Advanced Math/Ordinary differential equations
I am reviewing differential equations as I finished the course as an engineering major, but I am trying to refresh and I feel like I am taking it for the first time somehow, I can work through the normal exercises but not the proofs.
Book: Elementary differential equations by rainville/bedient Eight edition
Page 29 Q 36
From theorems 2.1 and 2.2 section 2.2 it follows that if F is homogeneous of degree k in x,y, F can be written in the form
F= x^k * phi(y/x) (A)
use A to prove that if F is homogeneous function of degree k in x,y
x *dF/dx +y*dF/dy = kF
theorem 2.1 is just that ratio of same degree homo functions is 0
theorem 2.2 is that any zero degree homo function can be written as f(y/x)
I don't know I don't feel that the question is even asking me to prove a correct statement
I mean the thing is It's
F(x) = x^k phi(y/x)
for it to equal kF there must be some symmetry with dF/dx and dF/dy which i don't find, but more importantly I have no information to link the derivative to the original function
I mean no information about phi'(x)
I mean I can try working in the math, but I don't see the sense of which this should be true
well anyway here is what I tried
well if F is a homogeneous function then
Phi is also homogeneous with degree 0
I mean ok let's say u=y/x
we have F(x) = x^k phi(u)
now F(ax) = a^k x^k phi(u) =a^k F(x)
so phi must be homogeneous of degree 0 for this to hold
so F= x^k phi(y/x) = phi(y/x *x^k)= phi(y *x^k-1)
but this doesn't help cause all it does is
xdF/dx + ydF/dy = kF
x(y * k *x^k-2) *dphi/dx + y * (y * x^k-1 )dphi/dy = kphi(y *x^k-1)
which isn't something that's worth trying to work on. It's clear to me that this isn't the right track.
Though I just want to know whether I should aim for a different approach or is my intuition that the statement is false true?
I mean this question isn't important to our applications but it pissed me off so I have to get it out of my mind, and well it might have relevance in electromagnetism actually so might as well.
ANSWER: Your phrasing of this question is somewhat misleading -- maybe it is also something that is harming you understand this question, or maybe you understand but did not type it properly.
The question is asking you to show:
x ∂F/∂x + y ∂F/∂y = k F
That is not
the same as:
x *dF/dx +y*dF/dy = kF
I took the book you cite from the library to verify that this is, indeed, a question that requires partial derivatives (∂/∂x), not just derivatives (d/dx).
Based on what you are writing, I think the major thing that you don't seem to understand is that φ is a function
and that φ(y/x) is what you get when you plug y/x into that function. It is not some symbol φ times the fraction y/x. You have written things that seem to say x^k φ(y/x) = phi(y *x^k-1) and that is not correct.
For example, the left-hand side of this differential equation is homogeneous of degree zero:
( x^2 y + y^2 x ) / ( 3x^2 y + 2y^2 x ) = dy/dx
You can check that the left hand side could be called F(x,y) and F(cx,cy) = F(x,y) for all nonzero values of c. If you divide the top and bottom of each fraction by y^2, you get:
( x^2/y^2 + x/y ) / ( 3x^2/y^2 + 2x/y ) = dy/dx
The function on the left hand side is the same, just re-written, but now everything is in terms of the fraction x/y. So you can really write this as:
( (x/y)^2 + x/y ) / ( 3(x/y)^2 + 2(x/y) ) = dy/dx
and if you define the function φ appropriately, this is what you have:
φ(u) = (u^2+u)/(3u^2+2u)
So that the left-hand side of this DE is φ(y/x), i.e. u=y/x.
Anyway, now that the notation should be more clear based on that example, let's proceed.
You know F = x^k φ(y/x)
So ∂F/∂x is computed using the product rule (and chain rule, and power rule):
∂F/∂x = k x^(k-1) φ(y/x) + x^k (∂φ/∂x)(y/x) (-y/x^2)
Multiplying through by x gives you:
x ∂F/∂x = k x^k φ(y/x) - y x^(k-1) (∂φ/∂x)(y/x)
Doing the partial ∂F/∂y does not need the product rule, and gives:
∂F/∂y = x^k (∂φ/∂y)(y/x) (1/x)
If you multiply this by y, you get:
y ∂F/∂y = y x^(k-1) (∂φ/∂y)(y/x)
When you take the sum of these two things, the second term of ∂F/∂x cancels out with the ∂F/∂y. What you get is simply:
x ∂F/∂x + y ∂F/∂y = k x^k φ(y/x)
You should see that this is just k F.
---------- FOLLOW-UP ----------
QUESTION: I knew phi(y/x) is a function of y/x but in the book it says
f ( lamba * x, lambda * y) = lamba ^k f(x,y)
so I took it that
lambda is x^k and f(x,y) well f (x/y, constant)
is that incorrect reasoning?
that's how I got
phi(y/x) * x^k = phi(y/x * x ^k)
That reasoning is not completely incorrect. You are right that c (or you are saying λ instead) can be "pulled out" of the function:
F(λx,λy) = λ^k F(x,y)
That is true for F. However, what the book has noted is that φ is homogeneous with degree zero
, not degree k. You cannot move x^k into φ like that. (Furthermore, x^k is not a constant, so you should not be treating it like the constant λ.)