My Math Forum Differential Equation

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 23rd, 2007, 05:46 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, 12:00 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,467 Thanks: 2038 Re: Differential Equation z = x² so dz/dx = 2x and dy/dx = (dy/dz)(dz/dx) = 2x dy/dz, so d²y/dx² = 2dy/dz + 4x² d²y/dz². When x = z = 1, dy/dx = -2 so dy/dz = -1. Substituting into the differential equation gives 2dy/dz + 4x² d²y/dz² + (8x² - 2)dy/dz + 4yx² = 0, i.e., d²y/dz² + 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 = (x² + 1)e^(1-x²).

 Tags differential, equation

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post PhizKid Differential Equations 0 February 24th, 2013 10:30 AM cheyb93 Differential Equations 3 February 7th, 2013 09:28 PM mathkid Differential Equations 0 October 9th, 2012 08:01 AM tomislav91 Differential Equations 1 May 30th, 2012 07:28 AM tsl182forever8 Differential Equations 2 March 14th, 2012 03:12 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top