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February 27th, 2010, 06:01 PM  #1 
Member Joined: Jan 2010 Posts: 32 Thanks: 0  FirstOrder Differential Equations
Need help with below two FirstOrder 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 + xy²)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/y² Let u = y³ + 3x², so that du/dx = 3y²dy/dx + 6x = (3/2)(y³/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 y³ + 3x² = Cx^(3/2). 
March 1st, 2010, 03:17 AM  #3 
Member Joined: Jan 2010 Posts: 32 Thanks: 0  Re: FirstOrder 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 + xy²)dy/dx = 1 Let t = 1/x  2, so that x = 1/(t + 2) and dx/dt = 1/(t + 2)². (y/(t + 2))(1 + y²/(t + 2))dy/dt = dx/dt = 1/(t + 2)². y(y² + t + 2)dy/dt = 1. Let u² = y² + t, so that 2udu/dt = 2ydy/dt + 1. (u² + 2)(2udu/dt  1) = 2. 2u(u² + 2)du/dt  u² = 0. u = 0 or 2(u + 2/u)du/dt = 1. Integrating the latter equation gives u² + 4lnu = t + C. From that, it's possible to get u² in terms of t by using the Lambert W function. 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: FirstOrder 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.


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