Question

I'm trying to find a function f such that \[F=\nabla f\]
and this is how far I've gotten
\[\frac{\partial{P}}{\partial{y}}=e^xcosy\]
\[\frac{\partial{Q}}{\partial{x}}=e^xcosy\]
\[f_x=e^xsiny\]
\[f=e^xsiny+P(y)\]
\[f_y=e^xcosy\]
\[f=e^xsiny+q(y)\]
where from here?

Answer 1

Well i'm not sure what \nablaf is.. but think you want to get two functions, P(y) and Q(y). You could do so by taking those first two partials, multiplying the partial of (x or y) over, and then integrating to get a function of y.

Answer 2

@MathSofiya you typed the last line wrong. Q is a function of x.

Set f=f
\[e^x\sin (y)+P(y)=e^x\sin (y)+Q(x)\]\[P(y)-Q(x)=0\]So obviously this expression is constant with respect to both x and y, so they must both be simply an arbitrary constant.\[f(x,y)=e^x\sin (y)+C\]

Answer 3

I'm having trouble loading this page. I'm not able to see the equations written in Latex.

Answer 4

I'll restart my computer. I'll be right back.

Answer 5

First equation:
e^x sin(y)+P(y)=e^x sin(y)+Q(x)

Second equation:
P(y)-Q(x)=0

Third equation:
f(x,y)=e^x sin(y)+C

PS: why did you ask your question in Latex then?

Answer 6

much better. I'm able to see everything now.

Answer 7

I posted this question 4 hours ago

Answer 8

So for example, when we have \[f=x^2 -3xy+p(y)\]
\[f=-3xy+2y^2-8y+q(x)\]
would we set those equal to each other
\[x^2 -3xy+p(y)=-3xy+2y^2-8y+q(x)\]
and get
\[x^2+p(y)=2y^2-8y+q(x)\]
I know we did this problem together yesterday @Herp_Derp but I still have a question.
Do we assume that the variables 'y' belong to p(y)
and 'x' belongs to q(x)?
So therefore we have \[q(x)=x^2\]
and \[p(y)=2y^2-8\]?

Answer 9

*-8y* typo...

Answer 10

Yes. You must set functions of one variable equal to those of the same variable.

Answer 11

Thanks @Herp_Derp !