1
$\begingroup$

In this article about exact differential equations, for the fourth example, they have the differential equation:

$$ (5xy^2 - 2y) dx + (3xy^2 -x) dy = 0$$

And they say they integrating factor for this is $$ \mu(x,y) = x^a y^b$$

Now, I don't get how they arrived that this should be correct integrating factors for the problem. Of course, this does integrating factor does satisfy the requirement that the second-order partials commute but I can't really understand how I'd come up with this if it was not already told that this is the integrating factor.

If not, for what kinds of differential equations would this integrating factor work?


Similar post to this

I have seen this post already and it is not the same as mine because the question was not really regarding the integrating factor which I mentioned but rather alternative methods.

$\endgroup$
2
$\begingroup$

$$(5xy^2 - 2y) dx + (\color{red}{3xy^2 }-x) dy = 0$$ You made a little mistake the correct differential equation is: $$(5xy^2 - 2y) dx + (3x^2y -x) dy = 0$$ Multiply by $x$: $$(5x^2y^2 - 2yx) dx + (3x^3y -x^2) dy = 0$$ Rearrange some terms: $$(5x^2y^2dx + 3x^3ydy)-(ydx^2 +x^2 dy) = 0$$ $$(5x^2y^2dx + 3x^3ydy)-d(x^2y) = 0$$

The integrating factor is now obvious: $$\mu(x,y)=x^2y$$ $$(5x^4y^3dx + 3x^5y^2dy)-x^2yd(x^2y) = 0$$ $$d(x^5y^3)-x^2yd(x^2y) = 0$$ Then integrate: $$2x^5y^3-x^4y^2 = C$$

$\endgroup$
15
  • $\begingroup$ I don't understand the step after you introduced the integrating factor. Why is the second equation just "the preceding one multiplied by $\mu$"? $\endgroup$
    – Ian
    Oct 9 '20 at 15:25
  • $\begingroup$ Differentiate $x^5y^3$ and see what you get @ian Have i made a mistake Ian ? $\endgroup$
    – MtGlasser
    Oct 9 '20 at 15:26
  • 1
    $\begingroup$ Yes that makes more sense. +1. $\endgroup$
    – Ian
    Oct 9 '20 at 15:31
  • 1
    $\begingroup$ my bad, it is P(x,y) and Q(x,y) $\endgroup$
    – Buraian
    Oct 12 '20 at 6:06
  • 1
    $\begingroup$ I see what you mean. @Buraian $\endgroup$
    – MtGlasser
    Oct 12 '20 at 6:08
0
$\begingroup$

If for $(5xy^2-2y)~dx+(3xy^2-x)~dy=0$ , we should need another approach:

$(5xy^2-2y)~dx+(3xy^2-x)~dy=0$

$\dfrac{dx}{dy}=-\dfrac{3xy^2-x}{5xy^2-2y}$

Let $x=yu$ ,

Then $\dfrac{dx}{dy}=y\dfrac{du}{dy}+u$

$\therefore y\dfrac{du}{dy}+u=-\dfrac{3y^3u-yu}{5y^3u-2y}$

$y\dfrac{du}{dy}=-\dfrac{3y^3u-yu}{5y^3u-2y}-u$

$y\dfrac{du}{dy}=-\dfrac{5y^3u^2+3y^3u-3yu}{5y^3u-2y}$

$y\dfrac{du}{dy}=-\dfrac{5y^2u^2+3y^2u-3u}{5y^2u-2}$

Let $r=y^2$ ,

Then $\dfrac{du}{dy}=\dfrac{du}{dr}\dfrac{dr}{dy}=2y\dfrac{du}{dr}$

$\therefore2y^2\dfrac{du}{dr}=-\dfrac{5y^2u^2+3y^2u-3u}{5y^2u-2}$

$2r\dfrac{du}{dr}=-\dfrac{5ru^2+3ru-3u}{5ru-2}$

$(2-5ru)\dfrac{du}{dr}=\dfrac{5u^2}{2}+\dfrac{3(r-1)u}{2r}$

This belongs to an Abel equation of the second kind.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.