1. ## 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

2. 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 $h(x)$.

3. 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.

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 $h(x)$.

4. eye no, it was late and I was too tired to check, thanks