Question

Find an integrating factor of the form u(x, y)=P(x)Q(y) for y(1+5*ln abs(x))dx+4x*ln abs(x)*dy=0 and solve the equation.

Answer 1

@SithsAndGiggles

Answer 2

Hmm I can't say I've ever come across this sort of integrating factor before... Give me a moment to look up some info.

Answer 3

Okay.

Answer 4

Is there a special name for this sort of integrating factor? I've only ever seen textbooks that cover the \(\mu(x)\) and \(\mu(y)\) cases, not both variables at once.

Answer 5

I don't know... Let me check my textbook.

Answer 6

No, there's no special name.

Answer 7

An IF \(u(x,y)\) would give the equation
\[u(x,y)M(x,y)+u(x,y)N(x,y)\frac{dy}{dx}=0\]
that is exact, which would mean
\[\frac{\partial u}{\partial x}N-\frac{\partial u}{\partial y}M=u\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)\]
Since \(u(x,y)\) can't be zero, we can safely divide through by \(u\) (I'll use \(u_x\) to denote the partial derivatives):
\[\frac{u_x}{u}N-\frac{u_y}{u}M=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\]
and now you want to find a separable solution for \(u\), which you can try doing by letting
\[F(x)=\frac{u_x}{u}~~\text{and}~~G(y)=\frac{u_y}{u}\]
These equations give
\[\frac{\partial}{\partial x}\log x=F(x)~~\text{and}~~\frac{\partial}{\partial y}\log y=G(y)\]
so you have the IF
\[\begin{align*}\log u(x,y)&=\int F(x)~dx+\int G(y)~dy\\\\
u(x,y)&=e^{\cdots}\end{align*}\]
So you have
\[\begin{align*}F(x)4x\log|x|-G(y)y(1+5\log|x|)&=(1+5\log|x|)-(4\log|x|+4)\\\\
F(x)4x\log|x|-G(y)y(1+5\log|x|)&=\log|x|-3
\end{align*}\]
Does this make sense so far?

Answer 8

The integrating factor for this problem is u(x, y)=x^4*y^3.

Answer 9

I suppose one way to find \(G(y)\), since it's a function of \(y\) only, would be to solve for it:
\[\begin{align*}G(y)&=\frac{F(x)4x\log|x|-(\log|x|-3)}{y(1+5\log|x|)}\end{align*}\]
We'd like a constant \(C\) that eliminates the factor of \(1+5\log|x|\) in the denominator, so that
\[F(x)4x\log|x|-(\log|x|-3)=C(1+5\log|x|)\]
If we let \(C=3\), we get
\[\begin{align*}F(x)4x\log|x|-\log|x|+3=3+15\log|x|\\\\
F(x)4x\log|x|=16\log|x|\\\\
F(x)=\frac{4}{x}\end{align*}\]
which would then give
\[\begin{align*}G(y)&=\frac{16\log|x|-(\log|x|-3)}{y(1+5\log|x|)}\\\\
G(y)&=\frac{15\log|x|+3}{y(1+5\log|x|)}\\\\
G(y)&=\frac{3}{y}\end{align*}\]
and so
\[u(x,y)=\exp\left(\int\frac{4}{x}~dx+\int\frac{3}{y}~dy\right)=x^4y^3\]
as desired. It's nice to see there's a method for this type of IF...

Answer 10

@SithsAndGiggles

Answer 11

How does \[F(x)4x \ln \left| x \right|-G(y)y(1+5\ln \left| x \right|)\]

Answer 12

\[=(1+5\ln \left| x \right|)-(4\ln \left| x \right|+4)\]

Answer 13

That's from substituting into this equation,
\[\frac{u_x}{u}N-\frac{u_y}{u}M=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\]

Answer 14

Is there any other way to find the integrating factor?

Answer 15

Not that I could find... Before figuring this out, I've only ever worked with one-dimensional integrating factors.

Answer 16

I see. So after finding the integrating factor, which is x^4*y^3, what do we do?

Answer 17

Distribute it to the equation as usual, after checking to see that it makes the equation exact.

Answer 18

Thank you! I got the right solution!

Answer 19

You're welcome!