inverse product rule in DEs

Question

l · Answer

Works for a lot of them, apparently.$$y' = ax + by + c$$$$\Rightarrow y' - by = ax + c$$$$\Rightarrow  e^{-bx} - be^{-bx} y = e^{-bx}(ax+c)$$$$\Rightarrow (e^{-{bx}}y)' = e^{-bx}(ax+c)$$$$\Rightarrow  y = \frac{1}{e^{-bx}}\int e^{-bx}(ax+c) dx $$

l · Answer

So I just try to express the $y$ and $y'$ term in such a way that it turns into a product-rule form. Here's another one:$$x\cos x y' + (x\sin x + \cos x ) y = 1$$This one isn't immediately obvious but you've got to multiply both sides by $\sec^2 x$ before you begin noticing the product-rule form. Now I'm wondering if this trick is theoretically possible for the following:$$x=yy'-(y')^2$$

ganeshie8 · Answer

Looks you've just figured out a nice general method to solve linear differential equations !

kainui · Answer

A possible substitution maybe, but I don't know if it'll get you very far:
$$(yy')'-yy''=(y')^2 $$

kainui · Answer

maybe it suggests a substitution yy'=u

ganeshie8 · Answer

Unlike other two DEs which you messed with earlier,  x=yy′−(y′)2 is not linear.  So your general method using reverse product rule will not work

l · Answer

So is it theoretically impossible for me to find something to mess with in such a way that I get a product-rule form?

kainui · Answer

Depends on how creative you are, but I'd say it's impossible in the strict sense, yeah.

l · Answer

Well, that is a shame. Could you show me how you formally solve this kind of stuff then?

ganeshie8 · Answer

Yeah,  reverse product rule can be used in other situations too.  Your method of finding integrating factor and simply multiplying both sides may not work with the last de

kainui · Answer

I have no method for this, I would just make stuff up or use the wolfram alpha method.

kainui · Answer

Depends, do you want the answer "y" or do you want to have fun? That's basically how I approach these, lol

l · Answer

Integrating factor? So is that $e^{-bx}$ term the "integrating factor"?

kainui · Answer

Here, I'll just show you the method for integrating factor and then maybe that'll help?

l · Answer

Jeez I thought I'd invented something of my own, but turns out it's taken as always. -_-

l · Answer

The problem was originally posed in this manner:

Solve $x = py - p^2$ where $p$ denotes $dy/dx$.

Now is that a stupid way to express the problem, or does expressing it like that have some serious benefit or meaning?

kainui · Answer

What you're doing is similar but not completely the same it seems. Specifically integrating factor method works for differential equations of this form:

$$y'(x)+p(x)y(x)=q(x)$$
You could have another function of x multiplying y' but there's no reason just like in solving a quadratinc you can divide the entire equation by 'a' to make a=1 essentialy. Same idea.

Now the integrating factor they usually choose this, $\mu(x)$ so we will do the same. Multiply both sides of the equation by it.

$$\mu (x) y'(x)+\mu(x)p(x)y(x) = \mu(x)q(x)$$
Since $\mu(x)$ is our own construction we make it whatever we want. We make it this for our convenience, even though we are making a new differential equation it seems like maybe it might be more difficult but really it ends up being easier.

$$\mu'(x) \equiv \mu(x)p(x) $$

Now we plug in, yay
$$\mu(x)y'(x) + \mu'(x)y(x)=\mu(x)q(x)$$$$(\mu(x)y(x))' = \mu(x)q(x)$$
that's the reverse product rule step, now we basically have this solved and all there is to do is to determine what $\mu(x)$ is from the differential equation above, and plug and chug.

l · Answer

:(

And how would you go about solving the one I gave ya?

kainui · Answer

I already said

kainui · Answer

$\color{blue}{\text{Originally Posted by}}$ @Kainui
I have no method for this, I would just make stuff up or use the wolfram alpha method.
$\color{blue}{\text{End of Quote}}$

kainui · Answer

I'm just showing you what integrating factors are so you can compare and contrast with your related method. Methods are probably better called "tricks". You either learn from other people or make them up yourself. http://www.azquotes.com/picture-quotes/quote-mathematics-is-a-collection-of-cheap-tricks-and-dirty-jokes-lipman-bers-76-49-78.jpg

ganeshie8 · Answer

x=yy′−(y′)^2 

Differentiate and get
1 = (y′)^2 +yy''- 2y'y''

Eliminating y from above two eqns and substituting y'=u gives
dx/du = x/[u(u^2-1)] + u/(u^2-1)

This is a linear eqn which can be easily solved using your reverse product rule trick

l · Answer

Wow :)