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 February 23rd, 2019, 12:45 AM #1 Newbie   Joined: Feb 2019 From: India Posts: 4 Thanks: 0 A good differential equation problem Ques7 , math forum https://imgur.com/gallery/MVN7HR5 $\displaystyle \left(2xy + x^2y + \frac{y^3}{3}\right)dx + \left(x^2 + y^2\right)dy = 0$ I've been trying to solve this differential equation from 2 days and now I seek help. Last edited by skipjack; February 23rd, 2019 at 05:02 AM. Reason: Typo , attached link of question February 23rd, 2019, 03:02 AM #2 Newbie   Joined: Feb 2019 From: India Posts: 4 Thanks: 0 Yesss!!!! I solved it! Finally, will someone cross-check my work? Last edited by skipjack; February 23rd, 2019 at 10:52 AM. February 23rd, 2019, 07:16 AM #3 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2256 $\left(3x^2y + y^3\right)e^x = \text{C}$, where $\text{C}$ is a constant. Thanks from topsquark and idontknow February 23rd, 2019, 08:38 AM   #4
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 Originally Posted by skipjack $\left(3x^2y + y^3\right)e^x = \text{C}$, where $\text{C}$ is a constant.
Please can you show me the steps? I don't think mine are correct.

Skipjack, please tell me how it's done... 2 days non-stop; I'm dying.

Last edited by skipjack; February 23rd, 2019 at 10:51 AM. February 23rd, 2019, 08:45 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 715 Thanks: 96 https://en.m.wikipedia.org/wiki/Exac...ntial_equation The type of the equation has solution like $\displaystyle C=F(x,y)$ . Check the page and it is done . Last edited by idontknow; February 23rd, 2019 at 08:47 AM. February 23rd, 2019, 10:56 AM #6 Global Moderator   Joined: Dec 2006 Posts: 21,020 Thanks: 2256 Differentiating $x^2y$ would account for the $2xy\hspace{2px}dx$ and $x^2\hspace{1px}dy$ terms in the equation. Differentiating $y^3/3$ would account for the $y^2\hspace{1px}dy$ term at the end of the equation. That leaves two terms unaccounted for, which very conveniently happen to be $x^2y\hspace{2px}dx$ and $y^3/3\hspace{2px}dx$. Hence letting $u = x^2y + y^3/3$ gives $u\hspace{2px}dx + du = 0$, which can be solved by using $e^x\!$ as an integrating factor. That means the original equation becomes exact if $e^x\!$ is used as an integrating factor, and integrating it will then give $ue^x= (x^2y + y^3/3)e^x\,=$ constant. Choosing $\text{C}/3$ for the constant and multiplying by 3 gives $(3x^2y + y^3)e^x = \text{C}$. I left it to you to confirm that this equation implicitly defines $y$ as a function of $x$ if $x$ is real. Thanks from ProofOfALifetime and Shubhankar4 February 23rd, 2019, 01:08 PM   #7
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Quote:
 Originally Posted by Shubhankar4 Yesss!!!! I solved it! Finally, will someone cross-check my work?
How can we possibly cross check your work if you don't show us what you did?

-Dan February 23rd, 2019, 11:57 PM   #8
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 Originally Posted by topsquark How can we possibly cross check your work if you don't show us what you did? -Dan
I'm sorry; I can't add pics on this website because it's my first time using a forum, and I'm using a phone.

Last edited by skipjack; February 24th, 2019 at 06:01 PM. Tags differential, equation, good, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post dylanxelle Differential Equations 6 February 8th, 2018 03:26 PM petermojer Differential Equations 0 June 13th, 2014 03:32 PM sivela Differential Equations 1 January 21st, 2011 05:53 PM yungman Differential Equations 0 November 24th, 2009 09:56 PM yungman Differential Equations 0 December 31st, 1969 04:00 PM

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