# Differential equation question (I think)

1. Homework Statement

It's to do with mirages, but I don't think the physics context is too important. It's also possible that the solution doesn't involve differential equations, and my method is completely wrong. I've been given that:

##A = \frac{n(1+ay)}{ sqrt(1+(y')^2)}##

where y' is dy/dx. I have to show that

##y = -\frac{1}{a} + \frac{A}{na}## cosh( ##\frac{na}{A}##(x-x0))

A, n and a are real constants.

## The Attempt at a Solution

I tried rearranging it to get dy/dx = ... , by multiplying by the denominator then dividing by A, squaring both sides, subtracting 1 and then square rooting. I ended up with:

##y'=\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}##

And that's separable. So
##dx = \frac{1}{\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}}dy##
And then if I integrate both sides by doing substitutions like u =1+ay, I get ln of something. Nowhere near the show that result.
Checked the rearrangement so many times. What did I do?

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SteamKing
Staff Emeritus
Homework Helper

## Homework Statement

It's to do with mirages, but I don't think the physics context is too important. It's also possible that the solution doesn't involve differential equations, and my method is completely wrong. I've been given that:

##A = \frac{n(1+ay)}{ sqrt(1+(y')^2)}##

where y' is dy/dx. I have to show that

##y = -\frac{1}{a} + \frac{A}{na}## cosh( ##\frac{na}{A}##(x-x0))

A, n and a are real constants.

## The Attempt at a Solution

I tried rearranging it to get dy/dx = ... , by multiplying by the denominator then dividing by A, squaring both sides, subtracting 1 and then square rooting. I ended up with:

##y'=\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}##

And that's separable. So
##dx = \frac{1}{\sqrt{\frac{n^2 (1+ay)^2}{A^2} -1}}dy##
And then if I integrate both sides by doing substitutions like u =1+ay, I get ln of something. Nowhere near the show that result.
Checked the rearrangement so many times. What did I do?
It's not clear how your integration got LN (something). Please post your work to get this result.

Sorry. So I substituted u=1+ay, du/a = dy so that the integration becomes the integral of

##\frac{1}{a} \frac{1}{\sqrt{ \frac{n^2 u^2}{A^2} -1}} du##

Which is ln. I checked with wolfram alpha by replacing n^2/A^2 with just b, because it's just a constant, like the 1/a which is ignored in the integral. So I think it's still the same thing.

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epenguin
Homework Helper
Gold Member
From what you say you get x = LN(some function of y).
This is not at all a sign that you have done anything wrong - when you work on your result (invert the equation to get y = some function of x) this may very well have a cosh in it.

• whatisreality
I don't think I know how to do that. To invert the ln, wouldn't that involve e?

Oh, wait, there's a definition of cosh involving e! 1/2 e-x+ex?

Haven't actually done the integral yet, just assumed because of the ln it would be wrong. So I'll try that integral now...

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HallsofIvy
Homework Helper
Oh, wait, there's a definition of cosh involving e! 1/2 e-x+ex?
Well, 1/2 (e-x+ex)- the parentheses are important. If B= (1/2)(e-x+ ex) then e-x+ ex= 2B. Multiply on both sides by ex to get 1+ (ex)2= 2Bex. Let y= ex and that becomes the quadratic equation 1+ y2= 2By or y2- 2By+ 1= 0. Use the quadratic formula to find y and then take the logarithm to solve y= ex for x.

• whatisreality
Oh, forgot about the brackets. So my integral is equal to
1/a ln(##\sqrt{a^2x^2-1}+ax) + c

If a is the n^2/A^2.

And that integral is equal to x. So I don't think I end up with two terms of e to the power of something, just two es multiplied together.

Hardest 'show that' question I've ever been given!

From what you say you get x = LN(some function of y).
This is not at all a sign that you have done anything wrong - when you work on your result (invert the equation to get y = some function of x) this may very well have a cosh in it.
I think I still haven't got a cosh. Although I haven't posted my full working, as far as I can tell I don't get two e^something terms when I invert to find y. I'm fairly sure my integration is correct, checked it with wolfram alpha!

epenguin
Homework Helper
Gold Member
I'm not working on it but idea where to look: it might have to do with when you write your first √ you have to write ± so maybe you get two |solutions you have to combine in one? Or if you integrate to get a LN there a tricky thing about LN | | , maybe? - check your textbook. Maybe these two things are related.

• whatisreality
epenguin
Homework Helper
Gold Member
Forget most of what I said. We haven't seen working and if we had I dare say someone would have spotted quite soon.

Now I had time to look at it (then was cut off by 48 h Internet blackout) it is looking like a simple error in calculation. The OP probably overlooked a square rooting at some point - he would have got this log result integrating an inverse quadratic, instead the problem gives an inverse square root of a quadratic whose standard integral is an inverse cosh.

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• whatisreality
HallsofIvy
Homework Helper
Since the problem is to show that $y = -\frac{1}{a} + \frac{A}{na} cosh( naA \frac{na}{A} (x-x0))$ is a solution to the given differential equation, I don't understand why you are integrating at all! Just differentiate y, the substitute the given y and its derivative into the equation and show that the equation is satisfied.

• whatisreality
epenguin
Homework Helper
Gold Member
Well yes people should always do that when they read physics instead of the so-called 'solving', they will get through college them faster and with less distraction from the physics. (Seriously).
You

• whatisreality
Since the problem is to show that $y = -\frac{1}{a} + \frac{A}{na} cosh( naA \frac{na}{A} (x-x0))$ is a solution to the given differential equation, I don't understand why you are integrating at all! Just differentiate y, the substitute the given y and its derivative into the equation and show that the equation is satisfied.
Yes, I did try that first.

So dy/dx is -sinh(na/A (x-x0) ) from the given equation, so y'2 is then sinh2(...), and sub into equation for A to get n(1+ay)/ sqrt (1+ sinh2(...) ) = n(1+ay) / sqrt ( cosh2(...) )
which is just n(1+ay) / cosh(...) = A. Then substitute the y? That'll be the step I forgot! Oops!

Edit: And the step which, having just completed, gave me the answer in a really quick and easy way! What a stupid mistake to make. Thanks for solving my problem! :D

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Forget most of what I said. We haven't seen working and if we had I dare say someone would have spotted quite soon.

Now I had time to look at it (then was cut off by 48 h Internet blackout) it is looking like a simple error in calculation. The OP probably overlooked a square rooting at some point - he would have got this log result integrating and inverse quadratic, instead the problem gives an inverse square root of a quadratic whose standard integral is an inverse cosh.
I got my ln result from integrating ## \frac{1}{ \sqrt{\frac{n^2(1+ay)^2}{A^2} -1}}##. Not that I actually know how to integrate that, I had to use wolfram alpha... Yeah, I suppose that might be the main problem really, I don't know where to even start with an integral like that!

Because even if I try and condense the thing under the square root into something manageable, which for me means replacing multiple constants with a single constant, I end up with
1 / ##\sqrt{a(1+by)^2 -1}##
Which isn't a standard kind of integral that I know...

Also an integral I no longer need to solve! :) I do feel very silly though!

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Forget most of what I said. We haven't seen working and if we had I dare say someone would have spotted quite soon.

Now I had time to look at it (then was cut off by 48 h Internet blackout) it is looking like a simple error in calculation. The OP probably overlooked a square rooting at some point - he would have got this log result integrating and inverse quadratic, instead the problem gives an inverse square root of a quadratic whose standard integral is an inverse cosh.
I know, I did realise afterwards that not posting workings must have made it more difficult. I do appreciate you replying despite that :) Thank you for helping me everyone! :D

epenguin
Homework Helper
Gold Member
I got my ln result from integrating ## \frac{1}{ \sqrt{\frac{n^2(1+ay)^2}{A^2} -1}}##. Not that I actually know how to integrate that, I had to use wolfram alpha... Yeah, I suppose that might be the main problem really, I don't know where to even start with an integral like that!

Because even if I try and condense the thing under the square root into something manageable, which for me means replacing multiple constants with a single constant, I end up with
1 / ##\sqrt{a(1+by)^2 -1}##
Which isn't a standard kind of integral that I know...

Also an integral I no longer need to solve! :) I do feel very silly though!
We were taught to do.these integrations at school. Somewhat pointless knowledge for most, I think. Or not even knowledge, at least I remember being asked for it or something like it a few months after school - similar things have some slight application in chemical kinetics - and I couldn't do it.

You should however be able to reduce it to standard form by substitution, otherwise you may not be able to find or recognize it in a table of standard forms. You only needed to make a new variable Y = n(1 + ay)/A and you will get to integrating 1/√(Y2 - 1), a standard form found in halfway decent books on calculus.