# Differential Equations - Linear Factor Proof

1. Feb 9, 2005

### DoubleMike

My class has recently done an intro to differential equations, and although I understand the method of solving simple equations, I want to know why the method of Linear Factors works. Unfortunately my book hasn't provided a proof for it. :grumpy:

Also in the final step where you integrate both sides of the equation:

$$\frac{d}{dx}[uy]=uq(x)$$
the book says to integrate each side in respect to the variable in them

So I would have $$\int\frac{d}{dx}[uy] dy= \int uq(x)dx$$
This doesn't make sense, considering each side has been multiplied by different differentials.

2. Feb 9, 2005

### dextercioby

No,you've missed something and the final expression is totally wrong.
$$\frac{d}{dx}(uy)=uq(x) \Rightarrow d(uy)=uq(x) dx$$

And now integrate.

Daniel.

3. Feb 10, 2005

### HallsofIvy

That is what I would call using an "integrating factor" to solve a linear equation.

You have a differential equation of the form
$$\frac{dy}{dx}+ p(x) y= q(x)$$
and multiply by some function u(x) such that
$$u(x)\frac{dy}{dx}+ u(x)p(x) y= \frac{d(u(x)y)}{dx}$$

It's easy to show that u(x) must satisfy u'= u(x)p(x), a separable equation.

I doubt that what you gave is, in fact, exactly what is in your book. (If it is it's a typo!)
What is correct is that
$$\int d[uy]= \int u(x)q(x)dx$$

Of course, the integral on the left is just u(x)y.

4. Feb 10, 2005

### DoubleMike

But the left side is missing the dy...

It seems suspect that you can multiply the dx as a fraction like that, then d[uy] really doesn't mean anything.

It's like performing math on notation... I'm sure it's just shorthand =/

5. Feb 10, 2005

### dextercioby

Yes.

Only to the ones which are unfamilar to methods involving differentials.

Of course ot does.It's the differential of the product u(x)y.

Daniel.

6. Feb 10, 2005

### DoubleMike

Correct me if I'm wrong, but $$\frac{d}{dx}[uy]$$ is just notation for the derivative of uy, the dx isn't really a differential at all.

If it was, then in reality you would be dividing by dx... Are you telling me that d[uy] = dy?

7. Feb 10, 2005

### dextercioby

That "dx" BEARS the name "differential of the independent variable".It's not a differential "stricto sensu".

As for the last,i'm not that naive to claim such thing...They obviously represent different objects.

Daniel.

8. Feb 10, 2005

### DoubleMike

So where does that leave us? Also, don't you need a dy to integrate the left side?

9. Feb 10, 2005

### dextercioby

You have a "genuine" differential in the LHS which u can integrate without any problem...,ain't it so...?

Daniel.

10. Feb 10, 2005

### DoubleMike

I thought that you needed to know in respect to which variable. I've just been mechanically integrating both sides, but I don't know why the left side doesn't need a dy.

As far as I'm concerned, the LHS is just an equation which happens to be the differential. It would still need a dy (it reminds me of the fundamental theorem of calculus for some reason...?)

11. Feb 10, 2005

### dextercioby

Well,what does the FTC say and why do you think it would apply here directly...?

Daniel.

12. Feb 11, 2005

### gokugreene

dexter, where did you learn all of your math from?

13. Feb 11, 2005

### dextercioby

About 99% from school,the rest individual study.Why?

Daniel.

14. Feb 11, 2005

### HallsofIvy

The left side: d[uy] doesn't need a left side because it is a differential all by itself:

$\int dx= x$, $\int dy= y$, $\int d(uy)= uy$, etc.

You could, if you like, say $\int\frac{d(uy)}{dx}dx= uy$ but most people prefer not to have that 'double' dx.