# Separable FODE

1. Feb 13, 2010

### PhysicsMark

1. The problem statement, all variables and given/known data
The following is an explanation from my tutorial. I do not understand it.

$${\frac{d}{dx}(y_1{{y_2}'}-y_2{{y_1}'})+P(x)(y_1{{y_2}'}-y_2{{y_1}'})=0$$

Overlooking for the moment that P(x) may be undefined at certain values of x(so-called-singular points of the equation), we recognize this equation to be a separable first-order differential equation for the function $$y_1{{y_2}'}-y_2{{y_1}'}$$ that can be integrated to give:

$$y_1{{y_2}'}-y_2{{y_1}'}= C_{12}(exp(-{\int{P(x')dx'}})$$

$$y_1{{y_2}'}-y_2{{y_1}'}= C_{12}\phi[P]$$

Where $$C_{12}$$ is an integration constant that depends, possibly, on the choices of functions y_1 and y_2, and phi[P] is a functional, that is, a function that depends upon another function, in this case, P(x). The expression on the left hand side is called the Wronskian of the functions y_1 and y_2.

I do not understand how they separate this equation. Here is what I do. I first separate like this:
$${\frac{d}{dx}(y_1{{y_2}'}-y_2{{y_1}'})=-P(x)(y_1{{y_2}'}-y_2{{y_1}'})$$

Then I multiply both sides by dx. Next I divide both sides by $$y_1{{y_2}'}-y_2{{y_1}'}$$

Then I have:

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

What am I doing wrong here?

2. Feb 14, 2010

### rock.freak667

You can't just divide by y1y2'-y2y1' and have it cancel out from d/dx(y1y2'-y2y1')

$$\frac{d}{dx}(y_1y_2'-y_2y_1')+P(x)(y_1y_2'-y_2y_1')=0$$

$$\Rightarrow \frac{d}{dx}(y_1y_2'-y_2y_1')= -P(x)(y_1y_2'-y_2y_1')$$

now if we divide by y1y2'-y2y1' we will get

$$\frac{1}{y_1y_2'-y_2y_1'} \frac{d}{dx}(y_1y_2'-y_2y_1') = -P(x)$$

Now because the term on the left is a tad bit confusing, let's simplify it and let z=y1y2'-y2y1'. So our equation becomes

$$\frac{1}{z} \frac{dz}{dx}= -P(x) \Rightarrow \frac{1}{z} dz = -P(x)dx$$

now you can integrate both sides. I am sure you can get z isolated on the left side.

Then replace 'z'.

Alternatively what you can do is multiply both sides of the original differential equation by an integrating factor of e∫P(x) dx.