# Solve the following ODE .. #8

• Feb 11th 2011, 12:17 PM
Liverpool
Solve the following ODE .. #8
Hello

I have this problem

Problem:
Solve the following differential equation by using three different methods :

$(xy-1+y)dx+xdy=0$

Solution:

One method is finding the integrating factor which is $e^x$ which found by the formula $\displaystyle e^{\displaystyle \int \dfrac{\dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial x}}{N} \, dx}$

what are the other 2 methods??
• Feb 11th 2011, 12:26 PM
chisigma
With some algebraic steps the DE becomes...

$\displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x}$ (1)

... which is linear and can be solved with 'standard approach'...

Kind regards

$\chi$ $\sigma$
• Feb 11th 2011, 12:27 PM
Ackbeet
You could always do a series solution method!
• Feb 11th 2011, 12:48 PM
pickslides
If you had some I.C's then a numerical method like runge-kutta?
• Feb 11th 2011, 01:02 PM
chisigma
Quote:

Originally Posted by chisigma
With some algebraic steps the DE becomes...

$\displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x}$ (1)

... which is linear and can be solved with 'standard approach'...

Kind regards

$\chi$ $\sigma$

The 'standard approach' seems to lead to a solution like...

$\displaystyle y(x)= e^{-(x-\frac{1}{x})} \ \{\int \frac{e^{(x-\frac{1}{x})}}{x}\ dx + c \}$ (1)

... that contains a 'not very pleasant' integral (Thinking)...

Kind regards

$\chi$ $\sigma$
• Feb 11th 2011, 01:13 PM
Ackbeet
The substitution $u=xy-1$ renders the equation separable:

$du=x\,dy+y\,dx,$ and hence if we have

$(xy-1)\,dx+y\,dx+x\,dy=0,$ then we arrive at

$u\,dx+du=0.$
• Feb 11th 2011, 01:45 PM
General
Another one is by using inspection.

$(xy-1)dx+ydx+xdy=0$

$(xy-1)dx+d(xy)=0$

$\dfrac{d(xy)}{xy-1}+dx=0$ ... (1)

Actually, equation (1) is equivelant to the last equation in Ackbeet's replay.
• Feb 11th 2011, 02:36 PM
General
Quote:

Originally Posted by chisigma
With some algebraic steps the DE becomes...

$\displaystyle y^{'} + y\ (1+\frac{1}{x^{2}}) = \frac{1}{x}$ (1)

... which is linear and can be solved with 'standard approach'...

Kind regards

$\chi$ $\sigma$

You should get : $y'+\left(1+\dfrac{1}{x}\right) \cdot y = \dfrac{1}{x}$
• Feb 11th 2011, 08:53 PM
chisigma
Quote:

Originally Posted by General

You should get : $y'+\left(1+\dfrac{1}{x}\right) \cdot y = \dfrac{1}{x}$

Of course!... and the 'standard approach' leads to the 'confortable' solution...

$\displaystyle y(x)= c\ \frac{e^{-x}}{x} + \frac{1}{x}$ (1)

Kind regards

$\chi$ $\sigma$
• Feb 12th 2011, 04:56 AM
Liverpool
Thanks All.
• Feb 12th 2011, 05:19 AM
Ackbeet
You're welcome for my (this time, real!) contribution. (Wink)
• Feb 12th 2011, 05:21 AM
Liverpool
Thanks.