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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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Quote:
 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.

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