# Help with this DE problem

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• Jan 31st 2010, 03:22 PM
whitepenguin
Help 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

$y' = \frac{y}{3x - y^2}$

Please give me just a little hint on this.
Thank you
• Jan 31st 2010, 04:50 PM
Krizalid
Put $y=\sqrt xt$ and that'd turn the ODE into a separable one.
• Jan 31st 2010, 04:56 PM
whitepenguin
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, 05:29 PM
Krizalid
mm, i didn't know, i just made that substitution and it worked out! :)
• Jan 31st 2010, 05:48 PM
whitepenguin
So I guess I'm pretty dumb, coz I've been thinking about that since yesterday...(Crying)
• Jan 31st 2010, 05:49 PM
Krizalid
no, don't say that.

apply the substitution, what do you get?
• Jan 31st 2010, 06:43 PM
whitepenguin
Yes, I was be able to solve the problem using your substitution.
so.. the seprable form is this...

$\frac {6 - 2v^2}{v-3+v^2}dv = \frac {dx}{x}$

Thank you.

BTW.
I got another I.V.P where I found $c = \sqrt4$ , I end up with c= +-2
which one should I choose?
• Jan 31st 2010, 06:45 PM
Krizalid
$\sqrt4=2,$ and not $-2.$

did you stop to think how a positive function could deliver a negative value?
• Jan 31st 2010, 06:51 PM
whitepenguin
Quote:

Originally Posted by Krizalid

did you stop to think how a positive function could deliver a negative value?

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, 09:30 PM
snaes
general 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 $x^ay^b$. where "a" and "b" are constants.

Get equation in exact form:
$(M)dx+(N)dy=0$
$(-y)dx+(3x-y)dy=0$

Make this fraction =1

$\dfrac{M_y-N_x}{N\dfrac{a}{x}-M\dfrac{b}{y}}=1$

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 $x^ay^b$.
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, 09:32 PM
Krizalid
that's funny, you got the same result as me and i've never studied well exact equations. :D
• Feb 2nd 2010, 05:00 AM
Calculus26
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, 10:36 AM
whitepenguin
Quote:

Originally Posted by snaes
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 $x^ay^b$. where "a" and "b" are constants.

Get equation in exact form:
$(M)dx+(N)dy=0$
$(-y)dx+(3x-y)dy=0$

Make this fraction =1

$\dfrac{M_y-N_x}{N\dfrac{a}{x}-M\dfrac{b}{y}}=1$

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 $x^ay^b$.
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.

Hi, can you put it in a little detail? How do you get $3x-y^2$ to $3x - y$
• Feb 2nd 2010, 10:44 AM
whitepenguin
Quote:

Originally Posted by Calculus26
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

So you make this look like a linear eqn?
I thought it should be dy/dx?
Im confused tho, coz y is a function has x .... or may be i'm wrong
Can you give me an explain
Thanks
• Feb 2nd 2010, 02:13 PM
Calculus26
I just inverted the equation to get x as a fn of y to make the solution a simple linear ODE.
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