My Math Forum  

Go Back   My Math Forum > College Math Forum > Differential Equations

Differential Equations Ordinary and Partial Differential Equations Math Forum


Thanks Tree2Thanks
  • 1 Post By harley05
  • 1 Post By harley05
Reply
 
LinkBack Thread Tools Display Modes
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.
harley05 is offline  
 
March 29th, 2015, 05:36 PM   #2
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,634
Thanks: 2620

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.
v8archie is offline  
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.
harley05 is offline  
April 27th, 2015, 08:23 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 20,464
Thanks: 2038

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.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Differential Equations

Tags
ode, solution



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
what is the pOH of a solution that has a pH of 3.847? girlbadatmath Chemistry 1 December 19th, 2014 03:49 AM
Why my solution is different from sample solution? rain Real Analysis 1 July 17th, 2013 12:28 PM
Solution Anyone ? Wannalearn Algebra 8 February 29th, 2012 09:50 PM
my correct solution but another solution? davedave Calculus 1 January 31st, 2012 02:48 PM
trying to make a solution (liquid solution...) problem... TreeTruffle Algebra 2 March 27th, 2010 01:22 AM





Copyright © 2019 My Math Forum. All rights reserved.