Question

find the integrating factor of (x²y + y²)dx + (y³–x³)dy = 0

Answer 1

find the integrating factor of (x²y + y²)dx + (y³–x³)dy = 0

Solution:

Let's try find integrating factor from the equation:

$m\cdot(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})=N\frac{\partial m}{\partial x}-M\frac{\partial m}{\partial y}$ , where

$m(x,y)$ - integration factor;

$M(x,y)=x^2y+y^2$

$N(x,y)=y^3-x^3$

$m\cdot(x^2+2y-(-3x^2))=(y^3-x^3)\frac{\partial m}{\partial x}$ $-(x^2y+y^2)\frac{\partial m}{\partial y}$

$m\cdot(4x^2+2y)=(y^3-x^3)\frac{\partial m}{\partial x}-(x^2y+y^2)\frac{\partial m}{\partial y}$

This is a partial differential equation and there is no an easy way to solve it except the following three special cases:

1)If the expression:

$\frac{(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{N}$

is a function of $x$ only, then $m$ is also a function of $x$ only and $m(x)$ is given by:

$m(x)=\exp{\left(\int\frac{(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{N}dx\right)}$ .

2) If the expression:

$\frac{(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})}{M}$

is a function of $y$ only, then $m$ is also a function of $y$ only and $m(y)$ is given by:

$m(y)=\exp{\left(\int\frac{(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})}{M}dx\right)}$ .

Let's check these cases:

$\frac{(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{N}=\frac{4x^2+2y}{y^3-x^3}$ isn't a function of $x$ only .

$\frac{(\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y})}{M}=\frac{-(4x^2+2y)}{x^2y+y^2}$ isn't a function of $y$ only .

So we can conclude that $m$ is a function of $x$ and $y$ and it cannot be found using the technique described above.

3) If $m(x,y)$ can be expressed as $m(x,y)=m(\omega(x,y))$ , where $\omega$ is known function then integrating factor can be found from the following equation:

$\frac{1}{m}\frac{dm}{d\omega}=\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{N\frac{\partial \omega}{\partial x}-M\frac{\partial \omega}{\partial y}}$ (*)

The condition of the task says nothing about the form of the function $m(x,y)$. And I couldn't find such a function $\omega(x,y)$ to exclude $x$ and $y$ from (*). So it is highly likely that the problem statement is incomplete or contains an error.