## 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.

1. phi

Can you post a screen shot, photo, or type in the exact problem? (and their answer)

2. anonymous

ydy/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)/(33-2x^2)).

3. phi

is 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 $\frac{1}{2}(y^2+1)^{-2} 2y \ dy = x \ dx$

4. phi

notice 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

5. phi

ok, I do get the book's answer

6. anonymous

Thank 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!

7. anonymous

It took a while, but I got it. Your help got me unstuck. Thanks again!

8. phi

excellent