[SOLVED] find integrating factor and solve the equation

• Mar 27th 2010, 04:32 PM
nameck
[SOLVED] find integrating factor and solve the equation
y dx + x ln x dy = 0 ; x > 0

my integrating factor is x..

so.. multiply with DE,

xy dx + x^2 ln x dy = 0

let M = xy ; N = x^2 ln x

dM/dy = x ; dN/dx = x + 2x ln x

the problem is.. i didn't get the exact equation after multiply the integrating factor.. i've double checked my integrating factor.. but.. still can't get the exact equation...
• Mar 27th 2010, 06:10 PM
Sudharaka
Quote:

Originally Posted by nameck
y dx + x ln x dy = 0 ; x > 0

my integrating factor is x..

so.. multiply with DE,

xy dx + x^2 ln x dy = 0

let M = xy ; N = x^2 ln x

dM/dy = x ; dN/dx = x + 2x ln x

the problem is.. i didn't get the exact equation after multiply the integrating factor.. i've double checked my integrating factor.. but.. still can't get the exact equation...

Dear nameck,

Take the integrating factor as $\frac{1}{x}$ and try.

• Mar 27th 2010, 06:15 PM
nameck
Quote:

Originally Posted by Sudharaka
Dear nameck,

Take the integrating factor as $\frac{1}{x}$ and try.

yes.. it works!! (Bow)
but.. why is 1/x and not x?
• Mar 27th 2010, 06:32 PM
Sudharaka
Quote:

Originally Posted by nameck
yes.. it works!! (Bow)
but.. why is 1/x and not x?

Dear nameck,

It just occured me that this integrating factor would work. If somebody could tell us a mathamatical way of getting the integrating factor for any first order differential equation that would be helpful is'nt??
• Mar 27th 2010, 07:24 PM
Prove It
If I am given a DE like this, I usually convert it to a first order linear if possible.

$y\,dx + x\ln{x}\,dy = 0$

$y + x\ln{x}\,\frac{dy}{dx} = 0$

$\frac{dy}{dx} + \frac{1}{x\ln{x}}\,y = 0$.

Now this is first order linear. So the integrating factor is

$e^{\int{\frac{1}{x\ln{x}}\,dx}} = e^{\int{\frac{1}{u}\,du}}$ after making the substitution $u = \ln{x}$

$= e^{\ln{u}}$

$= e^{\ln{(\ln{x})}}$

$= \ln{x}$.

So multiplying through by the integrating factor gives:

$\ln{x}\,\frac{dy}{dx} + \frac{1}{x}\,y = 0$

$\frac{d}{dx}(y\ln{x}) = 0$

$y\ln{x} = \int{0\,dx}$

$y\ln{x} = C$

$y = \frac{C}{\ln{x}}$.