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OBEYDUR RAHMAN · Aug 16, 2009, 01:50 AM

Hi
Would you help me to tell about integrating factor

galactus · Aug 16, 2009, 05:19 AM

Post a specific problem and we can show you.

Also, you can Google it and find plenty.

But here is an explanation:

Sometimes we can't separate variables, so we can use the integrating factor that Euler developed.

A first-order DE is called linear if we can write it as $\frac{dy}{dx}+p(x)y=q(x)$ ... [1]

Where p(x) and q(x) may be constants.

We can make the observation that if we define ${\mu}={\mu}(x)$ by ${\mu}=e^{\int p(x)dx}$ , then

$\frac{d{\mu}}{dx}=e^{\int p(x)dx} \cdot \frac{d}{dx}\int p(x)dx={\mu}p(x)$

Therefore:

$\frac{d}{dx}[{\mu}y]={\mu}\frac{dy}{dx}+\frac{d{\mu}}{dx}y={\mu}\frac{ dy}{dx}+{\mu}p(x)y$ ... [2]

If we multiply [1] through by ${\mu}$ , and then simplify using [2], we get:

${\mu}\frac{dy}{dx}+{\mu}p(x)y={\mu}q(x)$

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

This can be solved by integrating both sides to get:

${\mu}y=\int {\mu}q(x)dx+C \;\ or \;\ y=\frac{1}{\mu}\left[\int {\mu}q(x)dx+C\right]$

There ya' go. A general explanation. If you want a specific example, please post a problem you are having trouble with and I can try to guide you through it. Okey-doke?
It's not as bad as it looks.