Question

Another basic Differential Equation, problem prompt posted below shortly.

Answer 1

\[\text{Show that the differential equation is exact and then solve the exact equation.}\]

\[[\tan(x) − \sin(x)\sin(y)]dx + [\cos(x)\cos(y)]dy = 0\]

Answer 2

(Oh, lol, nvm)

Answer 3

So what's troubling ya about this one? :)

Answer 4

Nothing yet, just want to put it up in case I have an issue, that way I'll be able to possibly get help quick. I just worry about not figuring it out and nobody being around to potentially help. Working through it momentarily.

Answer 5

Alrighty, np.

Answer 6

Alright, this looks like an exact equation where the current body of the problem is an exact differential of some function; we assume that function to be continuous/differentiable at its first partials, so its second partials should be equal to one another, e.g.\[f_{xy}=f_{yx}\]

Answer 7

Or, more familiar, \[N_{x}-M_{y}=0\]

Answer 8

\[N_{x}=-\sin(x)\cos(y); \ \ \ M_{y}=-\sin(x)\cos(y).\]

Answer 9

Thus, the equation is exact. Alright, here's where I start getting hazy on remembering what to do, lol, just one second...

Answer 10

Oh, okay, this is the integrate/take derivative of thing repeatedly, one sec.

Answer 11

\[\int\limits_{}^{}N_{x}\ dx=\int\limits_{}^{}-[\sin(x)\cos(y)]dx;\]

Answer 12

I'm allowed to factor out the cos(y), I think, in this situation?

Answer 13

Wait, you tried to integrate cos(x)cos(y), right? Then you need to integrate that with respect to y.

Answer 14

Hm...maybe youre using a method different than Im used to :)

Answer 15

IDK, lemme check I'm doing things right, lul, don't want to go through all the work to find out I fundamentally was doing things wrong in the first place.

Answer 16

Unless theres a way of doing this I'm unaware of, you need to choose to integrate
\(tan(x) - sin(x)sin(y)\) with respect to x or \(cos(x)cos(y)\) with respect to y

Answer 17

Yeah, I'm going backwards, I even integrated the wrong thing to begin with, lol. Yeah, you're right.

Answer 18

\[\int\limits_{}^{}[\tan(x)-\sin(x)\sin(y)]dx=\int\limits_{}^{}\tan(x)dx-\int\limits_{}^{}\sin(x)\sin(y)dx\]

Answer 19

\[\int\limits_{}^{}\tan(x)=\ln|\sec(x)|+c, \ \text{or:} -\ln|\cos(x)|+c\]

Answer 20

\[\int\limits_{}^{}\sin(x)\sin(y)dx=\sin(y)\int\limits_{}^{}\sin(x)dx=\sin(y)[-\cos(x)]+c\]

Answer 21

Wait a sec, both of those should have a function w.r.t. y

Answer 22

Except its not quite + C this time.

Answer 23

Yeah, haha

Answer 24

\[\ln|\sec(x)|-\sin(y)\cos(x)+g(y)=f(x,y)\]

(not sure about what I set it equal to, one minute)

Answer 25

Now I take the partial with respect to y and set it equal to the second term from the original problem IIRC.

Answer 26

Yes, in order to solve for g'(y) and eventually g(y) itself.

Answer 27

\[\frac{\partial}{\partial y}\bigg(\ln|\sec(x)|-\sin(y)\cos(x)+g(y)\bigg)=-\cos(y)\cos(x)+g'(y)\]

Answer 28

\[-\cos(y)\cos(x)+g'(y)=N=\cos(x)\cos(y).\]So g(y) is some arbitrary constant c, but somewhere in there, I clearly dropped a negative sign.

Answer 29

Yeah, in the integration the sign shouldve changed.

Answer 30

Youforgot to write the negative in the 2nd integral, so even though you integrated it correctly, you had the wrong integral to start with.

Answer 31

Oh, yeah, I see where.

Answer 32

Yay for g(y) being 0? Lol.

Answer 33

Alright, so at least the negative didn't affect the overall integral.

Yeah, makes things easy but also feels kinda lame/trivial, heh

Answer 34

Alright, now that I know that, I integrate the second part with respect to y and get some terms plus an unknown function of x, I take the partial of it with respect to x, and so on and so forth. One moment.

Answer 35

No, youre done, lol.

Answer 36

Wait wat lol

Answer 37

So you integrated with respect to x and got:
\(ln|secx| + sin(y)cos(x) + g(y)\)
Once we found that g(y) = 0, our answer simply is
\(ln|secx| + sin(y)cos(x) = c\)

Thats all you need really.

Answer 38

Alright, just gimme a sec. Trying to make sure I get this.

Answer 39

If you do it the other way, youll get the same result. You can integrate the 2nd part with respect to y and do the partial blah blah with respect to x, but it's simply re-doing your work to get the same answer. Typically, you choose which one is easier to ingrate and go from there, doing the process you did to get an answer. It doesnt need to be done twice.

Answer 40

Yeah, I got it, makes sense now.

Answer 41

Now I just need to put it either in the form of an implicit or explicit solution, like-oh, you did that, too, lol.

Gonna see if I can make it explicit, one sec.

Answer 42

Leave it. The final solution is a function equal to c, that is how it should look.

Answer 43

Alright, cool. Thanks you! Going to go through a couple more problems momentarily.

Answer 44

No problem :)

Answer 45

Alright, nevermind, going to go eat dinner really quick, well, wait, maybe; how long do you think you're going to be on? If you're planning on getting out of soon, I'll just stay here and get them done now while there are people around that can help.

Answer 46

Um, another 2-3 hours. Until the starbucks im at closes >.>

Answer 47

Hah. Alright, I'll be back in about 30-45 minutes, just need to go to McDonald's or some place not crazy pricey before they close, I'll bbl.

Answer 48

Kk.