Recent content by Gummy Bear

could somebody at least give me something to go off of? This isn't making any sense.

Homework Statement (a) In Example 1.18, assume that a is less than b (so that k is less than 1) and find y as a function of x. How far does the rabbit run before the dog catches him? (b) Assume now that a=b, and find y as a function of x. How close does the dog come to the rabbit? Homework...

-ln(cos(c-y))+d That's what I keep getting. Maybe that's right, I just assumed the answer was supposed to be -ln(cos(x+c))+d

This makes no sense. The inverse is -tan(c-p)

Doesn't this give y=∫1/(1+p2)dp=arctan(p)+c The way I understood it is that both methods should get you the same answer. I know that the answer should be -ln(cos(x+c))+d

Because I'm not sure where these R's and Q's are coming from. I know that typically in the second method y'=p and y''=p*(dp/dy)

Could you show me what it would look like? For example, my first equation looks like p'=1+p2 after reduction of order.

I'm not following. What do you mean by P(p)=P(y')=y? And transforming it?

Ya, I figured that out after asking. But how do I check that both -x3/3+1 and 1 work?

Oh, so now I say because y(0)=1 0/3+c2=1 so, c2=1? Therefore, y=-x3/3+1 What do you mean don't use the same constant? Are you referring to y=∫0dx=c1? What do I do with the y=∫0dx=c1 anyway?

Wow. I got it mixed up with the problem out of the book. It's been a long day. You're right.. So all I have to do is: y=∫0dx=c y=∫(-x2)dx=(-x3)/3+c Therefore, y=∫(-x2)dx=(-x3)/3+c is the answer? Thanks for helping me by the way.

Homework Statement Solve y''=1+(y')2 in two ways, for x is missing and y is missing. Homework Equations Integration, and reduction of order. The Attempt at a Solution First method: (this is correct) y''=1+(y')2 let y'=p and y''=p' p'=1+p2 p'/(1+p2)=1 ∫p'/(1+p2)dx=∫1dx arctan(p)=x+c solve...

I'm confused, you're substituting back for p before solving for p? Also, y'(0)≠0 y'(0)=1. Does that mean my equation should be: y'2+x2y'=1 And from there solve for y?

Homework Statement (x2+2y')y''+2xy'=0 with initial conditions of y(0)=1 and y'(0)=0 Homework Equations reduction of order, and exact equations.. The Attempt at a Solution let y'=p and y''=p' Giving: (x2+2p)p'+2xp Let M=2xp and N=(x2+2p) So, Mp=2x and Nx=2x Therefore, they are...

Why won't anyone help me?! I need to know how to solve this problem before class!