My Math Forum 2 exact differential equation

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 July 16th, 2013, 06:14 AM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 2 exact differential equation Q.1. I am supposed to find the exactness of the following equation and thereby calculate its solution:- $(x^2dy-y^2dx)/(x-y)^2=0$ My question is should I solve it in the form $[x^2.dy/(x-y)^2] + [-y^2dx/(x-y)^2]= 0$or in the form$x^2dy - y^2dx= 0$ ? Q.2. I don't know how to solve the following problem $xdx + ydy=( y.dx - x.dy)/ x^2 + y^2$ This isn't an exact equation. So I have to transform it to an exact one. For that I need to calculate the integrating factor. But I just can't calculate it. Please help!
 July 16th, 2013, 09:41 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: 2 exact differential equation Q1.) In which form is the equation exact?
 July 16th, 2013, 11:24 PM #3 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: 2 exact differential equation $[x^2.dy/(x-y)^2] + [-y^2dx/(x-y)^2]= 0$ This form is the exact one.
 July 17th, 2013, 05:58 PM #4 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: 2 exact differential equation And multiplying both sides by $(x- y)^2$ would destroy that. Recall that every first order equation has a "multiplying factor", function of both variables which, when you multiply both sides of the equation by it, the equation becomes "exact". So of course, multiplying or dividing by a function can make an "exact" equation "non exact".
 July 17th, 2013, 10:46 PM #5 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: 2 exact differential equation Q.1 The two forms of the equation $(x^2dy-y^2dx)/(x-y)^2=0$ give 2 different solutions. That's why I want to know which one is the correct form to solve. According to the answer page of my book, I should 1st transform the above equation to $x^2dy - y^2dx= 0$ and then find its solution. But I doubt if this will be correct or not. So please tell me if I should solve it in the form $x^2dy - y^2dx= 0$ or in the form $[x^2.dy/(x-y)^2] + [-y^2dx/(x-y)^2]= 0$. That's what I want to know. Q.2 Here I am not able to calculate the integrating factor which turns the inexact equation into an exact one. It's not like I don't know how to calculate the integrating factor, in general. But in this particular problem, the usual process of calculating the integrating factor(as mentioned in my book) isn't working. So I don't know what to do. The fact is I am studying Economics all by myself, without any teacher's help. In Economics, there is of course Mathematics. So I am studying Math with the help of my book and this site. You people have helped me a lot and no matter how many times I say “thanks" to you all it will be of lesser value. So I am posting too much problems here in the hope of getting help. I hope I am not bothering you people too much.
 July 18th, 2013, 06:44 AM #6 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: 2 exact differential equation The difficulty is that the the first form is NOT "exact". I should have checked that myself. But $x^2dx+ y^2dy$ is exact- it is the differential of $\frac{x^3}{3}+ \frac{y^3}{3}$. This is a case where $(x- y)^2$ was the "integrating factor".
July 18th, 2013, 07:38 AM   #7
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Re: 2 exact differential equation

Quote:
 Originally Posted by HallsofIvy The difficulty is that the the first form is NOT "exact". I should have checked that myself. But $x^2dx+ y^2dy$ is exact- it is the differential of $\frac{x^3}{3}+ \frac{y^3}{3}$. This is a case where $(x- y)^2$ was the "integrating factor".
From where did $x^2dx + y^2dy$ come from? If it's about question 1 then the correct form is $x^2dy - y^2dx$.

 July 18th, 2013, 07:54 AM #8 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: 2 exact differential equation Again $[(x^2dy)/(x-y)^2 ] + [(-y^2dx)/(x-y)^2]$ is exact. Here$N= x^2/(x-y)^2 and M= (-y^2)/(x-y)^2$. partial derivative of M w.r.t. y = partial derivative of N w.r.t. x= (-2xy)/(x-y)^3 Because of this equality, the equation is exact. This is how my book teaches me to examine the exactness of an equation. Please correct me if I am wrong anywhere.
 July 18th, 2013, 08:30 AM #9 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: 2 exact differential equation Yes, the partials are equal and so: $\frac{x^2}{(x-y)^2}\,dx-\frac{y^2}{(x-y)^2}\,dy=0$ is exact. So what is your first step in obtaining the solution? edit: The equation is also separable...
 July 18th, 2013, 09:26 AM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: 2 exact differential equation Q2.) We are given (presumably): $x\,dx+y\,dy=\frac{y\,dx-x\,dy}{x^2+y^2}$ My first step would be to get the ODE in the form: $M(x,y)\,dx+N(x,y)\,dy=0$ and so multiplying though by $x^2+y^2$ gives: $$$x^2+y^2$$x\,dx+$$x^2+y^2$$y\,dy=y\,dx-x\,dy$ $$$\(x^2+y^2$$x-y\)\,dx+$$\(x^2+y^2$$y+x\)\,dy=0$ The test for exactness reveals that this is inexact. So, we look at: $\frac{(2xy-1)-(2xy+1)}{$$x^2+y^2$$y+x}=-\frac{2}{$$x^2+y^2$$y+x}$ This is not a function of $x$ alone, so we next look at: $\frac{(2xy+1)-(2xy-1)}{$$x^2+y^2$$x-y}=\frac{2}{$$x^2+y^2$$x-y}$ This is not a function of $y$ alone, so computing an integrating factor presents a difficulty. So, I suggest we go back to the original: $x\,dx+y\,dy=\frac{y\,dx-x\,dy}{x^2+y^2}$ and write it in the form: $x+yy'+\frac{xy'-y}{x^2+y^2}=0$ $2x+2yy'-2\frac{1}{1+$$\frac{x}{y}$$^2}\cdot\frac{y-xy'}{y^2}=0$ Integrating, we find the implicit solution: $x^2+y^2-2\tan^{\small{-1}}$$\frac{x}{y}$$=C$

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