Question

(ODE) Problem dealing with an exact equation, posted below in a moment.

Answer 1

\[(e^{x}+y)dx+(2+x+ye^{y})dy=0.\]\[\text{Show that the differential equation is exact and then solve the}\]\[\text{exact equation subject to the given condition.}\]

Answer 2

I'm aware of how to demonstrate that it's exact, just confused about solving it subject to the given condition. Could you help me on this?

Answer 3

@gleem

Answer 4

I'll write out what I tried to do to solve it next, but I messed up:

Answer 5

\[\int\limits_{}^{}N_{x} \ dx=\int\limits_{}^{}(e^{x}+y)dx=e^{x}+yx+g(y)\]

Answer 6

\[\frac{\partial}{\partial y}\bigg(e^{x}+yx+g(y)\bigg)=x+g'(y)=N(x,y). \]

Whoops, in my post before this one, that's supposed to be an M, not an N.

Answer 7

\[x+g'(y)=2+x+ye^{y}; \ \ \ g'(y)=2+ye^{y}\]

Answer 8

And now I get stuck, because I don't think this is right so far/I don't know how to ove on from here.

Answer 9

Shouldn't find \[N _{y} = \int\limits_{}^{} (2+x+ye ^{y})dy\] next?

Answer 10

I thought so, too, lol, I did the same on another problem and somebody told me not to. Are you sure about that last part? You could be right, I just don't know.

Answer 11

?

Answer 12

Alright, one second.

Answer 13

So how would you phrase your final answer? I just want to at least get the general idea of how to mechanically do this by witnessing somebody else do it.

Answer 14

disregard the previous post Made a mistake above couldn't delete it.
\[N _{y}=2y+xy+e ^{y}(y-1)\]one can see \[f(x) =e ^{x}\]and \[g(y)=2y+e ^{^{y}}(y-1)\]so\[F(x,y) =2y+ xy+e ^{x}+e ^{y}(y-1)\]

Answer 15

@Mendicant_Bias The process of solving is very similar to finding potential functions for vector fields, if you recall what you've (probably by now) learned in vector calculus.

Answer 16

^Yeah, I'm just bad at both of them, lol. The answer is this, posting it in a moment.

Answer 17

http://i.imgur.com/HyEgcSd.png

Answer 18

Where did the three come from?

Answer 19

"... subject to the given condition." I think you omitted it.