Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 23rd, 2007, 06:46 PM #1 Newbie   Joined: Dec 2006 Posts: 9 Thanks: 0 Differential Equation Using the substitution x=z^(1/2), transform the differential equation d^2y/dx^2 + (4x - 1/x)dy/dx + 4yx^2 = 0 into one relating y and z. hence find y in terms of x given that dy/dx = -2 and y = 2 when x=1. March 24th, 2007, 01:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 21,128 Thanks: 2337 Re: Differential Equation z = x� so dz/dx = 2x and dy/dx = (dy/dz)(dz/dx) = 2x dy/dz, so d�y/dx� = 2dy/dz + 4x� d�y/dz�. When x = z = 1, dy/dx = -2 so dy/dz = -1. Substituting into the differential equation gives 2dy/dz + 4x� d�y/dz� + (8x� - 2)dy/dz + 4yx� = 0, i.e., d�y/dz� + 2dy/dz + y = 0. Multiplying by e^z and integrating gives e^z dy/dz + e^zy = e (since dy/dz = -1 and y = 2 when x = z = 1). Integrating again gives e^z y = ez + e (since y = 2 when x = z = 1). Hence y = (x� + 1)e^(1-x�). Tags differential, equation 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 PhizKid Differential Equations 0 February 24th, 2013 11:30 AM cheyb93 Differential Equations 3 February 7th, 2013 10:28 PM mathkid Differential Equations 0 October 9th, 2012 09:01 AM tomislav91 Differential Equations 1 May 30th, 2012 08:28 AM tsl182forever8 Differential Equations 2 March 14th, 2012 04:12 PM

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