Answer to Question #179544 in Differential Equations for Vikram

Question

1/xy²+y⁴ is an integrating factor for the differential equation (x²y+y²)dx+(y³-x³)dy=0.

y1/x

Answer 1

$\displaystyle \textbf{\textsf{NO}}\\ (x^2y+y^2)\mathrm{d}x+(y^3-x^3)\mathrm{d}y=0\\ \frac{1}{xy^2+y^4}\left((x^2y+y^2)\mathrm{d}x+(y^3-x^3)\mathrm{d}y\right)=0\\ \frac{x^2y+y^2}{xy^2+y^4}\mathrm{d}x+\frac{y^3-x^3}{xy^2+y^4}\mathrm{d}y=0\\ \begin{aligned} \frac{\partial}{\partial y}\left(\frac{x^2y+y^2}{xy^2+y^4}\right) &= \frac{\partial}{\partial y}\left(\frac{x^2+y}{xy+y^3}\right) \\&= \frac{(xy + y^3)(1) - (x^2 + y)(x + 3y^2)}{(xy + y^3)^2} \\&= \frac{xy + y^3 - x^3 - xy - 3x^2y^2 - 3y^3}{(xy + y^3)^2} \\&= \frac{xy - 2y^3 - x^3 - xy - 3x^2y^2}{(xy + y^3)^2} \end{aligned}\\ \begin{aligned} \frac{\partial}{\partial x}\left(\frac{y^3 - x^3}{xy^2+y^4}\right) &= \frac{\partial}{\partial x}\left(\frac{y^3 - x^3}{xy^2+y^4}\right) \\&= \frac{(xy^2 + y^4)(-3x^2) - (y^3 - x^3)(y^2)}{(xy + y^3)^2} \\&= \frac{y(-3x^3y - 3x^2y^3 - y^4 + x^3y)}{(xy + y^3)^2} \\&= \frac{-2x^3y^2 - 3x^2y^4 - y^5}{(xy + y^3)^2} \end{aligned}\\ \textsf{Since}\,\,\,\frac{\partial}{\partial x}\left(\frac{y^3 - x^3}{xy^2+y^4}\right) \neq \frac{\partial}{\partial y}\left(\frac{x^2y+y^2}{xy^2+y^4}\right) \\ \therefore\textsf{The integrating factor is not}\,\,\frac{1}{xy^2+y^4}$