Tips on Solving Differential EquationsDate: 10/27/2003 at 07:19:52 From: David Subject: Differential equations I'm teaching myself differential equations with a great book (Differential Equations - With Applications and Historical Notes, George F. Simmons). Although the book is great, the questions often go beyond the methods discussed in the text. I have reached a total impasse with the following questions, which I suspect means that I'm missing an entire (or several entire) approaches: xdy = (x^5 + x^3*y^2 + y)dx (y + x)dy = (y - x)dx xdy = (y + x^2 + 9y^2)dx I've checked that they aren't exact and tried IF e^(Integral P dx) and similar IF for y. If you can help me I'll be very grateful !! Date: 10/28/2003 at 11:18:48 From: Doctor Korsak Subject: Re: Differential equations Hello David, Integrating factors can be difficult to discover, so there are various tricks you can try, like the exp(integral P(x)dx) factor. That is a good one to use for linear differential equations of first order, but none of your three cases is linear. I would be curious to know what methods are covered up to where you are in the book for the above examples. In my old college book, A First Course in Differential Equations by Rudolf E. Langer, some other tricks mentioned are: Collecting terms into groups, and looking for something "recognizably transformable into a differential by some suitable multiplier". Changes of variables: x = f(s,u) y = g(s,u). Equations with homogeneous coefficients, for which the substitution x = s y = su or x = su y = u turns the equation into one with separable variables, eg. y^2dx + (x^2 - xy + y^2)dy = 0 becomes uds + (s^2 + 1)du = 0. Equations with linear coefficients: (a1*x + b1*y + c1)dx + (a2*x + b2*y + c2)dy = 0 Let s = x + h, u = a2*x + b2*y + c2, then the equation becomes (A(s - h) + Bu + C)ds + udu = 0 after collecting terms into expressions represented by A, B, C. If A = 0, the variables are separable; if A =/= 0, let -Ah + C = 0, and that leads to an equation with homogeneous coefficients of degree 1. Replacement by a pair of differential equations: Given P(x,y)dx + Q(x,y)dy = 0, let x and y be functions of t: dx/dt = Q(x,y) dy/dt = -P(x,y), or using an integrating factor dx/dt = k(x,y)Q(x,y) dy/dt = -k(x,y)P(x,y) if that helps to solve for x in terms of t and y or y in terms of x and t. Then elimination of t results in an equation relating x and y. If you are still having trouble I can help further--let me know. - Doctor Korsak, The Math Forum http://mathforum.org/dr.math/ Date: 10/28/2003 at 11:57:02 From: David Subject: Thank you (Differential equations) Dr. Korsak, Thanks for the assistance - I'll try the methods given and see how I get on. Also, thanks for leaving me with a challenge and not just solving the problems. Cheers, David |
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