Understanding Separable First-Order Differential Equations

[PhysicsMark]

Homework Statement

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?

 

[rock.freak667]

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

Let's do it your way

\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}.