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# Advanced Math/Ordinary differential equations

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QUESTION: Hello,
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)

Answer
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.)

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