Question

solve
x - y ln y dx + y ln x dx + x ln y dy - x ln x dy = 0

Answer 1

i was stuck :(
i am trying to integrate:
\[\int\limits \ln (vx) dv\]

Answer 2

attempt?

Answer 3

Show your prior working.

Answer 4

\(\int \ln (vx) dv\)

here, x is a constant,
so its just like
\(\int \ln (ax)dx\)

Answer 5

and ofcourse, you will need uv rule

Answer 6

\[
\begin{split}
\int \ln(vx)dv
&= \int \ln(vx)\frac {xdv}x \\
&= \int \ln(vx)\frac {d(xv)}x
\end{split}
\]At this point it is pretty simple.

Answer 7

x dv = d(xv) ?? :O

i though d(xv) = x dv+ v dx

Answer 8

Well, I treated \(x\) like a constant...

Answer 9

So that means \(v~dx = 0\).

Answer 10

oh...you still need uv rule

Answer 11

You can use product rule on a constant, it's just that the derivative of a constant is 0 so only one term actually remains.

Answer 12

\(dx =0\).

Answer 13

i don't get the part where xdv=d(vx)

Answer 14

Well: \[
d(xy) = \frac{d(xv)}{dv} dv
\]This is just a definition thing. Anyway, since \(d(xv)/dv = x\) we get: \[
d(xy) = x~dv
\]

Answer 15

In general: \[
dy = \frac {dy}{dx}dx
\]

Answer 16

You are probability familiar with:\[
\frac{dy}{dx} = \lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x} = \lim_{\Delta x\to 0}\frac{y(x+\Delta x)-y(x)}{\Delta x}
\]Basically, \(d\) in this case is notation convenience for \(\Delta \) under some limit.

Answer 17

To avoid having to write out the variable of differentiation, we can define differentials as: \[
dy = \frac{dy}{dx}dx
\]and \(x\) can be any variable in this case. When we use substitution, or chain rule, we are just changing the variable of differentiate: \[
\int f(g(x))g'(x)~dx = \int f(g)~dg, \quad dg = g'(x)dx
\]

Answer 18

simpler solution please? sorry i still don't get that :(

Answer 19

you know uv rule of integration ?
can you solve your integral with that rule ?