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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 crosscheck 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: 20,931 Thanks: 2207 
$\left(3x^2y + y^3\right)e^x = \text{C}$, where $\text{C}$ is a constant.

February 23rd, 2019, 08:38 AM  #4  
Newbie Joined: Feb 2019 From: India Posts: 4 Thanks: 0  Quote:
Skipjack, please tell me how it's done... 2 days nonstop; 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: 600 Thanks: 88  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: 20,931 Thanks: 2207 
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. 
February 23rd, 2019, 01:08 PM  #7 
Math Team Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 928 Math Focus: Wibbly wobbly timeywimey stuff.  
February 23rd, 2019, 11:57 PM  #8 
Newbie Joined: Feb 2019 From: India Posts: 4 Thanks: 0  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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