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

3. Originally Posted by matheagle
To be exact, the term in front of dx must be $\displaystyle f_x$
and the term in front of dy must be $\displaystyle f_y$, WHERE the bottom line is to find that $\displaystyle f(x,y)$.
Ignore the silly N and M in must books, it's foolish.
Next from Clarabell's theorem $\displaystyle f_xy=f_yx$
You mean $\displaystyle f_{xy}= f_{yx}$
when you have more than one symbol after _ or ^, you need { } around them.

and in our case that's correct, we get $\displaystyle e^x\cos y$ in both cases.
So, it's exact. Hence integrate the $\displaystyle f_x$ wrt x
and $\displaystyle f_y$ wrt y to get this $\displaystyle f(x,y)$ and when you integratse, don't just add +C
in the first case it's $\displaystyle g(y)$ and the second it's $\displaystyle h(x)$.

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