xy^3+y}$
$=\dfrac {-8xy^2-4-8xy^3e^y}{2xy^4e^y+2xy^3+y}$
$=\dfrac {-4}y\Bigg[\dfrac{2xy^2+1+2xy^3e^y}{2xy^2+1+2xy^3e^y}\Bigg]$
$=\dfrac {-4}y f(y)$
If $e^{\int -(4/y)dy}=e^{-4\log y}=e^{\log (1/y^4)}=1/y^4$
Multiplying by the If we get
$\Bigg(2xe^y+\dfrac {2x}y+\dfrac 1{y^3}\Bigg)dx +\Bigg(x^2e^y-\dfrac {x^2}{y^2}-\dfrac {3x}{y^4}\Bigg)dy=0$
Which is exact.
$\int Mdx=\int(2xe^y+\dfrac {2x}y+\dfrac 1{y^3})dx$
$=x^2e^y+\dfrac {x^2}y+\dfrac x{y^3}$
$\int $ (Terms in N free from x)dy =0
$\therefore$ The complete solution is $x^2e^y +\dfrac {x^2}y+\dfrac x{y^3}=c$