New to Nonlinear Diff. Equations? Ask Here!

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(Sorry for the mistakes first thread using hand held device)

Hello,
I was working on Harold T. Davis
Introduction to Nonlinear Differential and Integral Equations

I saw this following equations

Printed page of a nonlinear differential equations textbook showing equations (3) f(x,y,α,β)=0, (4) ∂f/∂x + (∂f/∂y)(dy/dx) = 0, and (5) ∂²f/∂x² + 2 ∂²f/∂x∂y (dy/dx) + ∂²f/∂y² (dy/dx)² + ∂f/∂y d²y/dx² = 0, with handwritten marks.


1-So equation 4 came as a result of chain rule applies on equation. 3 ?

2- how did equation 5 come about? And specially the "2" in second term.

3-What kind of calculus I should study, so I know this sort of stuff and I can save some time.

Thank you.
 
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The point of this sort of book is for you to explore how you understand calculus

1-So equation 4 came as a result of chain rule applies on equation. 3 ?

... you can answer this for yourself: what happens when you apply the chain rule to eq3?

2- how did equation 5 come about? And specially the "2" in second term.
... have you tried just differentiating eq4 wrt x? What happens when you do?

3-What kind of calculus I should study, so I know this sort of stuff and I can save some time.
... differential calculus. "Differential equations" is the usually subject heading from 2nd year college.
 
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Simon Bridge said:
The point of this sort of book is for you to explore how you understand calculus

1-So equation 4 came as a result of chain rule applies on equation. 3 ?

... you can answer this for yourself: what happens when you apply the chain rule to eq3?

2- how did equation 5 come about? And specially the "2" in second term.
... have you tried just differentiating eq4 wrt x? What happens when you do?

3-What kind of calculus I should study, so I know this sort of stuff and I can save some time.
... differential calculus. "Differential equations" is the usually subject heading from 2nd year college.
3.with product rule I get

Handwritten chain rule expression on lined paper: ∂²f/∂x² + ∂²f/∂y∂x · dy/dx + ∂f/∂y · d²y/dx², with the image rotated so the symbols appear upside down.


do I need to take Y' as well and add it?
 
I do not know what you are trying to show... the "3" could refer to my point 3, which has nothing to do with the product rule... or it could refer to eq3, which does not get to your pic using the product rule.
 
I actually tried to differentiating eq4 wrt x. I got this(from your point 2). Sorry about the confusion.
Handwritten chain rule expansion on lined paper, written upside down: d²f/dx² + (∂²f/∂y∂x)(dy/dx) + (∂f/∂y)(d²y/dx²).
(about your point 3)
I learned ODE/PDE from Keryzsig's "advance engineering mathematics".
How ever I also have Tenenbaum/Pollard's "ordinary differential equation", which does not have such topic covered.
 
knockout_artist said:
I actually tried to differentiating eq4 wrt x. I got this(from your point 2). Sorry about the confusion.
(about your point 3)
I learned ODE/PDE from Keryzsig's "advance engineering mathematics".
How ever I also have Tenenbaum/Pollard's "ordinary differential equation", which does not have such topic covered.

One thing you are missing (and the book is treating implicitly) is that ##\frac{\partial f}{\partial x}## and ##\frac{\partial f}{\partial y}## are functions of two variables.
 
Now I get this

Handwritten chain-rule derivation on lined paper: the second derivative d²f/dx² equals ∂²f/∂x² + ∂²f/∂y∂x · dy/dx + ∂f/∂y · d²y/dx², with f_x(x,y) annotated; below, the result ∂f/∂y∂x + ∂²f/∂y² · dy/dx + ∂f/∂y · (dy/dx)² with f_y(x,y) annotated.


even if I add them I don't get equation 5.
 
That looks close. Aren't you just missing that last term in ##y_{xx}##?

Just on my phone so hard to see exactly what you are missing!
 
Here's a breakdown in shorthand notation, starting with:

##(f_x) + (f_y)( y_x)##

And differentiating with respect to ##x## gives:

##(f_{xx} + f_{xy} y_x) + (f_{yx} + f_{yy} y_x)(y_x) + (f_y)(y_{xx}) = f_{xx} + 2f_{xy} y_x + f_{yy} (y_x)^2 + f_y y_{xx}##
 
Thank you, for the reply.
On LHS term in the middle seems like differentiated wrt to y, rather then x.
 
knockout_artist said:
Thank you, for the reply.
On LHS term in the middle seems like differentiated wrt to y, rather then x.

Which term? Remember that ##f_x## and ##f_y## are functions of two variables, so the chain rule applies to them as it does to ##f## in the first place.
 
PeroK said:
Which term? Remember that ##f_x## and ##f_y## are functions of two variables, so the chain rule applies to them as it does to ##f## in the first place.
I see that.
Thank you very much for your help.