# Integrating factor strategy

1. Apr 11, 2005

### Naeem

Q. Motivate the Integrating factor strategy for U ( "Mew" ) of y

I know how to prove it for "Mew" of x but how to do for "mew" of y

Maybe something like this.

Mdx (x.y) + Ndy ( x, y ) = 0

Assume this is differentiable so let us multiply by "mew" of x on both sides to make it exact.

Then M ( tilda ) the left term and N ( tilda ) equal to the right term

Then may be, Find partial with respect to x in the M terms. and partial with respect to y in the N terms. Is this idea/approach correct.

2. Apr 12, 2005

### saltydog

Naeem, no offense but this is not clear and the notations is awkward. Perhaps if you specify a specific problem we can help you.

3. Apr 12, 2005

### Naeem

Well, I need to come up with the following final formula used for finding the Integrating factor, for a linear differential equation.

e ^ Integral Nx-My / M = Greek Letter U ( Mew) (y)

This is the formula used to find the Integrating factor, with respect to y in Linear differential equation.

4. Apr 12, 2005

### Hurkyl

Staff Emeritus
Just using 'u' is acceptable. However, you can create a Greek mu (that's how it's spelled) with the following incantation, if you remove the spaces:

& m u ;

That will be turned into the symbol μ. (Yes, I know the default font doesn't render it very well. If you want, you might write your post in the Times New Roman font -- it does Greek characters well)

5. Apr 13, 2005

### Andrew Mason

I am not sure what you are asking. Perhaps this will help:

For a first order differential equation put into the form:

$$\frac{dy}{dx} + P(x)y = Q(x)$$

you want to find a function $\mu(x)$ such that:

$$\mu\frac{dy}{dx} + \mu P(x)y = \mu Q(x)$$ where:

$$\mu\frac{dy}{dx} + \mu P(x)y = \frac{d}{dx}(\mu y)$$

This reduces to:

$$\mu P(x)y = y\frac{d\mu}{dx}$$

$$\frac{d\mu}{dx} = \mu P(x)$$

Dividing by $\mu$ and integrating both sides:

$$\int \frac{1}{\mu}d\mu = \int P(x) dx$$

$$ln\mu = \int P(x) dx$$

So the general solution for $\mu$ is:

$$\mu = \pm e^{\int P(x) dx}$$

AM