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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 February 27th, 2010, 06:01 PM #1 Member   Joined: Jan 2010 Posts: 32 Thanks: 0 First-Order Differential Equations Need help with below two First-Order differential equation problems: Problem# 1) xy (1+ xy^2) (dy/dx) = 1 Problem# 2) 2(dy/dx) = (y/x) - (x/y^2) Appreciate your help so much. Thank you. February 28th, 2010, 07:18 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 # 1)� xy(1 + xy�)dy/dx = 1 This can be solved using a special function, but can you check it's typed correctly before I explain? # 2)� 2dy/dx = y/x - x/y� Let u = y� + 3x�, so that du/dx = 3y�dy/dx + 6x = (3/2)(y�/x - x) + 6x = (3/2)(u/x - 3x - x) + 6x = (3/2)u/x. Hence x^(-3/2)du/dx - (3/2)x^(-5/2)u = 0. Integrating gives x^(-3/2)u = C, i.e., u = Cx^(3/2), so y� + 3x� = Cx^(3/2). March 1st, 2010, 03:17 AM #3 Member   Joined: Jan 2010 Posts: 32 Thanks: 0 Re: First-Order Differential Equations Thanks a ton. Yes, I problem # 1) is typed correctly. Thanks for solving problem# 2) March 1st, 2010, 04:55 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 Note that what I posted for #2 included a hidden assumption that x > 0. You can easily modify it to cover x < 0. I don't think it's possible for y to be differentiable at x = 0. # 1)� xy(1 + xy�)dy/dx = 1 Let t = 1/x - 2, so that x = 1/(t + 2) and dx/dt = -1/(t + 2)�. (y/(t + 2))(1 + y�/(t + 2))dy/dt = dx/dt = -1/(t + 2)�. y(y� + t + 2)dy/dt = -1. Let u� = y� + t, so that 2udu/dt = 2ydy/dt + 1. (u� + 2)(2udu/dt - 1) = -2. 2u(u� + 2)du/dt - u� = 0. u = 0 or 2(u + 2/u)du/dt = 1. Integrating the latter equation gives u� + 4ln|u| = t + C. From that, it's possible to get u� in terms of t by using the Lambert W function. More on this later. The equation u = 0 leads to the solutions y = �?(2 - 1/x), where x < 0 or x > 1/2. March 1st, 2010, 10:29 AM #5 Member   Joined: Jan 2010 Posts: 32 Thanks: 0 Re: First-Order Differential Equations Thanks so much for your help! Just a quick question though... How did you know to use the substitution t = 1/x - 2 ? Do we just use guess and check or is there some sort of strategy? March 1st, 2010, 04:43 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,978 Thanks: 2229 I don't know of a simple strategy, but it's often worth trying various substitutions in a slightly generalized form, such as x = at^b, and then choosing values of a, b that seem to help. A slight change to the original equation may affect its difficulty substantially, but equations that are set as exercises or in examinations tend to be ones that can be done without excessive difficulty. Tags differential, equations, firstorder Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jakeward123 Differential Equations 6 March 17th, 2011 03:23 PM riotsandravess Differential Equations 4 December 6th, 2010 08:20 AM diegosened Differential Equations 1 May 16th, 2010 09:58 AM pksinghal Differential Equations 0 December 31st, 1969 04:00 PM diegosened Differential Equations 1 December 31st, 1969 04:00 PM

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