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February 23rd, 2019, 12:45 AM   #1
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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
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February 23rd, 2019, 03:02 AM   #2
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Yesss!!!!
I solved it! Finally, will someone cross-check my work?

Last edited by skipjack; February 23rd, 2019 at 10:52 AM.
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February 23rd, 2019, 07:16 AM   #3
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$\left(3x^2y + y^3\right)e^x = \text{C}$, where $\text{C}$ is a constant.
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February 23rd, 2019, 08:38 AM   #4
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Quote:
Originally Posted by skipjack View Post
$\left(3x^2y + y^3\right)e^x = \text{C}$, 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.
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February 23rd, 2019, 08:45 AM   #5
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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.
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February 23rd, 2019, 10:56 AM   #6
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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.
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February 23rd, 2019, 01:08 PM   #7
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Quote:
Originally Posted by Shubhankar4 View Post
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
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February 23rd, 2019, 11:57 PM   #8
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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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