Solving equation seemingly incomplete gamma function

[Bliman]

Hi,
I need to solve this equation. I am following the book Ordinary differential equations with applications from Andrews.
It is this : xdx-ydy=y^2(x^2-y^2)dy
I found the integrating factor which was e^((-2/3)y^3)
With that I found the function f(x,y)=(1/2)x^2e^((-2/3)y^3)+g(y)
Now I need to find g(y)
with that I got integral of y^4e^((-2/3)y^3)-ye^((-2/3)y^3
But I just don't know how to solve that.
In an integral calculator I get an incomplete gamma function (never seen that).
And in the back of the book the solution is stated as (1/2)ln(x^2-y^2)=(1/3)y^3+C
Could anyone help me?

 

[blamocur]

I did not use integrating factors, but substitution [imath]u = x^2 - y^2[/imath] produced a solution which is close to the answer from the book.

 

[Bliman]

I did not use integrating factors, but substitution [imath]u = x^2 - y^2[/imath] produced a solution which is close to the answer from the book.

Thanks for the answer. Could you show me how you did that? Do you first put it in dy/dx form? Than I got dy/dx= -x/(y^4-x^2y^2-y). Do you do the substitution then?

 

[blamocur]

Thanks for the answer. Could you show me how you did that? Do you first put it in dy/dx form? Than I got dy/dx= -x/(y^4-x^2y^2-y). Do you do the substitution then?

No, I simply noticed that the left hand side looks like [imath]d (x^2-y^2)[/imath]

 

[Bliman]

But can you tell me if you solved it? And precisely how?

 

[blamocur]

I got an answer slightly different from the book, but I did not spend enough time to see where the discrepancy comes from.
There is a reason I haven't posted my solution : https://www.freemathhelp.com/forum/threads/read-before-posting.109846/

 

[Cubist]

with that I got integral of y^4e^((-2/3)y^3)-ye^((-2/3)y^3
But I just don't know how to solve that.
In an integral calculator I get an incomplete gamma function (never seen that).

The integral of this IS a (very) simple expression. However, I don't know how to do the integral myself without prior knowledge (without seeing the solution myself)! I'll leave it to the other helpers here to give you hints. (I used used Wolfram Alpha to find the answer to this integral, and it was an alternative that wasn't in terms of the incomplete gamma function).

And in the back of the book the solution is stated as (1/2)ln(x^2-y^2)=(1/3)y^3+C

I got an answer slightly different from the book,

I think that your back of the book solution is correct. I differentiated it, and put this back into the question, and the equation was satisfied. Perhaps the answer that blamocur found was a different form, but equivalent

 

[blamocur]

The integral of this IS a (very) simple expression. However, I don't know how to do the integral myself without prior knowledge (without seeing the solution myself)! I'll leave it to the other helpers here to give you hints. (I used used Wolfram Alpha to find the answer to this integral, and it was an alternative that wasn't in terms of the incomplete gamma function).





I think that your back of the book solution is correct. I differentiated it, and put this back into the question, and the equation was satisfied. Perhaps the answer that blamocur found was a different form, but equivalent

You are right. Somehow I assumed that [imath]d(x^2) = x dx[/imath]

 

[BigBeachBanana]

You are right. Somehow I assumed that [imath]d(x^2) = x dx[/imath]

I used your [imath]u[/imath] substitution and was able to match the book's answer. It's a separable DE. Nice hint!

You are right. Somehow I assumed that [imath]d(x^2) = x dx[/imath]

You probably forgot to multiply the RHS by 2.

 

[Cubist]

You are right. Somehow I assumed that [imath]d(x^2) = x dx[/imath]

Don't worry, I made a few mistakes myself before I proved the answer myself

The integral of this IS a (very) simple expression. However, I don't know how to do the integral myself without prior knowledge (without seeing the solution myself)! I'll leave it to the other helpers here to give you hints.

Can any of the helpers suggest a method to integrate the following (without just "spotting" the answer)...
[math] \int \left( (y^4 - y)e^{-(2/3)y^3} \right) dy[/math]
--
The best hint that I can think of, for spotting the answer, is to differentiate this...
[math]\dfrac{d}{dy}\left( f(y) e^{-(2/3)y^3} \right)[/math]...and then think about an f(y) that would make the answer equal to [imath] (y^4 - y)e^{-(2/3)y^3} [/imath]

 

[blamocur]

"Spotting" also happens when one tries integration by parts of [imath]\int y e^{-(2/3)y^3} dy[/imath]

 

[Bliman]

Ok I tried (I already tried that) with integration by parts. And then I get this solution.
(1/2)x^2e^((-2/3)y^3)-(1/2)y^2=c
I seem to be closer but not close enough. Can anyone tell me if I am on the right track?
Can't get a track anymore of how many pages I spend on this one .

 

[Cubist]

The best hint that I can think of, for spotting the answer, is to differentiate this...
[math]\dfrac{d}{dy}\left( f(y) e^{-(2/3)y^3} \right)[/math]...and then think about an f(y) that would make the answer equal to [imath] (y^4 - y)e^{-(2/3)y^3} [/imath]

@Bliman did you try to differentiate the above, and if so, then what did you get?

 

[Cubist]

It's very difficult to help you properly without seeing your full work (not just a summary of it). You can upload an image of any work done with paper/ pen if that's easier for you.

I found the integrating factor which was e^((-2/3)y^3)

I haven't checked that your integrating factor is correct. Did you rearrange the question into the following form?
[math]\dfrac{dx}{dy} + P(y)x = Q(y)[/math]I have only tried for a couple of minutes, but I'm struggling to do this. Please show your work for this (if this is wrong, then obviously there's no point going further)

EDIT: What page of the book (that you mentioned in post#1) did this come from?

--

You might be better off following the suggestion in post#2, unless you've been told to use a specific method.

 

[Bliman]

It's very difficult to help you properly without seeing your full work (not just a summary of it). You can upload an image of any work done with paper/ pen if that's easier for you.


I haven't checked that your integrating factor is correct. Did you rearrange the question into the following form?
[math]\dfrac{dx}{dy} + P(y)x = Q(y)[/math]I have only tried for a couple of minutes, but I'm struggling to do this. Please show your work for this (if this is wrong, then obviously there's no point going further)

EDIT: What page of the book (that you mentioned in post#1) did this come from?

--

You might be better off following the suggestion in post#2, unless you've been told to use a specific method.

Thanks for the help.
It came from page 35
I have written it a bit more properly and I have uploaded it. Here are the links.

CCF-000023 hosted at ImgBB

Image CCF-000023 hosted in ImgBB

ibb.co

CCF-000024 hosted at ImgBB

Image CCF-000024 hosted in ImgBB

ibb.co

 

[Cubist]

Thanks for posting your work. I spotted a missing minus, highlighted in red below...



...I hope this will help.

I'm not familiar with this method of solving a DE. I'll have to look for a copy of your book in my local library since I couldn't find any scans of the relevant pages online.

 

[BigBeachBanana]

Thanks for the help.
It came from page 35
I have written it a bit more properly and I have uploaded it. Here are the links.

CCF-000023 hosted at ImgBB

Image CCF-000023 hosted in ImgBB

ibb.co

CCF-000024 hosted at ImgBB

Image CCF-000024 hosted in ImgBB

ibb.co

I don't think the problem is meant to solve through integrating factors and partial differentiation. . It took me 4 lines to write out the solution.
Use the suggested substitution: [imath]u=x^2-y^2 \implies \frac{du}{dy} = x\,\frac{dx}{dy} -2y[/imath]

Your new equation is in terms of [imath]u[/imath] and [imath]y[/imath], and it's seperable.

 

[Cubist]

I suspect that @Bliman has stopped responding because after I pointed out the sign mistake it led them to the answer (they logged off from the website a few mins after I posted this)..

View attachment 33941

Propagating this simple fix to the end of their work gives...

and finishing off is easy from here to obtain the "back of the book" answer

I don't think the problem is meant to solve through integrating factors and partial differentiation. . It took me 4 lines to write out the solution.
Use the suggested substitution: [imath]u=x^2-y^2 \implies \frac{du}{dy} = x\,\frac{dx}{dy} -2y[/imath]

Your new equation is in terms of [imath]u[/imath] and [imath]y[/imath], and it's seperable.

The post#2 method seems MUCH simpler. However, I suspect that the book(see post#1) intended the reader to use the longer method. I think that Bliman would have been following previous worked examples in the book's pages. I think Bliman really needed help to complete the problem in the way that they had shown their work. But we may never know for sure

 

[Bliman]

I suspect that @Bliman has stopped responding because after I pointed out the sign mistake it led them to the answer (they logged off from the website a few mins after I posted this)..

Propagating this simple fix to the end of their work gives...
View attachment 33945
and finishing off is easy from here to obtain the "back of the book" answer


The post#2 method seems MUCH simpler. However, I suspect that the book(see post#1) intended the reader to use the longer method. I think that Bliman would have been following previous worked examples in the book's pages. I think Bliman really needed help to complete the problem in the way that they had shown their work. But we may never know for sure

Thank you very much for your help. I finally found it. And yes this was a segment in the book that touched on this subject also see here. https://www.mathsisfun.com/calculus/differential-equations-exact-factors.html . It was talking about this.
I still had to work to find the answer but finally got it which is satisfying . Now I can go on after many pages. How simple something is when you finally got it.

 

[Cubist]

Thank you very much for your help. I finally found it. And yes this was a segment in the book that touched on this subject also see here. https://www.mathsisfun.com/calculus/differential-equations-exact-factors.html . It was talking about this.
I still had to work to find the answer but finally got it which is satisfying . Now I can go on after many pages. How simple something is when you finally got it.

Thanks for letting us know that you got the final answer, and also thank you for the link!

Here's another resource that might be useful (I found it via Google) see page 4 (labelled page 52) under the heading "Exact type"...
https://www.maths.gla.ac.uk/~cc/2x/2005_2xnotes/2x_chap5.pdf
...sometimes reading several different descriptions of the same topic can help to reinforce it