## Differential Equations - Linear Factor Proof

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:

$$\frac{d}{dx}[uy]=uq(x)$$
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.
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

## Differential Equations - Linear Factor Proof

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 =/

Blog Entries: 9
Recognitions:
Homework Help
 Quote by DoubleMike But the left side is missing the dy...
Yes.

 Quote by DoubleMike It seems suspect that you can multiply the dx as a fraction like that,
Only to the ones which are unfamilar to methods involving differentials.

 Quote by DoubleMike then d[uy] really doesn't mean anything.

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

Daniel.
 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?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor 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.
 So where does that leave us? Also, don't you need a dy to integrate the left side?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor You have a "genuine" differential in the LHS which u can integrate without any problem...,ain't it so...? Daniel.
 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...?)
 Blog Entries: 9 Recognitions: Homework Help Science Advisor Well,what does the FTC say and why do you think it would apply here directly...? Daniel.
 dexter, where did you learn all of your math from?
 Blog Entries: 9 Recognitions: Homework Help Science Advisor About 99% from school,the rest individual study.Why? Daniel.
 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

 Similar discussions for: Differential Equations - Linear Factor Proof Thread Forum Replies Calculus & Beyond Homework 4 Calculus & Beyond Homework 3 Calculus & Beyond Homework 2 Calculus & Beyond Homework 0 Introductory Physics Homework 4