Question #140384
Q\Determine the following ODEs are exact or not? Why?
(y^3)dx+3x^3y^2dy=0
1
Expert's answer
2020-10-26T19:58:30-0400

The given ODE is

y3dx+3x3y2dy=0y^3 dx+3x^3y^2 dy = 0



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

δMδy=δNδx\frac{\delta M}{\delta y}=\frac{\delta N}{\delta x}

Here M=y3 and N=3x3y2Here \space M= y^3 \space and \space N = 3x^3y^2


Now,δMδy=3y2Now , \frac{\delta M}{\delta y}= 3y^2 ( partially differentiating M w.r.t 'y' keeping 'x' as constant)

again,δNδx=9x2y2again, \frac{\delta N}{\delta x}= 9x^2y^2 ( partially differentiating N w.r.t 'x' keeping 'y' as constant )


As,

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

\therefore The given ODE is not exact.


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