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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 March 29th, 2015, 04:51 PM #1 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 Ode for particular solution Hello, not sure if I'm going about this correctly, see below. 1) Find the general solution of: y'=(2x + cotan x)/3y^2 2) determine the integration constant using $\displaystyle y(π/2)$=0 3) present the particular solution subject to this initial condition in explicit form. Answer:Part (1) dy/dx = 1/3(2x + cot x)y^2 separating: 3dy/y^2 = 2x + cot x Integrating: -3/y = x^2 + ln(sin x) +C General solution is: y = -3/(x^2 + ln(sin x)+C) Answer: Part 2 do I just plug $\displaystyle π/2$ into the above general solution when y = 0? 0 = -3/(π/2^2 + ln(sin π/2) + C) Thanks in advance. Thanks from Yury Stepanyants Last edited by skipjack; March 29th, 2015 at 06:11 PM. March 29th, 2015, 05:36 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra I think you have separated your variables wrongly. Your method looks sound though. Last edited by skipjack; March 29th, 2015 at 06:12 PM. March 29th, 2015, 06:21 PM #3 Member   Joined: Apr 2014 From: australia Posts: 68 Thanks: 32 Thanks for pointing out my error; Answer: Part (1) dy/dx = (2x+cotx(/(3y^2) separating: 3y^2dy = (2x + cotx)dx Integrating: y^3 = x^2 +ln(sinx) +C General solution is: y = [(x^2+ln(sinx)+C)]^1/3 Answer: Part 2 do I just plug $\displaystyle π/2$ into the above general solution when y =0? 0 = [($\displaystyle π/2$^2+ln(sin$\displaystyle π/2$)+C)]^1/3 Thanks in advance. Thanks from Yury Stepanyants Last edited by skipjack; April 27th, 2015 at 08:22 AM. April 27th, 2015, 08:23 AM #4 Global Moderator   Joined: Dec 2006 Posts: 21,034 Thanks: 2271 You're nearly right. If we assume that 0 < x < $\pi$ and y is real, y = (x² + ln|sin(x)| + C)^(1/3). One needs C = -$\pi$²/4 for y to be zero when x = $\pi$/2. It's advisable (in general) to determine the constant of integration before determining the solution in explicit form. Although it doesn't matter for this problem, the way the problem is divided into parts suggests they expect you to do that. Tags ode, solution 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 girlbadatmath Chemistry 1 December 19th, 2014 03:49 AM rain Real Analysis 1 July 17th, 2013 12:28 PM Wannalearn Algebra 8 February 29th, 2012 09:50 PM davedave Calculus 1 January 31st, 2012 02:48 PM TreeTruffle Algebra 2 March 27th, 2010 01:22 AM

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