ODE Reduction in Order

Factual Questions
1

First, yes this is homework, but no I am not looking for someone to just give me the answers. Getting the answers is easy enough with Matlab/Maple/Wolfram Alpha, etc. I just can’t figure out how to solve these two problems by reducing them to a first order ODE. As I understand it,

  1. The general solution y=y1+y2 where y1 and y2 are each solutions, given y1 is a solution then y2=u(x) * y1

  2. Find the first and second derivative of y2 and then plug into the original ODE

  3. Now set w=u’ and the second order ODE has been reduced to a first order.

  4. Solve for w. Then integrate to find u.

  5. Now, take the original (given, or obvious by inspection solution) y1 and multiple by u to solve the ODE and find y

That’s a rough outline of my plan. However, I can’t find an “obvious” y1 solution and none is given. Here are the ODEs…

A: yy’’=3(y’)^2
B: y’’=1+(y’)^2

I’ve tried solutions of the form y=x^m, y=e^x, etc. Any advice would be appreciated.

2

For B, if I understand your notation, it looks like there’s no y, just y’ and y’’. So couldn’t you reduce it to order 1 by simply setting w = y’ ?

3

For the first one: I thought I had an idea how to do this, but I didn’t. Let me get back to you on this one.

I don’t think you need your whole procedure for the second one. There’s a way to reduce that one in order without “guessing” a solution. ETA: which Thudlow Boink described without being as cagey as I was being.

By the way:

The general solution of a linear ODE is the arbitrary linear combination of two solutions y[sub]1[/sub] and y[sub]2[/sub]. The ODEs you have here aren’t linear.

Hope this helps! Feel free to ask follow-up questions.

4

For the first one: try a power law. This is the form you should “guess” when the equation is homogeneous in both the independent and the dependent variable. (Homogeneous means that if you rescale the variable by a factor lambda, you get the same equation back.)

5

… and upon further reflection, I realized that the method described in the OP doesn’t work unless the ODE is linear. You’ll need a different technique to reduce the order of the ODE in question A. The fact that the ODE is homogeneous (as I mentioned in my previous post) is useful.

6

While we’re at it, if you’re setting up a problem for an eventual numerical solution, you can always reduce the order of a system of differential equations to 1, by increasing the number of equations in the system. Just define a new variable for y’, and another one for y’’, and so on as necessary. Though this doesn’t appear to be what the OP is looking for in this case.

7

First, sorry it has taken me so long to respond. My phone is giving me issues posting from mobile so I have to actually be at a computer so I couldn’t respond as quickly as i would have liked.

Second, thanks very much for the help guys. I think my fundamental misconception was that I was only thinking of the linear case and assumed I needed to either be given a solution y1 or obtain one via some other means such as inspection. Then I could do the y2=u(x)*y1 and y=y1+y2.

I ended up following the advice I got here for

B: y’’=1+(y’)^2

by setting w=y’. Then I could solve via separation of variables to obtain y=c1 - ln(cos(x+c2))

where c1, c2 are arbitrary constants that could be obtained via initial conditions.

For A,

A: y*(y’’)=3(y’)^2

I set w=y’, and, this is the part I kept messing up, I saw in my book where they then say that y’’=(dw/dy)w (as opposed to simply y’’=w’). Once you plug this is and get y(dw/dy)w=3w^2 you can use separation of variables to obtain y=(2x+C)^-0.5 where C is an arbritary constant obtained by using the initial conditions.

Thanks again for the help!

8

That’s almost the same as the answer I got. Keep in mind, though, that any second-order ODE should have two arbitrary constants in its general solution, not just the one that you have here. So your answer isn’t quite complete.

9

I think this is referring to a trick you can do if the ODE is *homogenous *(i.e. x does not appear explicitly anywhere in the equation). In that case you can treat the equation as if y was the independent variable, and y’ and y’’ are functions of y.

So when the book is doing that transformation, instead of defining w as being equal to y’ as function of x, w is y’ as a function of y, so derivatives with respect to x have to be converted to derivatives with respect to y through the chain rule, so:

dw/dx = dw/dy * dy/dx = dw/dy * w

which explains the weirdness of the transformation.

10

This is why I prefer to use Liebnitz notation all the way. When you put the "d"s in explicitly, everywhere, everything is so much clearer.

11

But then you run into other problems, like if you are evaluating the function f at two different variables u, and v, do you name the derivative df/du, or df/dv? Lagrange notation resolves that ambiguity.

12

Spelling nitpick: All the modern sources I’ve seen refer to him as “Leibniz,” but I’ve seen older references to “Leibnitz” (with the t). Upon researching, I found the New World Encyclopedia’s explanation that

IIUC the “z” in German sounds like “tz,” so the pronunciation would be the same either way. However, “ei” and “ie” are quite distinct in German, and Leibniz’s name is unambiguously spelled with “ei.”

I’m not sure what kind of situation you’re describing. Can you provide some context where this would be an issue?

13

Say g(x) = f(v(x)) + f(u(x)). Find dg/dx

dg/dx(x) = df/dv (v(x)) * dv/dx (x) + df/du (u(x)) * du/dx (x)

But df/dv and df/du here are the same function (the derivative of f with respect to its parameter) but Leibniz notation gives it different names depending on what letter we assign to the parameter. Lagrange notation would call it just f’ in both locations.

In my experience people who want to think of the variables as representing actual physical quantities varying with each other prefer Leibniz notation, while people who want to think of the functions as first class objects and the parameters as arbitrary prefer Lagrange or other notations.