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 April 11th, 2016, 03:08 AM #1 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.
April 11th, 2016, 04:09 AM   #2
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 Originally Posted by kkg 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.

Last edited by skipjack; April 11th, 2016 at 09:17 AM.

April 11th, 2016, 06:27 AM   #3
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 Originally Posted by Country Boy 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.

 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.

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