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I HAVE SOLVED THE EXERCISE. I'M LIKELY TO WRITE AN ANSWER LATER

The original question follows:

Find the value of b for which the differential equation below is exact. $$(ye^{7xy}+x)dx+(bxe^{7xy})dy=0$$ Then solve it using that value of b. An implicit solution is fine. My attempt uses $$M=ye^{7xy}+x\text{ and } N= bxe^{7xy}$$

$ \frac{M_y-N_x}{M}= -\frac{7y}{b}+\frac{1}{bx}-\frac1x +7x$
I can use $b= \frac{1}{1+7x}$ to drop out the x's, but it reintroduces x's, in the term $\frac{7y}{b}=\frac{7y}{\frac{1}{1+7x}}$

Also, if b contains x's, when we take the partial, $N_x$ should I introduce $b_x$? I'm working on dividing the original equation by b and will see what happens. Then I'll take $M_y$, which is unaffected by b(x). $ \frac{M_y-N_x}{M}$ hopefully will have b only in x terms.

Some context- I'm a tutor, with a B.S. in Math. I don't have precise resources because this is for a student. I did find an essentially identical problem in the Boyce, Diprima and Meade textbook, 11th edition, titled "Elementary Differential Equations and Boundary Value Problems". Apparently, it comes from Section 2.6, page 89, problem number 12. This is a first semester introductory course to ODE's.

Also, I'm having trouble googling these type of exercises because I get primarily simpler exact equations. Is there a name for exact ODE equations based on $\frac{M_y-N_x}{M}$ or it's counterpart?

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    $\begingroup$ With $b=1$ this is exact without integrating factor. $\endgroup$ Feb 12 at 10:54
  • $\begingroup$ hmmmm, really? Wow, I've 4 pages of overcomplicating it. My student kept insisting on using $\frac{M_y-N_x}{M}$ or the counterpart as part of finding the integrating factor. I'll work it. Can one sentence count as an answer? LOL. $\endgroup$
    – nickalh
    Feb 12 at 11:00
  • $\begingroup$ Comparing similar terms, one gets $...=\frac12d(x^2)+e^{7(xy)}d(xy)$. The mentioned formula is for integrating factors depending on $x$, while structurally here an integrating factor depending on $xy$ is needed. $\endgroup$ Feb 12 at 11:07
  • $\begingroup$ Ohh, I caught a mistake. I think my numerator cancels to zero now. $\endgroup$
    – nickalh
    Feb 12 at 11:09
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    $\begingroup$ I totally forgot the partial integral applied to each coeffecient. I have the correct answer. Thanks so much for your help. You helped me catch two mistakes. Should I write up my answer and post an answer to my own question? $\endgroup$
    – nickalh
    Feb 12 at 11:42

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