Differential Equations - Linear Factor Proof

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.

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

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

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

dextercioby

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

And now integrate.

Daniel.

HallsofIvy

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

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

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
[tex]\int d[uy]= \int u(x)q(x)dx[/tex]

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

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 =/

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.

DoubleMike

Correct me if I'm wrong, but [tex]\frac{d}{dx}[uy][/tex] 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?

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.

DoubleMike

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

dextercioby

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

Daniel.

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...?)

dextercioby

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

Daniel.

gokugreene

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

dextercioby

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

Daniel.

HallsofIvy

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

[itex]\int dx= x[/itex], [itex]\int dy= y[/itex], [itex]\int d(uy)= uy[/itex], etc.

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