- #1

- 45

- 3

4xy'+y=x

Am I missing something? I got 4xyy'+y=x instead.

Very rarely have I ever required starting a thread in a forum to resolve something as seemingly trivial as this but I just want to verify whether this is a typo or not.

Here are the steps I took:

The original is: (y-x)dx+(4xy)dy = 0

From there I got:

(4xy)dy=(x-y)dx

then

(4xy)dy/dx = x-y

then

(4xy)y'+y=x or 4xyy'+y=x