Question

Find an integrating factor of the form u(x, y)=P(x)Q(y) for (a*cosxy-y*sinxy)dx+(b*cosxy-x*sinxy)dy=0 and solve the equation.

Answer 1

\[F(x)(b \cos xy-x \sin xy)-G(y)(a \cos xy-y \sin xy)\]

Answer 2

What's the partial derivative of (a*cosxy-y*sinxy) with respect to y?

Answer 3

Is it -sinxy?

Answer 4

\[\begin{align*}
F(x)(b\cos xy-x\sin xy)\quad&\\
-G(y)(a\cos xy-y\sin xy)&=(a\cos xy-y\sin xy)_y-(b\cos xy-x\sin xy)_x
\end{align*}\]
\[\frac{\partial}{\partial y}(a\cos xy-y\sin xy)=-ax\sin(xy)-\sin xy-xy\cos xy\\
\frac{\partial }{\partial x}(b\cos xy-x\sin xy)=-by\sin xy-\sin xy-xy\cos xy\]

Answer 5

Sorry about the random set of parentheses around the \(xy\)...

Answer 6

That's fine.

Answer 7

Do those derivatives make sense? It's mostly chain and product rules.

Answer 8

Yes, they do make sense.

Answer 9

I have a question. What's the prerequisite of Complex Variables?

Answer 10

Hmm, that depends... If it's some sort of rigorous course on complex analysis, you'll need to be familiar with a bit of real analysis and a lot of calculus. If it's to do with differential equations, it's mostly a matter of knowing Euler's formula, so trigonometry and a bit of linear algebra. Or if you mean complex variables in general, it's essentially an extension of algebra and deals with equations in two variables.

Answer 11

So after learning Partial Differential Equations, should I learn Complex Variables?

Answer 12

Anyway...
\[\begin{align*}
F(x)(b\cos xy-x\sin xy)\quad&\\
-G(y)(a\cos xy-y\sin xy)&=(-ax\sin xy-\sin xy-xy\cos xy)\\&\quad-(-by\sin xy-\sin xy-xy\cos xy)\\\\

F(x)(b\cos xy-x\sin xy)\quad&\\
-G(y)(a\cos xy-y\sin xy)&=(by-ax)\sin xy\\\\
F(x)&=\frac{(by-ax)\sin xy+G(y)(a\cos xy-y\sin xy)}{b\cos xy-x\sin xy}
\end{align*}\]
Hmm... might take a while to figure out this "constant" function we need to eliminate the appropriate variables...

Answer 13

How did you get (by-ax)sinxy?

Answer 14

You subtract the partial derivatives. The \(-\sin xy\) and \(-xy\cos xy\) terms disappear, and I factored the common \(\sin xy\).

Answer 15

I have to get going, but I'll get back to this when I have the time.

Answer 16

So what should the "constant" function be?

Answer 17

Finally got it after some effort. The integrating factor, I mean, not the \(C\) function. I don't think that approach will work here.

Instead, what I did was assume
\[\large\begin{cases}
F(x)=\alpha x^\gamma\\\\
G(y)=\beta y^\delta
\end{cases}\]
which leads to getting the IF,
\[\large u(x,y)=e^{ax+by}\]

Answer 18

I'd post the details now, but there's a lot of writing involved... I'll return to this another time if you need an in-depth explanation.

Answer 19

How did you get F(x) and G(y)? Can you please show your work?

Answer 20

Guesswork, mostly... I started off assuming \(\large F(x)=\alpha x^\gamma\) and \(\large G(y)=\beta y^\delta\) and tried to see if I could find constants \(\alpha,\beta,\gamma,\delta\) that would work.

Answer 21

Substituting those "solutions" into the equation:
\[\large\begin{align*}
\alpha x^\gamma(b\cos xy-x\sin xy)\quad&\\
-\beta y^\delta(a\cos xy-y\sin xy)&=(by-ax)\sin xy\\\\

\alpha bx^\gamma\cos xy-\alpha x^{\gamma+1}\sin xy\quad&\\
-\beta ay^\delta\cos xy+\beta y^{\delta+1}\sin xy&=(by-ax)\sin xy\\\\

(\alpha bx^\gamma-\beta ay^\delta)\cos xy\quad\quad\quad&\\+(\beta y^{\delta+1}-\alpha x^{\gamma+1})\sin xy&=(by-ax)\sin xy
\end{align*}\]
Matching up sines and cosine gives
\[\large\begin{cases}
\alpha bx^\gamma-\beta ay^\delta=0\\
\beta y^{\delta+1}-\alpha x^{\gamma+1}=by-ax
\end{cases}\]
The first equation needs to have \(\gamma=\delta=0\) so that we can remove the \(x\) and \(y\). Solving for \(\alpha\) then gives
\[\alpha=\frac{\beta a}{b}\]
and subbing into the second equation gives
\[\beta y-\frac{\beta a}{b} x=by-ax\]
which gives means \(\beta=b\) and \(\dfrac{\beta a}{b}=\alpha~~\iff~~\alpha=a\). Hence
\[\large\begin{cases}
F(x)=\alpha\\\\
G(y)=\beta
\end{cases}\]
and the IF becoems
\[\large\log u(x,y)=\int \alpha ~dx+\int\beta~dy~~\iff~~u(x,y)=e^{ax+by}\]

Answer 22

Oh and I never replied to your question about your complex-number course. I wouldn't know how to advise you on what courses to take. It all really depends on what you want to do for a living. I've never taken a course on PDEs, but I have had a semester of complex analysis, and it had nothing to do with differential equations. Like I mentioned before, it was all calculus, but I couldn't quite see much practical application to it.

Answer 23

There's actually another way of solving this without finding the integrating factor. I know the instructions require finding the IF, but I deemed it worth mentioning:
\[\begin{align*}(a \cos xy-y \sin xy)+(b \cos xy-x \sin xy)y'&=0\\\\
(a+by')\cos xy&=(y+xy')\sin xy\\\\
(a+by')\cos xy&=\frac{d}{dx}[-\cos xy]\\\\
a+by'&=-\frac{\dfrac{d}{dx}[\cos xy]}{\cos xy}\\\\
a+by'+\frac{\dfrac{d}{dx}[\cos xy]}{\cos xy}&=0\\\\
\int\left(a+by'+\frac{\dfrac{d}{dx}[\cos xy]}{\cos xy}\right)~dx&=C\\\\
ax+by+\ln(\cos xy)&=C\end{align*}\]

Answer 24

Thank you so much for the extra info and the guide on math courses.