1. ## Separable equation

I have another problem with separable equations, can anyone help?

I need to know how I end up with PI and tan in the answers to these differential equations, its just not making much sense to me at the minute!

dy/dx = 1+y^2/2+4x

which satisfies y(1) = 1

so this becomes

1/1+y^2 dy = 1/2+4x dx

??
Am I right so far?

then does the LHS
become
tan-1 and do you take the log of the RHS?
can anyone help me through the stages??

I have the solution
i just need to see how this works! especially with PI coming into it, how??

tan(PI/4+1/4ln(|2+4x/6|)).....
Any help would be gr8!

2. Originally Posted by dankelly07
I have another problem with separable equations, can anyone help?

I need to know how I end up with PI and tan in the answers to these differential equations, its just not making much sense to me at the minute!

dy/dx = 1+y^2/2+4x

which satisfies y(1) = 1

so this becomes

1/1+y^2 dy = 1/2+4x dx

??
Am I right so far?

then does the LHS
become
tan-1 and do you take the log of the RHS?
can anyone help me through the stages??

I have the solution
i just need to see how this works! especially with PI coming into it, how??

tan(PI/4+1/4ln(|2+4x/6|)).....
Any help would be gr8!
$\displaystyle \frac{dy}{1 + y^2} = \frac{dx}{4x + 2}$

Integrating both sides gives
$\displaystyle tan^{-1}(y) = \frac{1}{4} ln(4x + 2) + C$

I'm not clear on this: Are you asking how to do the integral for the RHS?

-Dan

3. Same problem posted and solved here.

Does really make sense to change the numbers?

4. Originally Posted by Krizalid
Same problem posted and solved here.

Does really make sense to change the numbers?
(Growls) A very good question.

Please review the rule on double posting. (Technically this isn't one, however you are making yourself a tempting target for ThePerfectHacker...)

-Dan

5. sorry about that, I'm just having trouble with a similar kind of problem... well I get the integration and the separating, I was just wondering really how tan and PI get involved in the equation? I understand tan-1 now because
the integral of 1/1+y^2 = tan-1, but still dont understand where Pi is coming from...

6. Originally Posted by dankelly07
sorry about that, I'm just having trouble with a similar kind of problem... well I get the integration and the separating, I was just wondering really how tan and PI get involved in the equation? I understand tan-1 now because
the integral of 1/1+y^2 = tan-1, but still dont understand where Pi is coming from...
You now know that $\displaystyle \tan^{-1}(y) = \frac{1}{4} \ln |4x + 2| + C$.

You're given y(1) = 1. You plug that in to get the value for C:

$\displaystyle \tan^{-1}(1) = \frac{1}{4} \ln |4 + 2| + C$

$\displaystyle \Rightarrow \frac{\pi}{4} = \frac{1}{4} \ln(6) + C$

$\displaystyle \Rightarrow \pi = \ln(6) + 4C$

$\displaystyle \Rightarrow C = \frac{\pi - \ln(6)}{4}$.

Therefore:

$\displaystyle \tan^{-1}(y) = \frac{1}{4} \ln |4x + 2| + \frac{\pi - \ln(6)}{4}$

$\displaystyle \Rightarrow \tan^{-1} (y) = \frac{1}{4} \left( \ln |4x + 2| + \pi - \ln (6) \right) = \frac{1}{4} \left( \ln\left| \frac{4x + 2}{6}\right| + \pi \right) = \frac{\pi}{4} + \frac{1}{4} \ln\left| \frac{2x + 1}{3}\right|$.

If you're working at this level, where the $\displaystyle \pi$ comes from really shouldn't be an issue .....