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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. 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: 7,675 Thanks: 2655 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. Last edited by skipjack; April 27th, 2015 at 08:22 AM. 
April 27th, 2015, 08:23 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,935 Thanks: 2209 
You're nearly right. If we assume that 0 < x < $\pi$ and y is real, y = (x² + lnsin(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. 

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