2 Attachment(s)

Another "mixed partials" question

The question and my work are both attached as PDF files.

Basically, in my work I only found the partial derivatives to be different and therefore did not proceed because of this but apparently this is wrong. Could someone help me figure out what I am doing wrong?

Any help would be greatly appreciated!

Thanks in advance!

Re: Another "mixed partials" question

This time, your $\displaystyle \partial N/\partial x$ should have been -3. You were asked to find an integrating factor **in order to make it** exact. It doesn't appear as though you did that. I'd suggest an integrating factor of the form $\displaystyle h=y^{n}.$ You might ask how I found that. I started by assuming an integrating factor of the form $\displaystyle h(x^{m}y^{n}),$ and then using the exactness condition to solve a differential equation for $\displaystyle h.$

Re: Another "mixed partials" question

Could you please elaborate about how I find the an integrating factor to make equations exact because I am breaking my head with this and I am still confused with what you said (I've been rereading it and watching YouTube videos, etc)? Does your way always work? (Assuming that an integrating factor does exist).

Assuming, I followed the procedure correctly, the method here: Integrating factors 1 - YouTube did not work for me for this problem by the way.

Re: Another "mixed partials" question

Here's the basic idea, and this will work for a wide variety of first-order ODE's, but certainly not all first-order ODE's. You multiply through the DE by $\displaystyle h(x^{m}y^{n}),$ and then assert the exactness condition. In your case, you have

$\displaystyle (y-3y^{4})\,dx=(y^{3}+3x)\,dy,$ or

$\displaystyle (y^{3}+3x)\,dy+(3y^{4}-y)\,dx=0,$ and hence

$\displaystyle h(x^{m}y^{n})(y^{3}+3x)\,dy+h(x^{m}y^{n})(3y^{4}-y)\,dx=0.$

Asserting the exactness condition yields that

$\displaystyle h'(x^{m}y^{n})(mx^{m-1}y^{n})(y^{3}+3x)+h(x^{m}y^{n})(3)$

$\displaystyle =h'(x^{m}y^{n})(nx^{m}y^{n-1})(3y^{4}-y)+h(x^{m}y^{n})(12y^{3}-1).$

So you whittle things down and simplify, etc., etc., etc. You will often, at some point, have an option to choose one of the exponents, $\displaystyle m$ or $\displaystyle n$, in order to simplify things greatly. That's a bit of an art. The three options I would look at first are n=0, m=0, or n=m. Your goal is to solve this first-order differential equation for $\displaystyle h.$ Then that's your integrating factor.