Hello

problem:

Solve :

It must be solved using inspection

which is , ... etc

But this equation is bothering me

I tried as much I can, and it seems impossible!

any help??

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- Feb 9th 2011, 03:25 AMLiverpoolSolve the following ODE .. #7
Hello

problem:

Solve :

It must be solved using inspection

which is , ... etc

But this equation is bothering me

I tried as much I can, and it seems impossible!

any help?? - Feb 9th 2011, 04:48 AMAckbeet
y = 0 is a perfectly good solution.

- Feb 9th 2011, 05:16 AMLiverpool
I did not say that I want a particular solution.

- Feb 9th 2011, 05:37 AMProve It
You would probably need an integrating factor.

- Feb 9th 2011, 05:42 AMLiverpool
I can find this.

But the problem is that it should be solved using inspection. - Feb 9th 2011, 07:06 AMAckbeet
Well, aside from the trivial solution that I gave you before, nothing is obvious. It's not linear in either x or y, it's not exact, homogeneous, or separable. And I don't see any obvious substitution. All of this means that, unless you can manipulate the equation to the point where you can do a direct integration, you may not be able to find a general solution at all. I certainly see no possibility for a solution

*by inspection*, except for what I have already given.

I tried a quotient rule with a function of two variables thus: make

Carrying out the differentiation and comparing to the original DE yields the two equations

and

Unfortunately, it's impossible to make those two equations work, as far as I can tell. You're close, actually, very close. I get

But plugging this into the second equation above yields

which is off by a minus sign and a fraction of 1/2.

This approach might be made to work, but I'm not seeing how. - Feb 9th 2011, 08:41 AMGeneral

Re-arrange:

Taking common factors:

Or:

Devide by :

Or:

Integrate. - Feb 9th 2011, 08:44 AMAckbeet
- Feb 9th 2011, 08:49 AMGeneral
Practice makes perfect =)

- Feb 9th 2011, 08:51 AMAckbeet
I guess so! Incidentally, integration of your brilliant equation there yields a 1-parameter family of solutions, for which the solution y = 0 is a singular solution.

Cheers. - Feb 11th 2011, 11:06 AMLiverpool
Thanks !!

- Feb 11th 2011, 11:14 AMAckbeet
Well, you're certainly welcome for my contribution, but you can obviously see that General did the heavy lifting.

- Feb 11th 2011, 11:54 AMLiverpool
Actually, the "thanks" for General not for you xD

but thank you anyway :p - Feb 11th 2011, 11:57 AMAckbeet
- Feb 11th 2011, 11:58 AMLiverpool
I See.

Thanks all.