Find a P(x,y) function that makes the equation below an "exact differential equation."
C CREZI New member Joined Apr 27, 2020 Messages 2 Apr 27, 2020 #1 Find a P(x,y) function that makes the equation below an "exact differential equation."
MarkFL Super Moderator Staff member Joined Nov 24, 2012 Messages 3,021 Apr 27, 2020 #2 Hello, and welcome to FMH! What we would require is: [MATH]\pd{P}{y}(x,y)=\pd{}{x}\left(xe^{xy}+2xy+\frac{1}{x}\right)=(xy+1)e^{xy}+2y-\frac{1}{x^2}[/MATH] Can you proceed?
Hello, and welcome to FMH! What we would require is: [MATH]\pd{P}{y}(x,y)=\pd{}{x}\left(xe^{xy}+2xy+\frac{1}{x}\right)=(xy+1)e^{xy}+2y-\frac{1}{x^2}[/MATH] Can you proceed?
