
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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November 14th, 2010, 09:59 AM  #1 
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Differential Equations I need assistance with part 1, please. For part 2, I just need it to be checked. ================================================== ================================================== =============== [color=#0080FF][2][/color] [color=#0080FF][1][/color] [color=#0080FF][1] + [2][/color] [color=#0080FF][1][/color] when 0 = A when t = 0 
November 14th, 2010, 12:30 PM  #2 
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Re: Differential Equations
No help? Is it too long? 
November 14th, 2010, 01:31 PM  #3 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Differential Equations
Today is my sabbath, so I don't help anyone except myself! But seriously... The first one in the second part is correct. That's all I had time to look at. 
November 14th, 2010, 03:46 PM  #4  
Senior Member Joined: Oct 2010 Posts: 126 Thanks: 0  Re: Differential Equations Quote:
 
November 14th, 2010, 06:11 PM  #5 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2159 
(1) dy/dx + y/x = y³ (i) Multiplying by the integrating factor 2/(x²y³) gives 2(x(dy/dx) + y)/(xy)³ = 2/x², which can be integrated to give 1/(xy)² = 2/x + C, etc. What was that about not being able to use an integrating factor? (ii) Differentiating 1/y² = 2z w.r.t. x gives (2/y³)(dy/dx) = 2dz/dx, so (2/y³)(y³  y/x) = 2dz/dx, i.e., 2(1  1/(xy²)) = 2dz/dx, so dz/dx  2z/x = 1. (iii) Let 1/y² = 2z, so that dz/dx  2z/x = 1 and z = 1/8 when x = 1. Dividing by x² and integrating gives z/x² = 1/x + 7/8, so 1/(xy)² = 2/x + 7/4. In hindsight (to avoid division by zero), I'll assume x > 0. By rearranging, one gets y = 2/?(8x  7x²), so I'll also assume x < 8/7. 
November 14th, 2010, 06:15 PM  #6  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Differential Equations Quote:
, do some substitutions..... iii) So, solve at first. Conditions, , so, , solution is or  
November 14th, 2010, 06:31 PM  #7  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Quote:
dz/dx2z/x=x² not the const "1"  
November 14th, 2010, 07:21 PM  #8  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Differential Equations Quote:
Conditions, yields  
November 14th, 2010, 07:23 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,823 Thanks: 2159 
Oops, I integrated 1 instead of 1/x². I've amended my working above to change the x to 1/x and recalculated the constant.

November 14th, 2010, 07:30 PM  #10  
Senior Member Joined: Oct 2010 From: Changchun, China Posts: 492 Thanks: 14  Re: Quote:
 

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