How to Solve a First and Second Order ODE for y(x) and u=dy/dx

In summary: Yes, I understand that part. But what is f(x) in terms of F(x)?f(x)=F'(x)In summary, the conversation discusses differentiating and solving a second order ODE for y(x). It can be rewritten as a first order ODE for u=dy/dx, which can then be integrated to find the general solution with two arbitrary constants. The conversation also touches on the frustration of using mathematical symbols in online posts.
  • #1
Rake-MC
325
0
The Problem

You are given: http://img530.imageshack.us/img530/4468/88346476ca9.jpg

Where http://img408.imageshack.us/img408/209/53113174nt5.jpg is constant (taken as B).

a) Differentiate both sides to produce a second order ODE for y(x)
b) Show that it can be written as a first order ODE for u=dy/dx
http://img229.imageshack.us/img229/5861/81200662fv5.jpg
c) Find the general solution for part b), you should have two arbitrary constants.

Homework Equations



The fundamental theorem of calculus:
0241b99e934431d67c700892a79033cc.png


The Attempt at a Solution



a) Using fundamental theorem of calculus, d2y/dx2 = Bf(x)
=> d2y/dx2 = B(1 + (dy/dx)2)1/2

b) Let u = dy/dx => du/dx = B(1 + (dy/dx)2)1/2 = B(1 + u2)1/2

c) Not a clue.. I have actually got no idea where to start for this. I would solve du/dx first, but wouldn't that just result in what I was given, the equation for part a)?

Thanks in advance.
 
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  • #2
The differential equation for u is separable. Separate it and integrate both sides. Then once you have u, integrate again to get y.
 
  • #3
(a) and (b) looks correct.

(c)Just do as it says. You will not end up with what you have in part a, if you denote dy/dx with u. You just have to find the GS.
 
  • #4
Okay, thanks for the fast replies.

I separated and integrated both sides of the ODE for u like this:

http://img91.imageshack.us/img91/5970/84788939vg3.jpg

Think it's right so far, but I need to get theta back into terms of u. I don't exactly know how I'm going to go about this..
After that I'm going to need to integrate again I think, however my equation won't be in the form of a differential equation will it? How will I know what to integrate in terms of.
Thanks.
 
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  • #5
Gosh, that's so hard to read!

You have [itex]du/dx= B\sqrt{1+ u^2}[/itex] and then
[itex]\sqrt{1+ u^2}du= Bdx[/itex]

Well, that's wrong right there. You have to divide both sides by [itex]\sqrt{1+ u^2}[/itex], not multiply.
 
  • #6
Wow. I cannot believe that I made such a blatant error myself. Blimey!

Okay, I shall withdraw myself for sometime and make sure that next time I put something up here it isn't so horribly flawed haha...

Also, so that it isn't so hard to read next time, it would be great if someone could show me how to type all those math symbols in normal posts (I'm pretty new here).

Cheers
 
  • #7
If you click the mouse on some of those symbols, like the ones in Hall's post, a window should pop up with the TeX in it. You then enclose the stuff between [ tex] and [ /tex] (without the space). Like this,
[tex]
\int \frac{1}{\sqrt{1+ u^2}}du= \int Bdx
[/tex]
 
  • #8
I did the tex thing, but for some reason when I write more than one line of it I get a page load error. Been trying different combinations for about 40 minutes so I decided I'll try again later and post this for now.

http://img81.imageshack.us/img81/96/99642138ji4.jpg

The problem I have now is getting theta back in terms of u.

Thanks

EDIT: Hang on... theta = arcsinh(u)?
if it's that simple I'm going to be kicking myself!
 
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  • #9
Ok, arcsinh(u)=B*x+C. u=y'. You still have to get back to y. Sometimes the tex thing can be frustrating. But I'm not using tex, and you understand me, right?
 
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  • #10
so [tex] y = \int sinh(Bx+C)dx = \int sinh(Bx)dx + \int Cdx = \frac{1}{B}cosh(Bx) + Cx + D
[/tex]

The only thing I'm not sure about here is whether or not I'm allowed to take the C out of the sinh function, I think I am allowed to because it's an arbitrary constant. However if I'm not allowed to, I might need a helping hand in doing the integral because I'm not sure how I'd go about it.

Pretty sure this is correct though, and I do have the correct number of constants!
 
  • #11
You aren't allowed to take C outside of the sinh. You have to live with that. Can't you still integrate sinh(Bx+C) without making that mistake? Just substitute u=Bx+C and integrate.
 
  • #12
Okay hopefully this is the last time..


[tex] dx = \frac{du}{B} [/tex]

[tex] y = \int sinh(u)dx
\int \frac{1}{B} sinh(u)du
= \frac{1}{B} cosh(u) + D
= \frac{1}{B} cosh(Bx + C) + D
[/tex]

Thanks for being so patient, I know this is really frustrating!
 
  • #13
It's not half as frustrating as most. In fact, it's not frustrating at all. Yes, y=cosh(Bx+C)/B+D. If you have any doubts, put it back into the ODE and try it. The only thing frustrating here is that I have to keep scrolling my pages back and forth because of your oversized scans.
 
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  • #14
Referring to part (a) , I don't quite get it. If F'(x) = f(x), how do you solve the integral f(t) dt to get f(x)? Mine giving me a headstart to this? Thanks
 
  • #15
To get the integral of f(t) from 0 to x, you find an antiderivative F (such that F'=f) and evaluate F(x)-F(0), right? So the derivative of that integral is f(x).
 

FAQ: How to Solve a First and Second Order ODE for y(x) and u=dy/dx

What is the difference between first and second order ODEs?

A first order ODE involves only the first derivative of the unknown function, while a second order ODE involves the second derivative. In other words, a first order ODE can be written as y' = f(x,y), while a second order ODE can be written as y'' = f(x,y,y').

How are first and second order ODEs used in scientific research?

First and second order ODEs are used to model and study various physical phenomena, such as motion, chemical reactions, and population dynamics. They are also used in many engineering applications, such as designing control systems and analyzing electrical circuits.

What methods can be used to solve first and second order ODEs?

There are several methods for solving first and second order ODEs, including separation of variables, integrating factors, and series solutions. More complex ODEs may require numerical methods, such as Euler's Method or Runge-Kutta methods.

Can first and second order ODEs have multiple solutions?

Yes, first and second order ODEs can have multiple solutions. In fact, there can be an infinite number of solutions for a given ODE. This is because the general solution of a first or second order ODE contains an arbitrary constant, which can take on any value and result in a different solution.

How are initial conditions used in solving first and second order ODEs?

Initial conditions, also known as boundary conditions, are used to find the specific solution to an ODE that satisfies a given set of conditions. For first order ODEs, one initial condition is needed, while for second order ODEs, two initial conditions are needed (one for the function and one for its derivative).

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