written 6.1 years ago by | • modified 6.0 years ago |
$ M=x^2y-2xy^2\\ N=-(x^3-3x^2y)$
$ \frac{\partial M}{\partial y}=x^2-4xy$
$ \frac{\partial N}{\partial x}=-(3x^2-6xy)$
$\frac{\partial M}{\partial y}\neq\frac{\partial N}{\partial x}$
The given Equation is not an exact differential equation.
M and N are homogenous functions of degree 3
$I.F=\frac{1}{Mx+Ny}$
$=\frac{1}{(x^2y-2xy^2)x-(x^3-3x^2y)y}$
$=\frac{1}{(x^3y-2x^2y^2)-(x^3y-3x^2y^2)}$
$=\frac{1}{x^2y^2}$
Multiplying the given equation by the integrating factor, we get
$ \left( x^2y=2xy^2\right)\frac{1}{x^2y^2}dx-(x^3-3x^2y)\frac{1}{x^2y^2}dy=0$
$\left(\frac{1}{y}-\frac{2}{x} \right)dx-\left(\frac{x}{y^2}-\frac{3}{y} \right)dy=0 $
$\frac{\partial M}{\partial y}=\frac{-1}{y^2} $
$\frac{\partial N}{\partial x}=\frac{-1}{y^2}$
The given Equation is an exact differential equation
Solution is given by
$ \int Mdx+\int (Terms\ in\ N\ not\ containing\ x)dy=c$
y=constant
$\int \left( \frac{1}{y}-\frac{2}{x}\right)dx+\int{\frac{3}{y}dy}=c$
$ \frac{x}{y}-2\log x+3\log y=c$