"Verify that the given differential equation is exact; then solve it."

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

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- Feb 25th 2009, 04:28 PMjohnnymath345Stumped
"Verify that the given differential equation is exact; then solve it."

(*e*^(*x*) * sin*y*+ tan*y*)*dx*+ (*e*^(*x*) * cos*y*+*x** (sec*y*)^2)*dy*= 0 - Feb 25th 2009, 09:36 PMmatheagle
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 $\displaystyle h(x)$. - Feb 26th 2009, 07:42 AMHallsofIvy
You mean $\displaystyle f_{xy}= f_{yx}$

when you have more than one symbol after _ or ^, you need { } around them.

Quote:

and in our case that's correct, we get $\displaystyle e^x\cos y$ 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 $\displaystyle h(x)$.

- Feb 26th 2009, 07:56 AMmatheagle
eye no, it was late and I was too tired to check, thanks