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

Question

Solve:

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

I tried to solve the differential equation in the following manner, but I am unable to arrive at the correct solution. The correct solution is:

$$ 4(xy)^{\frac{1}{3}} - \frac{2}{3}(x/y)^{\frac{3}{2}} = C $$

Please help me identify my mistake in the solution.

RECTIFICATION:

solved here: https://math.stackexchange.com/questions/2588586/solve-y22x2ydx2x3-xydy-0 — bjorn93, Dec 27 '19 at 15:30
I don't know if that helps but you calculated IF to be $1/2xy^2$, but then you multiplied the DE by $1/2xy$. — NickD, Dec 27 '19 at 15:43

user577215664 · Answer 1 · 2019-12-27T16:40:28.033

1

$$(y^2+2x^2y)dx+(2x^3−xy)dy=0$$ $$(y+2x^2)ydx+(2x^2−y)xdy=0$$ Is not of the form $$f_1(xy)ydx+f_2(xy)xdy=0$$ So you can't use that method for this DE

Be carefull Soumee. $(y+2x^2)$ is a function $f(x,y)$ But it's not a function of the form $f(xy)$

Edit

The integrating factor $\mu(x,y)=x^{-5/2}y^{-1/2}$ proposed by @Soumee: $$y(ydx-xdy)+2x^2dxy=0$$ Multiply by IF $x^{-5/2}y^{-1/2}$ $$x^{-5/2}y^{-1/2}y(ydx-xdy)+2x^{-5/2}y^{-1/2}x^2dxy=0$$ $$-x^{-1/2}y^{1/2}d\frac {y}{x}+2x^{-1/2}y^{-1/2}dxy=0$$ $$-\left (\frac {y}{x} \right )^{1/2}d\left (\frac {y}{x} \right )+2(xy)^{-1/2}d(xy)=0$$ Integrate

edited Dec 27 '19 at 16:40

answered Dec 27 '19 at 15:44

user577215664

40,625

So, when the differential equation consists of terms of the form$ (xy)^k $ then only the DE can be solved using the method I have used? – Soumee Dec 27 '19 at 15:47
Soumee a function of $xy$ can be any function that content $xy$. For exampes $\sin(xy)$ is a function of $xy.$ Or another example is $ e^{3xy}$ You see what I mean ? @Soumee – user577215664 Dec 27 '19 at 15:48
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Yes, I get it .. – Soumee Dec 27 '19 at 15:49
@Soumee good. Now I think this DE can't be solved easily with exactness. The link provided in the comments gives you different methods for solving this DE – user577215664 Dec 27 '19 at 15:50
I think integrating factor can be assumed as $ x^h y^k $ and then be solved. – Soumee Dec 27 '19 at 16:09
I got $h = \frac {-5}{2}$ and $ k = \frac{-1}{2} $. Wait, I am uploading the answer. – Soumee Dec 27 '19 at 16:13
Ok I will look at it @Soumee – user577215664 Dec 27 '19 at 16:14
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I have updated my question. – Soumee Dec 27 '19 at 16:16
I have uploaded the complete solution. However, there is just a small mistake. – Soumee Dec 27 '19 at 16:28
I think it's correct now I updated my answer @soumee – user577215664 Dec 27 '19 at 16:37
@Soumee your integrating factor works fine. – user577215664 Dec 27 '19 at 16:40
But the problem is $ 4(xy)^{\frac{1}{3}} - \frac{2}{3}(x/y)^{\frac{3}{2}} = C $ is the answer. However, I have got $4(xy)^{\frac{1}{3}} - \frac{2}{3}(y/x)^{\frac{3}{2}} = C$ as the answer. – Soumee Dec 27 '19 at 16:52
the first exponent is $1/2$ not $1/3$ In the comment in the link provided the answer given by user is the same answer as yours. @Soumee are you talking aboutanother answer provided by your book ? – user577215664 Dec 27 '19 at 16:57
Sorry, I got $ 4(xy)^{\frac{1}{2}} - \frac{2}{3}(y/x)^{\frac{3}{2}} = C $ as the answer. Therefore now both the terms are incorrect! :( – Soumee Dec 27 '19 at 17:06
Did you check the link provided in the comments ? The answer given by use is the same as yours @soumee – user577215664 Dec 27 '19 at 17:07
Oh! yes, user170231 has the same answer as mine. What answer are you getting? – Soumee Dec 27 '19 at 17:11
Same answer as yours @Soumee with $1/2$ for the first exponent – user577215664 Dec 27 '19 at 17:11
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Thank you for your help. Then could the answer given in the text be wrong? – Soumee Dec 27 '19 at 17:14
YW @Soumee Sometimes it happens that answers provided in textbooks are wrong. The most important is that you know how to apply the methods – user577215664 Dec 27 '19 at 17:16
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Yes, true. Thanks. – Soumee Dec 27 '19 at 17:17

score 0 · Answer 2 · answered Dec 27 '19 at 18:51

A very simple solution: $$(y^2+ 2x^2y)dx + (2x^3 - xy)dy = 0$$ $$y(ydx-xdy)+ 2x^2(xdy +ydx) = 0$$ $$-y d\left ( \frac {y}x \right )+ 2d(xy) = 0$$ $$-\sqrt y d\left ( \frac {y}x \right )+ \frac 2 {\sqrt y}d(xy) = 0$$ An obvious integrating factor is $\mu(x)=1/\sqrt x$ $$-\sqrt {\frac y x} d\left ( \frac {y}x \right )+ \frac 2 {\sqrt {xy}}d(xy) = 0$$ It's integrable. $$\frac 13 \left ( {\frac y x} \right)^{\frac 32}-2 {\sqrt {xy}} = C$$

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

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