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 May 21st, 2017, 06:48 PM #1 Senior Member   Joined: Nov 2015 From: hyderabad Posts: 211 Thanks: 2 Non-Exact Differential Equation Hello All Please help me to solve this differential equation which is non-exact. $(x^2 y^2+y^4+2x)dx+2y(x^3+xy^2+1)dy=0$ Thank you
May 23rd, 2017, 08:59 AM   #2
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 Originally Posted by Lalitha183 Hello All Please help me to solve this differential equation which is non-exact. $(x^2 y^2+y^4+2x)dx+2y(x^3+xy^2+1)dy=0$ Thank you
Hello...Please anyone help me to solve this problem.
I'm not able to solve this in any of the ways

 May 23rd, 2017, 06:59 PM #3 Global Moderator   Joined: Dec 2006 Posts: 18,586 Thanks: 1489 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
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 Originally Posted by skipjack If $u = x^3 + xy^2\!$, $x^2u(1 + 3x^3)dx - x^3(1 + u)du = 0$, which is separable.
Can you explain it in detail... I'm not much aware of separable method

May 23rd, 2017, 08:27 PM   #5
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 Originally Posted by skipjack If $u = x^3 + xy^2\!$, $x^2u(1 + 3x^3)dx - x^3(1 + u)du = 0$, which is separable.
Quote:
 Originally Posted by Lalitha183 Can you explain it in detail... I'm not much aware of separable method
$x^2u(1+3x^3)dx-x^3(1+u)du=0$

$x^2u(1+3x^3)dx=x^3(1+u)du$

$\dfrac{1+3x^3}{x}dx = \dfrac{1+u}{u}du$

$\left(\dfrac 1 x + 3x^2\right)dx = \left(\dfrac 1 u + 1\right)du$

$\ln(x) + x^3 = \ln(u) + u + C$

and you can take it from there

May 23rd, 2017, 10:12 PM   #6
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 Originally Posted by romsek $x^2u(1+3x^3)dx-x^3(1+u)du=0$ $x^2u(1+3x^3)dx=x^3(1+u)du$ $\dfrac{1+3x^3}{x}dx = \dfrac{1+u}{u}du$ $\left(\dfrac 1 x + 3x^2\right)dx = \left(\dfrac 1 u + 1\right)du$ $\ln(x) + x^3 = \ln(u) + u + C$ and you can take it from there
I want to know how do we differentiate non-exact D.E from variable separable method ?

 May 23rd, 2017, 10:51 PM #7 Global Moderator   Joined: Dec 2006 Posts: 18,586 Thanks: 1489 If $x$ and $y$ have to be real, one can write $\ln|x| + x^3 = \ln|u| + u + \text{C}$. 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
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 Originally Posted by Lalitha183 I want to know how do we differentiate non-exact D.E from variable separable method ? Please help!!!
This is very strange. Usually "separable differential equations" are the very first kind of differential equation taught - well before methods for "exact" equations.

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
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 Originally Posted by romsek $x^2u(1+3x^3)dx-x^3(1+u)du=0$ $x^2u(1+3x^3)dx=x^3(1+u)du$ $\dfrac{1+3x^3}{x}dx = \dfrac{1+u}{u}du$ $\left(\dfrac 1 x + 3x^2\right)dx = \left(\dfrac 1 u + 1\right)du$ $\ln(x) + x^3 = \ln(u) + u + C$ and you can take it from there
So the final answer would be $xy^2+\log(x^2+y^2)=$constant?

And also I want to know how $x^2u(1+3x^3)dx-x^3(1+u)du=0$ 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: 18,586 Thanks: 1489 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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