Ordinary Differential Equations: Separation of Variables

  • #1
Please have mercy on my non-mathematical mind...

I am struggling so hard with Differential Equations.

This is my third time to take the class and I still feel like I am walking through the woods at midnight on an extremely dark night.

This is the problem that I am working on:

dy/dx = x / (1 + y4)

The instructions are to find (on paper) the general solution.

Well, I did that and got the following:

y + (y5 / 5) = (1/2) x2 + C1

Now, I am supposed to verify that the equation found is a solution to the differential equation.

Now, I know that the original equation can also be written as:

y' = x / (1 + y4)

And, I know that I should be able to take the derivative of the general solution, then plug in and solve.

But, I also think (whoa... this is where it gets sketchy), that the general solution should be have the terms of y on the LHS and collected.

So, what did I miss this time? Substitution by parts? Some trig identity?

Please help me. I so wish that I got this.

I am ready to go dig ditches somewhere if this fog doesn't lift for me.

TIA
 

Answers and Replies

  • #2
jmb
67
1
y + (y5 / 5) = (1/2) x2 + C1
That's fine!

But, I also think (whoa... this is where it gets sketchy), that the general solution should be have the terms of y on the LHS and collected.

Solving the differential equation and rearranging your answer into the form "y= something" are separate tasks. While you should always try and do both, sometimes you will come across situations where you can solve the equation but can't express the answer as "y= something". That's just the way it goes!

The good news is you don't need to write the answer as "y= something" in order to find the derivative. Just take the answer as you have it and differentiate the whole equation with respect to x, making use of the chain rule. You should then find that you can rearrange the resulting equation to give you y' without difficulty. The result should just be the differential equation you started with --- thus proving you found a correct solution

Have a go and if you get stuck, or just want us to check your answer, then post your working here and we can look through it for you...
 
  • #3
The good news is you don't need to write the answer as "y= something" in order to find the derivative. Just take the answer as you have it and differentiate the whole equation with respect to x, making use of the chain rule. You should then find that you can rearrange the resulting equation to give you y' without difficulty. The result should just be the differential equation you started with --- thus proving you found a correct solution

That is the next step for this problem and I was able to knock that out with implicit differentiation without a problem.

It was when I took that result and tried to plug it back into the original (seems kinda redundant... but I did so, and started to get a bunch of odd fractional "parts" (y and y').

Maybe I freaked out prematurely. I will continue to work the algebra out and see if I can complete the problem.

Thanks for the reply jmb!
 
  • #4
jmb
67
1
It was when I took that result and tried to plug it back into the original (seems kinda redundant... but I did so, and started to get a bunch of odd fractional "parts" (y and y').

'Plugging it back into the original' shouldn't be necessary, if you have done the differentiating and rearranging correctly you will already have ended up with the original equation!

Please post your working so that we can see where your mistake is...
 
  • #5
This is the "1(b)" problem where I started to stray:

"Calculate, by hand, using implicit differentiation, the derivative dy/dx of your general solution, and check that it satisfies the differential equation (??)."

So, I did the first part and then went to try to figure out what the "check that it satisfies the differential equation (??)." meant and that is where I decided to ask for help.

Because I know (or think I know) that if I have a particular solution, that I can use y(x) and y'(x) (and so on...) to substitute into the original D.E. and verify that the solution I find is valid?

---

Maybe this is a bad problem? (or at least the last part with ??).

We had a "proctor" last week to lead the class because our instructor had some commitment to honor. I don't think that this was discussed, but I could be wrong. I do have a touch of the adhd...
 
  • #6
jmb
67
1
Maybe this is a bad problem? (or at least the last part with ??).

The problem is fine.

This is the "1(b)" problem where I started to stray:

"Calculate, by hand, using implicit differentiation, the derivative dy/dx of your general solution, and check that it satisfies the differential equation (??)."

So, I did the first part and then went to try to figure out what the "check that it satisfies the differential equation (??)." meant

If you "did the first part" correctly then the rest should happen automatically! Please post your working to this part...
 
  • #7
1(a)
Given
dy/dx = x/(1+y4)

After Integrating
y + (y5/5) = (1/2)x2+C1

1(b)
dy/dx(1/5)(y5) +dy/dx(y) - (1/2)(x2)(d/dx) + C1

dy/dx(y4+1) = x

dy/dx = x/(y4+1)

Verify
y = (1/2)x2 - (1/5)(y5)

y' = (1/6)x3 - (1/30)(y6)

Plug N Pray
(1/6)(x3-(1/5)y6) = x/(((1/2)x2-(y5/5))4+1)

**This is as far as I have gotten, because I have another class that I am trying to finish HW in and I need to leave in 45 mins to be on time.**

I will be back on this around ~13:00 (GMT-6:00).

Thank you again jmb.
 
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  • #8
jmb
67
1
1(a)
Given
dy/dx = x/(1+y4)

After Integrating
y + (y5/5) = (1/2)x2+C1

Yup, as I said before this bit is fine.

1(b)
dy/dx(1/5)(y5) +dy/dx(y) - (1/2)(x2)(d/dx) + C1

As it stands this is wrong. I suspect a lot of the problem is your abuse of notation, and that you do have some idea of how to actually differentiate the equation (especially as you end up with the right answer). But if I had to mark this as an exam I'd be forced to give you no marks as it's so unclear as to not give enough assurance you know what you are doing!
If you have time, try to do it again making it more clear what you mean... [And if you don't understand what's wrong with this then please say as it may mean you don't understand what's happening correctly!]

dy/dx(y4+1) = x

This is correct, although as I say above, I have worries as to how you arrived at it!

dy/dx = x/(y4+1)

Yes, and so you have got the differential equation you started with (i.e. this is exactly the equation you solved by separation of variables in part 1a) after differentiating your answer --- this is your verification!

y = (1/2)x2 - (1/5)(y5)

y' = (1/6)x3 - (1/30)(y6)

This is just complete nonsense I'm afraid. You have differentiated the LHS with respect to x, and then integrated part of the RHS with respect to x and the other part with respect to y, that's just meaningless. It again makes me worry as to whether you really understand how to differentiate the equation...

In summary you have got to the right answer, but your working is really unclear and there's also some stuff (like the last bit above) that you've done completely wrong. If you are serious about passing this class I'd really strongly advise you to try and write it out again this time being much more accurate and careful with your working (showing more steps at this level wouldn't hurt). I (and I'm sure others on the forum) would be more than happy to look through it again and comment on it.
 
  • #9
Okay, well I apparently am not going to the other class today (I haven't completed that work either, but actually get the other subject (Statics). I spend way too much time in this class muddling my way through examples and searching the web for some obscure example that will enlighten me to the D.E. mentality.)

Anyway, I am always in this predicament. I honestly cannot understand how I let myself get like this. I keep fighting to keep up. I have no TV, no job, a wife and no kids. We both go to school, but I have hit some kind of roadblock that has plagued me for three whole semesters now.

I think a lot of it is that everything takes me so long to study and digest. The really messed up thing is that I am repeating my classes from the summer! Anyway, enough OT. I guess I need to go somewhere else to get that off my chest.

---

I honestly would love nothing more than to pass this class. Actually, if I don't I am pretty sure that I am going to have to leave college for good.

So...

I am going to post a link to the actual homework assignment (for clarity).

Yes, it is due today. But, right now I am so sick and tired of not getting it, I really couldn't care less if I got no points for this as long as I walk away with some kind of hope that I will be able to actually "learn" D.E. (I am saying this, because I really would surprise myself if there would be some kind of "blinding flash of genius" that would enable ME to complete this assignment on time. And, I am not so much concerned with "getting 'er done", as I am with "getting it for good". So, my loss for not asking sooner. Ignore the due date. My life is on hold for this one ;-) )

Homework #1, Pg 1
th_UAB_MA252_September_2009_HW1_01.jpg

Homework #1, Pg 2
th_UAB_MA252_September_2009_HW1_02.jpg

Homework #1, My work so far
th_UAB_MA252_September_2009_HW1_Work01.jpg


jmb, I will reply to your last post in just a minute. I am going to take a minute to clean this desk off and eat a quick lunch.
 
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  • #10
[And if you don't understand what's wrong with this then please say as it may mean you don't understand what's happening correctly!]

I actually don't know I did wrong there. I have only vaguely been able to understand D.E. so far. Math, in general, has not been my strongest subject.

Anyway, I couldn't conceive another way to implicitly differentiate that equation. To be honest, it does seem like cheating. So, I am sure that I was "skipping" a procedure somewhere. (Believe it or not, as "bad" as I am with math, I have been frequently admonished by teachers for leaving out things like the chain rule, or not writing out my substitutions - but a point here and there is not as bad as a blank last page :-) )

And, amazingly enough, I get the correct answers frequently. Very frustrating too, because some of the algebra I have written out to "show work" is often similar work that I skip and do in my head and caused me trouble on tests for a while (with time and manual execution).

But, I realize that mathematics curriculum focuses on demonstration of skill. So, I am open to criticisms that would make my work easier to read or validate my skills (However, time is always an issue for me on tests, so I do have limits to what I can change)

This is just complete nonsense I'm afraid. You have differentiated the LHS with respect to x, and then integrated part of the RHS with respect to x and the other part with respect to y, that's just meaningless. It again makes me worry as to whether you really understand how to differentiate the equation...

I knew that I wasn't going anywhere with that. I like the way you "polish" it ;-)

The frustration from realizing that I was LOST, again - was the impetus for me to join the forum while I was coming to grips with the fact that what I had just done "was complete nonsense". (I don't think I was that nice to me though...)
 
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  • #11
--- this is your verification!

Now, I did know that when I was doing the work.

There are two things that come to mind now that caused me some unease.

1. If you differentiate the function that you found to be the general solution, how could it ever be different from the original equation? (I am not being snippy or anything, I really have a hard time visualizing how that this case could ever be different.)

Is this a "proof validation" sort of exercise? Not that it matters, it just seemed kind of trivial and my natural distrust of "trivial" things kind of left me feeling like something wasn't right. (most likely, it was my incorrect execution use of a concept that once clicked in attempt #1 or #2 at this subject).

2. This obviously isn't the end of 1(b) though. I mean, the problem statement said "check." To me, that means substitution. That is where I got frustrated.

So, what is it that I am missing fundamentally here? I'm sure that my D.E. foundation in general is like a block of Swiss Cheese, but what concept am I glossing over?
 
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  • #12
4
0
I'm just starting off ODE too....

Here's my work up to a certain point...

dy/dx = x / (1+y^4)
multiply both sides by (1+y^4) and dx you get

(1+y^4) *dy = x dx
Integrate

I get y + y^5 = (1/2)x^2 + C
 
  • #13
jmb
67
1
OK, if you don't understand why what you've written is wrong, let's start there, since misusing notation often implies a misunderstanding...

You went from:

[tex]y + \frac{y^5}{5} = \frac{1}{2}x^2+C_1[/tex]

to:

[tex]\frac{dy}{dx}\left(\frac{1}{5}\right)\left(y^5\right) +\frac{dy}{dx}\left(y\right) - \left(\frac{1}{2}\right)\left(x^2\right)\left(\frac{d}{dx}\right) + C_1[/tex]

as supposedly your first step in differentiating the equation.

Can you explain what you meant by: [itex]\frac{dy}{dx}\left(\frac{1}{5}\right)\left(y^5\right)[/itex]?

What about [itex]\left(\frac{1}{2}\right)\left(x^2\right)\left(\frac{d}{dx}\right)[/itex]?

And why is [itex]C_1[/itex] still present?

Post your answer to those three questions and we'll work through things from there...
 
  • #14
jmb
67
1
1. If you differentiate the function that you found to be the general solution, how could it ever be different from the original equation? (I am not being snippy or anything, I really have a hard time visualizing how that this case could ever be different.)

Because the differential equation might not be that simple. If for example the differential equation were:
[tex]
\frac{d^2y}{dx^2}+y + \frac{dy}{dx}-2x^2=0
[/tex]
and I found a solution [itex]y(x)[/itex] then I could differentiate my answer to get [itex]dy/dx[/itex] and again to get [itex]d^2y/dx^2[/itex] and then I would have to add these three answers together and verify that the answer I got was [itex]2x^2[/itex].

But that's not what you had: your equation was already in the form "dy/dx = something", so if you find a solution and then take it's derivative to obtain "dy/dx" all you have to do is verify that you are back at the same equation.

YES this is just like integrating something and then differentiating it again and saying oh look I got the same thing! The point is just to verify that you didn't make any mistakes in the integration (or subsequent differentiation) --- it's a good check for errors. It really is that trivial. One way to prove that a technique to integrate something is successful is just to show that differentiation "undoes" the process.

2. This obviously isn't the end of 1(b) though. I mean, the problem statement said "check." To me, that means substitution. That is where I got frustrated.

No it really is the end! You have checked --- the only way you can't be happy is if you aren't happy with the rules you used for doing the differentiation, and in that case you would have to rederive those rules from first principles which is way beyond what the question is asking for.
 
  • #15
Can you explain what you meant by: [itex]\frac{dy}{dx}\left(\frac{1}{5}\right)\left(y^5\right)[/itex]?

I shouldn't have written the differential notation with the y, right? It seems redundant anyway now that you point it out.

Would this have been correct? [itex]\frac{d}{dx}\left(\frac{1}{5}\right)\left(y^5\right)[/itex]

What about [itex]\left(\frac{1}{2}\right)\left(x^2\right)\left(\frac{d}{dx}\right)[/itex]?

I knew this was not needed, but since I was unsure of my "method", I was "showing my work"?? (lol)

And why is [itex]C_1[/itex] still present?

Yeah, I am bad for that. The derivative of a constant = 0.
 
  • #16
35,134
6,884
EtherealMonke,
I saw your post 7 and thought it deserved some comment. In line with jmb's reply in post 13, you started with an equation. You differentiated both sides implicitly, but your next step is not an equation. You are apparently using dy/dx when you mean d/dx, and d/dx when you mean dy/dx.

dy/dx is a thing; namely, the derivative of y with respect to x. In contrast, d/dx is an action that is yet to be taken. More precisely, d/dx is an operator that signifies your intent to take the derivative of what is to the right of it.

For example,
[tex]\frac{dy}{dx}\left(\frac{1}{5}\right)\left(y^5\right) [/tex]
means the derivative of y with respect to x times 1/5 times y5. This error is similar to the one where a student interprets sin x as "sine times x."

Students who are new to calculus sometimes write things like this:
dy/dx (x2) = 2x. I understand what they mean to say, but what they should say is d/dx(x2) = 2x. A student who does not understand the difference will have an extremely hard time when he or she sees
[tex]\frac{dy}{dx}x^2[/tex]
and interprets this incorrectly to mean "take the derivative of x2." What this expression actually means is the product of some unspecified function and x2.
 
  • #17
I'm just starting off ODE too....

Here's my work up to a certain point...

dy/dx = x / (1+y^4)
multiply both sides by (1+y^4) and dx you get

(1+y^4) *dy = x dx
Integrate

I get y + y^5 = (1/2)x^2 + C

Hey Alfonso. Are you in the 5:30 PM class?

Your work is almost right. You missed a step when you integrated.
 
  • #18
EtherealMonke<i>y</i>,
I saw your post 7 and thought it deserved some comment. In line with jmb's reply in post 13, you started with an equation. You differentiated both sides implicitly, but your next step is not an equation.

I think I was going to put the equation in "homogeneous form", but then realized that it was not applicable. I remembered the eqn a(x)y' + p(x)y = b(x) and so, when I finalized the differentiation (in my mind, apparently) - I dropped the answer back in that form.

Was that the right form for the answer? I probably need to re-read the entire first two chapters in my D.E. book (by Trench).

You are apparently using dy/dx when you mean d/dx, and d/dx when you mean dy/dx.

dy/dx is a thing; namely, the derivative of y with respect to x. In contrast, d/dx is an action that is yet to be taken. More precisely, d/dx is an operator that signifies your intent to take the derivative of what is to the right of it.

For example,
[tex]\frac{dy}{dx}\left(\frac{1}{5}\right)\left(y^5\right) [/tex]
means the derivative of y with respect to x times 1/5 times y5. This error is similar to the one where a student interprets sin x as "sine times x."

I did know that as well. I wonder how long I may have been causing my professors grief over this incorrect usage of the notation? (I'm sorry...)

Students who are new to calculus sometimes write things like this:
dy/dx (x2) = 2x. I understand what they mean to say, but what they should say is d/dx(x2) = 2x. A student who does not understand the difference will have an extremely hard time when he or she sees
[tex]\frac{dy}{dx}x^2[/tex]
and interprets this incorrectly to mean "take the derivative of x2." What this expression actually means is the product of some unspecified function and x2.

Point taken (to the heart). I am not trying to make any mathematical endeavor harder on myself. I will cease that usage ASAP.

p.s. Who can fix my name for me? It should be EtherealMonkey. But I misfired the enter button when signing up (should have checked my work ;-)
 
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  • #19
jmb
67
1
I shouldn't have written the differential notation with the y, right? It seems redundant anyway now that you point it out.

Would this have been correct? [itex]\frac{d}{dx}\left(\frac{1}{5}\right)\left(y^5\right)[/itex]

Yes that's better though I would have preferred:
[tex]\frac{d}{dx}\left(\frac{1}{5}\;y^5\right)}[/tex]

It's really important to realise that there are two separate things here:
  • [itex]dy/dx[/itex] is the derivative of [itex]y[/itex] with respect to [itex]x[/itex]. It is a function. So what you originally wrote involved taking [itex]y^5/5[/itex] and multiplying it by the derivative of [itex]y[/itex], not taking the derivative of it.
  • [itex]d/dx[/itex] is an operator not a function. It means "take the derivative of the following with respect to [itex]x[/itex]. It is generally a good idea to surround the "something" with a single set of brackets to indicate the scope of what is being operated on. Also as a convention it only operates on stuff to the right hand side of it.

I knew this was not needed, but since I was unsure of my "method", I was "showing my work"?? (lol)

No you were right to show your work, but what you've written makes no sense because of the second point above (d/dx operates on stuff to the right of it), so to indicate that you are about to take the derivative of [itex]x^2/2[/itex] you should have written [itex]\frac{d}{dx}\left(x^2/2\right)[/itex].

Note that I am really not being pedantic here, your working as it was written was so obscure that as an examiner I could not have given you any marks for getting the right answer because there was no way to tell that it wasn't just a 'lucky guess'. Also if you aren't very happy with the meaning of the notation you will become easily confused with more complicated calculus (I'm afraid "sloppy notation leads to sloppy thinking or even sloppy understanding").

Anyway, back to the issue at hand, we now have:

[tex]
\frac{d}{dx}\left(\frac{y^5}{5}\right)+\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{2}\right)[/tex]

as your first line of working. Can you now add in an extra line showing how you got from this to:
[tex]\left(y^4+1\right)\frac{dy}{dx} = x[/tex]

Then, provided I'm happy you're clear about that step too, I think you should move on and try the next question of your assignment; since that really covers it for this one (unless you still have a conceptual problem with 1b).

[PS: Sorry Mark44 I realise you'd already pointed this out, but hadn't spotted your new post before I hit reply!]
 
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  • #20
Anyway, back to the issue at hand, we now have:

[tex]\frac{d}{dx}\left(\frac{y^5}{5}\right)+\frac{dy}{dx} = \frac{d}{dx}\left(\frac{x^2}{2}\right)[/tex]

as your first line of working. Can you now add in an extra line showing how you got from this to:
[tex]\left(y^4+1\right)\frac{dy}{dx} = x[/tex]

Actually, I cannot. I think that I had the chain rule in mind and separating one of the "y" variables out of the LHS to accomplish that. I can't actually provide the work though.

So... I guess I need to concentrate on that for a minute.

Also, reading what you wrote about the syntax of the "language" actually makes a lot of sense to me. I was an Air Traffic Controller for the US Navy and am fond of C++, C# and VBA. So, I am aware of the necessity of protocol in communication.

I cannot thank you enough for pointing these little details out though that have probably been keeping my professors wholly unsure of my ability. What's worse is that they may have thought that I didn't care. Thanks again for taking the time to point these things out.

I actually need to get ready to go to class now. I hope that I will not be the only one empty handed (i.e., I pray he gives us until Monday?)

Anyway, thanks again. I'm sure I will be back soon.
 
  • #21
jmb
67
1
Actually, I cannot. I think that I had the chain rule in mind and separating one of the "y" variables out of the LHS to accomplish that. I can't actually provide the work though.

So... I guess I need to concentrate on that for a minute.

OK, well when you next have time, write down the chain rule and then think about how you could apply that to the derivative of [itex]y^5/5[/itex] with respect to [itex]x[/itex]. Post your thoughts and we'll take it from there.
 
  • #22
OK, well when you next have time, write down the chain rule and then think about how you could apply that to the derivative of [itex]y^5/5[/itex] with respect to [itex]x[/itex]. Post your thoughts and we'll take it from there.

Hehe. Okay, I will have to investigate another method.

I did check this answer with Maple. But, I did the work by hand first on the piece of paper that I scanned and posted earlier.

Maybe it really was a guess because I don't see how I got that right.

(I may have done this problem before though - so I'm not gonna go betting the house on that ;-) )
 
  • #23
35,134
6,884
Let me help you out.
[quote Originaly posted by jmb]
[tex]\frac{d}{dx}\left(\frac{y^5}{5}\right)+\frac{dy}{d x} = \frac{d}{dx}\left(\frac{x^2}{2}\right)[/tex]
[/quote]
There are two derivatives we need to take here: one on the left side and one on the right.
For the first, d/dx(1/5*y5) requires the use of the chain rule. This could by written as d/dy(1/5*y5)*dy/dx. Hopefully you can do this in your head.
The second is simpler in that it is just d/dx(1/2*x2).
After differentiation, the equation we started with becomes
y4*dy/dx + dy/dx = x

On the left side, dy/dx is a common factor, so can be factored out, which gives us
(y4 + 1)*dy/dx = x.

At this point, it's a simple matter to get back to your original differential equation, which was
dy/dx = x/(1 + y4).

If I can give you some advice, keep plugging away at this. You sound like you're willing to put in the effort, and that's enough to get lots of people to help you on this forum, as you are seeing.

Mark

BTW, thanks for your service.
 

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