Answer to Question #140384 in Differential Equations for akram Ahmad

Question

Q\Determine the following ODEs are exact or not? Why?
(y^3)dx+3x^3y^2dy=0

Answer 1

The given ODE is

$$y^3 dx+3x^3y^2 dy = 0$$

The necessary and sufficient condition that $M\space dx + N \space dy = 0$ be exact is

$$\frac{\delta M}{\delta y}=\frac{\delta N}{\delta x}$$

$Here \space M= y^3 \space and \space N = 3x^3y^2$

$Now , \frac{\delta M}{\delta y}= 3y^2$ ( partially differentiating M w.r.t 'y' keeping 'x' as constant)

$again, \frac{\delta N}{\delta x}= 9x^2y^2$ ( partially differentiating N w.r.t 'x' keeping 'y' as constant )

As,

$$\frac{\delta M}{\delta y} \neq\frac{\delta N}{\delta x}$$

$\therefore$ The given ODE is not exact.