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1st order Differential equation help

Hi,How do I get the differential equation below into the form dy/dx P(x)y=Q(x) so that I can use e^INTEGRALofP(x) as the integrating factor? As there is no y on its own to figre out P(x).Thank you
E3B04046-DC84-4B67-A1B1-A935D796FFD9.png
(edited 1 year ago)
Reply 1
Avatar for Skiwi
Skiwi
13
I'm not sure if it's possible to do so in this question( i haven't actually tried but nothing leaps out immediately, someone else may be able to find a way?)
Have you tried solving it by separating the variables? It seems like the more natural thing to do
Agree. Separate variables.
Reply 3
Avatar for davros
davros
16
Original post by Rhys_M
Hi,How do I get the differential equation below into the form dy/dx P(x)y=Q(x) so that I can use e^INTEGRALofP(x) as the integrating factor? As there is no y on its own to figre out P(x).Thank you
E3B04046-DC84-4B67-A1B1-A935D796FFD9.png

Side note: it always helps if you upload a picture of the original question with any context, so that people can see if you've been given any additional info like what approach to take to solve the DE, or what method to use, or (if it's from a book) which chapter or section it comes from because it may be testing you on a particular method :smile:

Just at a glance it looks like writing u = e^y may help because then du/dx = (du/dy)(dy/dx) = e^y(dy/dx) and you get a simpler DE involving du/dx, u and x, but I may be misleading you here if that's not the intended approach :biggrin:
I just tried it and if you use separation of variables after your 2nd line, it works out really nicely :smile:

It's best to post the original question though if you are able to
(edited 1 year ago)
by inspection the LHS is the exact derivative of x2ey
Original post by the bear
by inspection the LHS is the exact derivative of x2ey


<Hobbit>: I don't think they know about exact derivatives, Bear.

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