Hi

I got a problem in DE too

I checked all the possible methods that I've learned so far includes Linear, Homogeneous,Bernoulli, and exact equation however none of them worked for me

Please give me just a little hint on this.

Thank you

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- Jan 31st 2010, 02:22 PMwhitepenguinHelp with this DE problem
Hi

I got a problem in DE too

I checked all the possible methods that I've learned so far includes Linear, Homogeneous,Bernoulli, and exact equation however none of them worked for me

Please give me just a little hint on this.

Thank you - Jan 31st 2010, 03:50 PMKrizalid
Put and that'd turn the ODE into a separable one.

- Jan 31st 2010, 03:56 PMwhitepenguin
Hi, can you tell me how you know to put it that way? Is there any general method for this kind of problem?

Thank you - Jan 31st 2010, 04:29 PMKrizalid
mm, i didn't know, i just made that substitution and it worked out! :)

- Jan 31st 2010, 04:48 PMwhitepenguin
So I guess I'm pretty dumb, coz I've been thinking about that since yesterday...(Crying)

- Jan 31st 2010, 04:49 PMKrizalid
no, don't say that.

apply the substitution, what do you get? - Jan 31st 2010, 05:43 PMwhitepenguin
Yes, I was be able to solve the problem using your substitution.

so.. the seprable form is this...

Thank you.

BTW.

I got another I.V.P where I found , I end up with c= +-2

which one should I choose? - Jan 31st 2010, 05:45 PMKrizalid
and not

did you stop to think how a positive function could deliver a negative value? - Jan 31st 2010, 05:51 PMwhitepenguin
Im not sure what you're asking

an you give me a little explain? ,why is it not negative?

I found the SOLN , and plug back to orginal problem, with both case c=+-2

,looks like both satify the problem.....

I dont really understand...

And what about uniqueness ... ???? I though they said IVP is unique soln, but... I found 2 C's....(Thinking) - Feb 1st 2010, 08:30 PMsnaesgeneral solution
Hi, i read over this post i think i know how to find a "general solution" for this probelm. Its actually a specific case for this problem, but it'll help you find an integrating factor in the form . where "a" and "b" are constants.

Get equation in exact form:

Make this fraction =1

By making "a" and "b" appropriot values this should make the fraction 1. Thereby giving you the values of "a" and "b"to fill in the integrating factor .

This should make the equation turn into an "exact differential equation"

Hope this helps!

*Note: I havent tried this for your specific problem, but this is the solution my professor has taught us and has worked for me on other problems. - Feb 1st 2010, 08:32 PMKrizalid
that's funny, you got the same result as me and i've never studied well exact equations. :D

- Feb 2nd 2010, 04:00 AMCalculus26
For another approach

http://www.mathhelpforum.com/math-he...c06ef73a-1.gif

rewrite

dx/dy = (3x-y^2)/y

dx/dy -(3/y)x = -y

use integrating factor 1/y^3

x/y^3 = 1/y + c

x = y^2 + cy^3 - Feb 2nd 2010, 09:36 AMwhitepenguin
- Feb 2nd 2010, 09:44 AMwhitepenguin
- Feb 2nd 2010, 01:13 PMCalculus26
I just inverted the equation to get x as a fn of y to make the solution a simple linear ODE.