# Stumped

• Feb 25th 2009, 05:28 PM
johnnymath345
Stumped
"Verify that the given differential equation is exact; then solve it."

(e^(x) * siny + tany)dx + (e^(x) * cosy + x * (secy)^2)dy = 0
• Feb 25th 2009, 10:36 PM
matheagle
Quote:

Originally Posted by johnnymath345
"Verify that the given differential equation is exact; then solve it."

(e^(x) * siny + tany)dx + (e^(x) * cosy + x * (secy)^2)dy = 0

To be exact, the term in front of dx must be $f_x$
and the term in front of dy must be $f_y$, WHERE the bottom line is to find that $f(x,y)$.
Ignore the silly N and M in must books, it's foolish.
Next from Clarabell's theorem $f_{xy}=f_{yx}$
and in our case that's correct, we get $e^x\cos y$ in both cases.
So, it's exact. Hence integrate the $f_x$ wrt x
and $f_y$ wrt y to get this $f(x,y)$ and when you integrate, don't just add +C
in the first case it's $g(y)$ and the second it's $h(x)$.
• Feb 26th 2009, 08:42 AM
HallsofIvy
Quote:

Originally Posted by matheagle
To be exact, the term in front of dx must be $f_x$
and the term in front of dy must be $f_y$, WHERE the bottom line is to find that $f(x,y)$.
Ignore the silly N and M in must books, it's foolish.
Next from Clarabell's theorem $f_xy=f_yx$

You mean $f_{xy}= f_{yx}$
when you have more than one symbol after _ or ^, you need { } around them.

Quote:

and in our case that's correct, we get $e^x\cos y$ in both cases.
So, it's exact. Hence integrate the $f_x$ wrt x
and $f_y$ wrt y to get this $f(x,y)$ and when you integratse, don't just add +C
in the first case it's $g(y)$ and the second it's $h(x)$.
• Feb 26th 2009, 08:56 AM
matheagle
eye no, it was late and I was too tired to check, thanks