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December 3rd, 2015, 04:11 PM   #1
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Separation of variables

I am not sure whether I did this correctly. If the separation of variables method doesn't work, I can use the exact method.
Please confirm or help me on my problem, thank you!!




Last edited by skipjack; December 9th, 2015 at 07:08 AM.
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December 3rd, 2015, 05:09 PM   #2
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It looks good but for one thing: $\displaystyle \int \frac{du}{u} = \ln | u | $.

-Dan

Last edited by skipjack; December 5th, 2015 at 03:03 AM. Reason: Re-write
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December 3rd, 2015, 05:14 PM   #3
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I see more problems with the question than with your solution. I think you forgot to cross out a $y$ in line 2.
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December 4th, 2015, 09:55 AM   #4
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Your original equation is given as $\displaystyle \frac{dy}{dx}= (x^2+ 1)(\tan(y))y'x$.

What does the y' mean on the right?

Last edited by skipjack; December 5th, 2015 at 03:04 AM.
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December 6th, 2015, 08:41 AM   #5
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I have the same problem with the question as well. I felt as if the problem was written incorrectly, but my professor assured me it's fine. The y' is y prime? I honestly think the problem is just messed up and I really don't want to lose any points on this one :/ If anyone sees this and can confirm my question/answer that would be awesome.

Last edited by skipjack; December 9th, 2015 at 07:02 AM.
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December 6th, 2015, 08:44 AM   #6
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Quote:
Originally Posted by Country Boy View Post
Your original equation is given as $\displaystyle \frac{dy}{dx}= (x^2+ 1)(\tan(y))y'x$.

What does the y' mean on the right?
Sorry forgot to quote you for the tag.
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December 9th, 2015, 06:43 AM   #7
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Quote:
Originally Posted by Njprince94 View Post
I have the same problem with the question as well. I felt as if the problem was written incorrectly, but my professor assured me it's fine. The y' is y prime? I honestly think the problem is just messed up and I really don't want to lose any points on this one :/ If anyone sees this and can confirm my question/answer that would be awesome.
Didn't you ask your professor what y' meant? Just saying "y prime" doesn't say anything! That's like saying "x is the lower case letter ex"!

Last edited by skipjack; December 9th, 2015 at 07:04 AM.
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December 9th, 2015, 07:20 AM   #8
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Quote:
Originally Posted by Njprince94 View Post
Please confirm or help me on my problem
The equation appears to end with $y'\!x\!$, but that doesn't appear in your work. Sometimes, $y'$ is used to mean $\dfrac{dy}{dx}$, but that wouldn't make sense here. Perhaps $y'\!x$ was intended to be $y/x$. Was anyone else also given this problem, but given some other explanation?
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