# Differential equation - integrating factor

• Jun 18th 2009, 01:59 PM
coobe
Differential equation - integrating factor
hello ! i need help bringing this equation into the form $\frac {y} {dy} = ....$

the equation is:

$\frac {dy} {dx} = \frac {y} {x-1} + x(x-1)$

edit: actually i have no clue what to do at all..... no real question given...
ok obviously i need to solve this Diff. Equ.
• Jun 18th 2009, 02:36 PM
pickslides
Quote:

Originally Posted by coobe
hello ! i need help bringing this equation into the form $\frac {y} {dy} = ....$

the equation is:

$\frac {dy} {dx} = \frac {y} {x-1} + x(x-1)$

edit: actually i have no clue what to do at all..... no real question given...
ok obviously i need to solve this Diff. Equ.

I think putting the equation in question to the from $\frac {y} {dy} = ....$ implies it is separable.

The bad news is your DE is not separable and you need to use the integrating factor method to solve it.

It should be written in the form

$\frac {dy} {dx} + f(x)\times y = g(x)$

therefore it should be

$\frac {dy} {dx} - \frac {1} {x-1}y = x(x-1)$
• Jun 18th 2009, 02:39 PM
pickslides
now have a look at this post. It shows you how to apply the integrating factor method.

http://www.mathhelpforum.com/math-he...equations.html
• Jun 19th 2009, 06:18 AM
HallsofIvy
Quote:

Originally Posted by coobe
hello ! i need help bringing this equation into the form $\frac {y} {dy} = ....$

the equation is:

$\frac {dy} {dx} = \frac {y} {x-1} + x(x-1)$

edit: actually i have no clue what to do at all..... no real question given...
ok obviously i need to solve this Diff. Equ.

First write it as $\frac{dy}{dx}- \frac{y}{x-1}= x(x-1)$.
That is a linear equation and, as you say, has an "integrating factor".
That means there exist some function u(x) such that multiplying the entire equation by u(x), the left side becomes an "exact derivative":
$u(x)\frac{dy}{dx}- u(x)\frac{y}{x-1}= \frac{d(u(x)y)}{dx}$
Do the differentiation on the right side, using the product rule, and compare:
$u(x)\frac{dy}{dx}- u(x)\frac{y}{x-1}= u(x)\frac{dy}{dx}+ \frac{du}{dx}y$

We must have $-u(x)\frac{1}{x-1}= \frac{du}{dx}$, a separable equation for u(x): $-\frac{dx}{x-1}= \frac{du}{u}$. Integrating both sides of that, ln|u|= -ln|x-1| so $u(x)= \frac{1}{x-1}$ is the integrating factor:
$\frac{1}{x-1}\frac{dy}{dx}- \frac{y}{(x-1)^2}= x$
$\frac{d(\frac{y}{x-1})}{dx}= x$.
Now integerate both sides of that.