Roark Posted March 18, 2003 Share Posted March 18, 2003 Hi, I'm just starting to do differential equations and I'm stuck on this one: (y^2 - x^2)dy - 2xydx = 0 Ans: x^2 +y^2 - cy = 0 How do I separate the variables? What's the intergrating factor? Thanks! Link to comment Share on other sites More sharing options...
blike Posted March 18, 2003 Share Posted March 18, 2003 hrm, i'll see if i can help after class Link to comment Share on other sites More sharing options...
Dave Posted March 21, 2003 Share Posted March 21, 2003 i've been playing with this one for about 15 minutes now and i can't seem to make any inroads into it myself, so i thought i'd check back here to check the formula is correct. the formula you've given strictly speaking isn't an actual differential equation, but i gathered the formula you want solved is: (y^2 - x^2)dy/dx - 2xy = 0 i've tried this and although i'm no beginner to differential equations, i'm no expert either i can't see any way of seperating the variables as it stands, and if you are a beginner to differential equations, i highly recommend you go and find some more examples, as this one appears to be quite difficult. on the other hand, i may have missed something extremely obvious. Link to comment Share on other sites More sharing options...
Roark Posted March 21, 2003 Author Share Posted March 21, 2003 That's the equation that's given in the book. Is this a valid approach? (y^2 - x^2)dy - 2xydx = 0 (y^2 - x^2)dy = 2xydx (y^2 - x^2)dy/dx = 2xy (y^2 - x^2)dy/dx = x -------------- 2y Intergrate both sides: (y^2 - x^2)/2y= 1/2x^2 + C Intergrate both sides: (y^2 - x^2)= yx^2 + C (somehow reduces to?) Ans: x^2 +y^2 - cy = 0 Link to comment Share on other sites More sharing options...
JaKiri Posted March 21, 2003 Share Posted March 21, 2003 That's invalid. Link to comment Share on other sites More sharing options...
Dave Posted March 21, 2003 Share Posted March 21, 2003 you're okay until you get to this line: Originally posted by Roark Intergrate both sides: (y^2 - x^2)/2y= 1/2x^2 + C and then you meet the problem, how do you integrate something with respect to x and y, which you can't in this particular instance. i'd get another question personally Link to comment Share on other sites More sharing options...
Aleph-Null Posted July 9, 2003 Share Posted July 9, 2003 Originally posted by Roark (y^2 - x^2)dy - 2xydx = 0 Thanks! dy / dx = y' y'(y^2 - x^2) - 2xy = 0 Is in the form of M(x, y) + N(x, y)y' = 0 My(x, y) = -2x Nx(x, y) = -2x My = Nx --> So, we know that this is an exact differential equation. Link to comment Share on other sites More sharing options...
JaKiri Posted July 9, 2003 Share Posted July 9, 2003 Originally posted by Aleph-Null dy / dx = y' y'(y^2 - x^2) - 2xy = 0 Is in the form of M(x, y) + N(x, y)y' = 0 My(x, y) = -2x Nx(x, y) = -2x My = Nx ---> So, we know that this is an exact differential equation. Go away and learn maths again. Link to comment Share on other sites More sharing options...
Aleph-Null Posted July 9, 2003 Share Posted July 9, 2003 Originally posted by MrL_JaKiri Go away and learn maths again. Are you to imply through your refutation that it is not an exact differential equation? Link to comment Share on other sites More sharing options...
Dave Posted July 9, 2003 Share Posted July 9, 2003 Originally posted by MrL_JaKiri Go away and learn maths again. Hmm, it appears that you may be wrong on this one. http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html I have to say I've not heard of an exact first order ordinary differential equation, but apparently it is p(x,y)dx + q(x,y)dy = 0 => p(x,y) = -2xy, q(x,y) = y^2 - x^2 The partial derivatives of each of these are -2x, and hence it's exact. Apparently the general solution is this, and the answer comes out as: -x^2 * y + y^3/3 = c where c is a constant. Well done Aleph-Null, this one had been bothering me for some time Link to comment Share on other sites More sharing options...
Roark Posted July 10, 2003 Author Share Posted July 10, 2003 This doesn't match the answer in the book. Did we figure that the book is wrong? Link to comment Share on other sites More sharing options...
Dave Posted July 10, 2003 Share Posted July 10, 2003 I would say the answer in the book is probably wrong, yes. If you differentiate the answer that they give, I don't think you get the equation quoted in the question because of that c that's knocking about. Link to comment Share on other sites More sharing options...
JaKiri Posted July 10, 2003 Share Posted July 10, 2003 Originally posted by Aleph-Null Are you to imply through your refutation that it is not an exact differential equation? Sorry dearest, most of the mathematics we get here tends to be a little off and I only subjected it to a cursory glance because of that. Link to comment Share on other sites More sharing options...
Aleph-Null Posted July 10, 2003 Share Posted July 10, 2003 Thats assuming dave did it correctly. Roarck, I'm sure your book has a section on exact differential equations. What text are you using? I have done the problem but I am hesitant to give out the answer before you've learned the section on exact diffy q. I'm :lame: I know. Link to comment Share on other sites More sharing options...
Roark Posted July 10, 2003 Author Share Posted July 10, 2003 It's from teach yourself calculus. The chapther is an introduction to diff eqs and the section is separation of variables. Here's a reply from Karl of Karl's calculus forum. It sounds like you are both on the same track but that track is down the line for me. I'll be getting to this in about six months. Thanks. Reply to Peter: It would take a lot more than 3 weeks to do this diff. eq. by separation of variables because this one can't be done that way. This one you have to first find the integrating factor that makes it exact. Then you apply the method for solving exact equations. There is no recipe for finding integrating factors. You have to do it by educated guessing. After about two or three guesses I found that the integrating factor for this one is 1/y^2. The equation becomes (1 - x^2/y^2)dy + (2x/y)dx = 0 Notice if you take the partial derivative of the left-hand factor with respect to x, it is equal to the partial derivative of the right-hand factor with respect to y. That is what it means for an equation to be exact. This means that I simply integrate either factor -- either the left dy or the right dx to get one step closer to the solution (whose form will be F(x,y) = C). I'll do the right. F(x,y) = x^2/y + g(y) Notice that instead of C as the constant of integration, I used g(y). Why? Because y is independent of x, and I integrated dx. So a function of y is a constant with respect to x. Now if I take the partial derivative of this solution with respect to y, it should be equal to the left-hand factor ¶F = -x^2/y^2 + g'(y) = 1 - x^2/y^2 ¶y You get some cancellation and find that g'(y) = 1 or equivalently g(y) = y. So the solution is F(x,y) = x^2/y + y = C Now multiply by y and you have x^2 + y^2 = Cy Hope this helps. Link to comment Share on other sites More sharing options...
Dave Posted July 11, 2003 Share Posted July 11, 2003 Hmm, I don't know. I may be wrong, I've never touched this subject before Link to comment Share on other sites More sharing options...
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