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May 21st, 2017, 06:48 PM  #1 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  NonExact Differential Equation
Hello All Please help me to solve this differential equation which is nonexact. $(x^2 y^2+y^4+2x)dx+2y(x^3+xy^2+1)dy=0$ Thank you 
May 23rd, 2017, 08:59 AM  #2 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 23rd, 2017, 06:59 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
If $u = x^3 + xy^2\!$, $x^2u(1 + 3x^3)dx  x^3(1 + u)du = 0$, which is separable.

May 23rd, 2017, 08:15 PM  #4 
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  
May 23rd, 2017, 08:27 PM  #5  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,602 Thanks: 816  Quote:
Quote:
$x^2u(1+3x^3)dx=x^3(1+u)du$ $\dfrac{1+3x^3}{x}dx = \dfrac{1+u}{u}du$ $\left(\dfrac 1 x + 3x^2\right)dx = \left(\dfrac 1 u + 1\right)du$ $\ln(x) + x^3 = \ln(u) + u + C$ and you can take it from there  
May 23rd, 2017, 10:12 PM  #6  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
Please help!!!  
May 23rd, 2017, 10:51 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
If $x$ and $y$ have to be real, one can write $\lnx + x^3 = \lnu + u + \text{C}$. If the equation isn't exact and one can't find an integrating factor that makes it exact, it's difficult to know whether some substitution will enable separation of variables. Once the variables are separated, the equation is exact of course. Last edited by skipjack; May 23rd, 2017 at 10:58 PM. 
May 24th, 2017, 05:51 AM  #8  
Math Team Joined: Jan 2015 From: Alabama Posts: 2,822 Thanks: 750  Quote:
A "separable differential equation" is simply one in which you can put all "x" terms, as well as dx, on one side of the equation, all "y" terms, as well as dy, on the other side, and integrate both separately. For example, $\displaystyle \frac{dy}{dx}= x^3y xy$ is "separable" because it can be written $\displaystyle \frac{dy}{dx}= (x^3 x)y$ and then as $\displaystyle \frac{dy}{y}= (x^3 x)dx$. Since all x and y terms have been "separated", we can integrate each separately: $\displaystyle \ln(y)= \frac{1}{4}x^4 \frac{1}{2}x^2+ C$. Last edited by skipjack; May 24th, 2017 at 08:33 AM.  
May 24th, 2017, 06:52 PM  #9  
Senior Member Joined: Nov 2015 From: hyderabad Posts: 206 Thanks: 2  Quote:
And also I want to know how $x^2u(1+3x^3)dxx^3(1+u)du=0$ is developed from $(x^2y^2+y^4+2x)dx+2y(x^3+xy^2+1)dy $ ? Thank you. Last edited by skipjack; May 24th, 2017 at 11:24 PM.  
May 25th, 2017, 01:41 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1418 
The original differential equation can be written as $2xdx + 2ydy = 2x^3ydy  2xy^3dy  x^2y^2dx  y^4dx = (2xydy  y^2dx)(x^2 + y^2)$. By inspection, that has $y = ix$ and $y = ix$ as (complex) solutions. For other solutions, let $u = x^2 + y^2\!$, then $du = 2xdx + 2ydy$, and so $du + (2xydy + y^2dx)u = 0$. By using $e^{xy^2}\!$ as an integrating factor, the equation becomes exact and integrating gives $ue^{xy^2} = \text{C}$ (where $\text{C}$ is a constant), so $x^2 + y^2 = \text{C}e^{xy^2}\!$. What you got was equivalent. So why did I include an extra factor of $x$ in my original substitution? Well, one can write $\text{C}xe^{x^3} = (x^3 + xy^2)e^{x^3 + xy^2}\!$. For certain values of $x^3 + xy^2\!$, that implies $x^3 + xy^2 = \text{W}(Cxe^{x^3})$, where $\text{W}$ denotes the Lambert W function. That leads to $\displaystyle y = \sqrt{\text{W}\!\left(\text{C}xe^{x^3}\right)\!/x  x^2}$ or $\displaystyle y = \sqrt{\text{W}\!\left(\text{C}xe^{x^3}\right)\!/x  x^2}$. This can be pursued further, but, as you weren't asking for $y$ as an explicit function of $x$, I leave that to you. 

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