My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Thanks Tree1Thanks
  • 1 Post By Country Boy
Reply
 
LinkBack Thread Tools Display Modes
April 11th, 2016, 03:08 AM   #1
kkg
Newbie
 
Joined: Apr 2016
From: UK

Posts: 10
Thanks: 0

Differential Equation Help

Hi, I'd really appreciate any help with this.

Obtain the solution of:

\[2\cot x\frac{\text{d}y}{\text{d}x} = (4-y^2)\]

for which y=0 at x=pi/3 giving your answer in the form \[\sec^2x=g(y)\]

Here's my attempt at a solution:

\[\int_{}^{}\frac{2}{4-y^2}\text{d}y=\int_{}^{}\tan(x)\text{d}x \]

\[\int_{}^{}\tan x\text{d}x = \ln|\sec x| +c \]

\[\int_{}^{}\frac{2}{4-y^2}\text{d}y = \frac{1}{2}\int_{}^{}\frac{1}{y+2}-\frac{1}{y-2} dy

\]
\[\frac{1}{2}(\ln|y+2|-\ln|y-2|) = \ln|\sec x| + C


\]
\[\ln|y+2|-\ln|y-2| = \ln|\sec^2x| +C


\]

Not sure where to go from there or even whether what I've done so far is correct. It'd be nice to know what the answer actually is as well so that I can see if I can get it right.

Thanks.

Last edited by skipjack; April 11th, 2016 at 09:16 AM.
kkg is offline  
 
April 11th, 2016, 04:09 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
Originally Posted by kkg View Post
Hi, I'd really appreciate any help with this.

Obtain the solution of:

\[2\cot x\frac{\text{d}y}{\text{d}x} = (4-y^2)\]

for which y=0 at x=pi/3 giving your answer in the form \[\sec^2x=g(y)\]

Here's my attempt at a solution:

\[\int_{}^{}\frac{2}{4-y^2}\text{d}y=\int_{}^{}\tan(x)\text{d}x \]

\[\int_{}^{}\tan x\text{d}x = \ln|\sec x| +c \]

\[\int_{}^{}\frac{2}{4-y^2}\text{d}y = \frac{1}{2}\int_{}^{}\frac{1}{y+2}-\frac{1}{y-2} dy

\]
\[\frac{1}{2}(\ln|y+2|-\ln|y-2|) = \ln|\sec x| + C


\]
\[\ln|y+2|-\ln|y-2| = \ln|\sec^2x| +C


\]

Not sure where to go from there or even whether what I've done so far is correct. It'd be nice to know what the answer actually is as well so that I can see if I can get it right.

Thanks.
You've done very well. Now I presume you know the "laws of logarithms", in particular, log(a)+ log(b)= log(ab) and log(a/b)= log(a)- log(b). So $\displaystyle \ln|y+ 2|- \ln|y- 2|= \ln\left|\frac{y+2}{y- 2}\right|= \ln C'|\sec^2(x)|$ where C= ln(C').

Now, recall that ln(x) is the "inverse" to $\displaystyle e^x$. That is [math]e^{\ln(a)}= a[math] so taking the exponential of both sides $\displaystyle \frac{y+ 2}{y- 2}= C' \sec^2(x)$. We can drop the absolute value signs because C' itself can be taken to be positive or negative.
Thanks from kkg

Last edited by skipjack; April 11th, 2016 at 09:17 AM.
Country Boy is offline  
April 11th, 2016, 06:27 AM   #3
kkg
Newbie
 
Joined: Apr 2016
From: UK

Posts: 10
Thanks: 0

Quote:
Originally Posted by Country Boy View Post
You've done very well. Now I presume you know the "laws of logarithms", in particular, log(a)+ log(b)= log(ab) and log(a/b)= log(a)- log(b). So $\displaystyle \ln|y+ 2|- \ln|y- 2|= \ln\left|\frac{y+2}{y- 2}\right|= \ln C'|\sec^2(x)|$ where C= ln(C').

Now, recall that ln(x) is the "inverse" to $\displaystyle e^x$. That is $\displaystyle e^{\ln(a)}= a$ so taking the exponential of both sides $\displaystyle \frac{y+ 2}{y- 2}= C' \sec^2(x)$. We can drop the absolute value signs because C' itself can be taken to be positive or negative.
Hi, thanks for the help!

Can I just ask why is it that C=ln(C') and it isn't just + C?

Last edited by skipjack; April 11th, 2016 at 10:17 AM.
kkg is offline  
April 11th, 2016, 10:36 AM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 20,975
Thanks: 2224

As y = 0 at x = $\pi$/3, C' = -1/4.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
differential, equation



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Differential equation max233 Calculus 4 March 26th, 2016 03:21 AM
Gompertz equation - differential equation Sonprelis Calculus 6 August 6th, 2014 10:07 AM
Show that an equation satisfies a differential equation PhizKid Differential Equations 0 February 24th, 2013 10:30 AM
Help with a differential equation pepeatienza Differential Equations 1 May 13th, 2008 01:14 PM
Differential Equation Help! jwade456 Differential Equations 1 May 8th, 2008 12:14 PM





Copyright © 2019 My Math Forum. All rights reserved.