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anonymous
 one year ago
I'm trying to do question 11, p. 181. The answer is in the book, but I don't see how to get there.
anonymous
 one year ago
I'm trying to do question 11, p. 181. The answer is in the book, but I don't see how to get there.

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phi
 one year ago
Best ResponseYou've already chosen the best response.0Can you post a screen shot, photo, or type in the exact problem? (and their answer)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ydy/dx=x((y^4)+2y^2)+!). They want a particular solution for y=1 when x=4. The books's answer is y = sqrt(((2x^2)31)/(332x^2)).

phi
 one year ago
Best ResponseYou've already chosen the best response.0is the problem \[ y \frac{dy}{dx} = x ( y^4 +2y^2 +1) \] ? If so, put the y's on one side and x on the other: \[ \frac{y}{y^4 +2y^2 +1} \ dy = x \ dx\] the denominator is a perfect square, so write it as \[ \frac{y\ dy }{(y^2+1)^2} = x \ dx\] or \[ \frac{1}{2}(y^2+1)^{2} 2y \ dy = x \ dx\]

phi
 one year ago
Best ResponseYou've already chosen the best response.0notice if we let u= y^2+1 then du = 2y dy and the left side is equivalent to \[ \frac{1}{2} u^{2} du \] which we can integrate

phi
 one year ago
Best ResponseYou've already chosen the best response.0ok, I do get the book's answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you, phi! I will try to finish now. You have been a great help! I had some of those ideas, but I could not quite put it all together. I hope I can get the rest of it on my own. Many thanks. I've been stewing over that problem for a while!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It took a while, but I got it. Your help got me unstuck. Thanks again!
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