john.johnson
New member
- Joined
- Dec 5, 2014
- Messages
- 9
Could someone tell me if have the right answer for this questions?
\(\displaystyle y'\, =\, \dfrac{2x\, -\, e^y}{e^y x \, -\, 1}\)
dy/dx of xe^y-y=x^2-2
Y= (2x-e^y)/(e^yx-1)
Hopefully I typed this correctly...
\(\displaystyle y'\, =\, \dfrac{2x\, -\, e^y}{e^y x \, -\, 1}\)
dy/dx of xe^y-y=x^2-2
Y= (2x-e^y)/(e^yx-1)
Hopefully I typed this correctly...
Last edited by a moderator: