Question

r^2R'' + R'r = 0

Answer 1

reduction of order

Answer 2

@Kainui @Astrophysics

Answer 3

how about we if could treat it as a quadratic equation?

Answer 4

\(m = \frac{dy}{dx}\\ m^2 = \frac{d^2y}{dx^2} \)

Answer 5

so
\( r^2m^2 + mr = 0\)

Answer 6

m will be 0 and -1/r

Answer 7

@ganeshie8

Answer 8

R = Ae^(0) + Be^(-1) ?

Answer 9

so general form is \(y = Ae^{m1x} + Be^{m2x}\) where \(\Delta > 0\)

Answer 10

interesting...

(dy/dx)^2 is not same as d^2y/dx^2

should that be a problem ?

Answer 11

no. it says R''. not (R')^2 right?

Answer 12

R'' means second derivative I believe..

Answer 13

do you?

Answer 14

that solution no right

Answer 15

it's R(r) = E + Fln(r)

Answer 16

@Mimi_x3 I didn't get it:(

Answer 17

hey i was refering to ur second reply from top...

Answer 18

im sure it is a typo or something..

Answer 19

well I just read this article. http://www.intmath.com/differential-equations/7-2nd-order-de-homogeneous.php

Answer 20

My answer is based on that.

Answer 21

i think that works only if the coefficients are constants

here r is the dependent variable right ?

Answer 22

this is not homogeneous i suppose :-\

Answer 23

well not separable variables

Answer 24

yes. I agree with that.

Answer 25

using reduction of order, we get something like this :

\(r^2 R'' + R'r = 0\)

let \(t=R' \)
that means \(t'=R''\) and the de becomes
\(r^2t' + tr = 0\)
which is separable

Answer 26

okay.

Answer 27

In reduction of order the r term disappears right

Answer 28

why, im letting \(R' = t(r)\)
so the de changes from (R'', R', R, r) variables to (t', t, r)

Answer 29

I wasn't even looking at the substitution haha, I haven't used it for this kind of problem so I was kind of thinking if you have a solution, then y=y1v and when you take the derivative and plug it back in the v term usually always disappears

Answer 30

\(y=vx\) substitution works nicely for first order homogeneous equations

Answer 31

once we reduce the order, we can try any of those first oder de tricks

Answer 32

Ok I see

Answer 33

but the de we got after reducing the order is much simpler than that...
it is separable !

Answer 34

using reduction of order, we get something like this :

\(r^2 R'' + R'r = 0\)

let \(t=R' \)
that means \(t'=R''\) and the de becomes
\(r^2t' + tr = 0\)
which is separable

@Mimi_x3 pretty sure you can finish it off...

Answer 35

@ganeshie8 It is a nice solution. thanks for sharing

Answer 36

was confused about this all the time :3

Answer 37

can't we tell by looking at it, R is going to be in Ar^n form.

Answer 38

because if
\(R = Ar^n \\ R' = Anr^{n-1}\\ R'' = An(n-1)r^{n-2}\)

Answer 39

\(r^n\) is cancelled out after substitution. rest is \[n(n-1) + n = 0\]

Answer 40

\(n^2 = 0\) that means y is not a variable.

Answer 41

R = constant. would that be correct?

Answer 42

and this leads to my previous answer. It was also a constant

Answer 43

because if
\(R = e^{ar} \\ R' = ae^{ar}\\ R'' = a^2e^{ar}\)

euler thought of this idea first, but it works only for constant coefficeint homogeneous differential equations like :
\[R''-5R'+6R=0\]

when you plugin \(R =e^{ar}\) above, you get a quadratic using which we can solve the value of \(a\)

Answer 44

that method doesn't work in our case because we don't have constant coefficeints...

Answer 45

I see

Answer 46

\(\color{red}{r^2}R'' + R'\color{red}{r}=0\)

the coefficients, \(\color{red}{r^2}\) and \(\color{red}{r}\) are variables here. so we need to try something else..

Answer 47

okay. I have no background in non-homogeneous DE. So I believe what you said is right.:

Answer 48

since we're talking about constant coefficient second order odes, below might be a nice review

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-9-solving-second-order-linear-odes-with-constant-coefficients/

i like that professor, he makes everything look so simple...

Answer 49

\(r^2 R'' + R' r = 0\)
\(r R'' + R' = 0 = (rR')'\)
is it not just that? or do i not understand the problem?

Answer 50

that will do haha !

Answer 51

yes. he is amazing professor. I like him too:)

Answer 52

i became his fan after watching his lecture on laplace transforms
he starts with the discrete version, power series, then goes ahead and derives laplace transform formulas...

Answer 53

mine was complex numbers

Answer 54

lochana's method looks like it works @ganeshie8, that's what I came here to do was his method, I think you're confusing using \(R=e^{ar}\) with choosing \(R=r^n\)

Answer 55

It doesn't cover all the solutions though, since it's second order we gotta somehow find another one so even though his method is right, it's not enough to solve it for this special case it seem cause the unfortunate repeated root.

Answer 56

how do we know the solutions are of form \(r^n\) ?

Answer 57

we need two independent solutions somehow, but there is no guarantee that we get them using the method of constant coefficents..

Answer 58

It's just a method like using \(y=e^{ax}\), it's an initial guess that ends up giving two solutions to a second order differential equation most of the time. Unfortunately this one ends up giving us a solution \(n^2=0\) so it doesn't do it.

If the question had been slightly changed to:

\[r^2 R'' + 2rR'=0\]

Then his solution would have worked perfectly well.

Answer 59

\(n(n-1) + n = 0\) gives me \(n=0\)
so one independent solution is \(R = c\). i reach dead end

Answer 60

Yeah, I'm not disagreeing with you on that

Answer 61

You said: "that method doesn't work in our case because we don't have constant coefficeints... "

And I'm saying, "that method doesn't work in our case because we have a repeated root..."

That's all haha.

Answer 62

i didn't like that substitution because there is no reason to assume that the independent solutions of a de are polynomials... its just waste of time, you might get lucky once in a while... but its a bogus method.. my opinion anyways :)

Answer 63

constant coefficents case is different, we have euler to backup :)

Answer 64

It's just as bogus as any other technique in DE lol. \[R(r)=r^n\] solves
\[ar^2R''+brR'+cR=0\]
for constants a, b, c as long as

\[an(n-1)+bn+c=0\] has two distinct roots, this is a really common and simple method, and is especially useful when doing things in spherical coordinates.

Answer 65

that works only if the solutions are polynomials
we don't really need to do all that if we already know that the solutions are polynomials. we can simply eyeball the degree of polynomial

Answer 66

\(ar^2R''+brR'+cR=0\)

if the solution is of form \(r^n\), it is easy to guess that \(n=0\). i don't buy that method yet...

Answer 67

if you do this as a euler cauchy or that throwaway i posted above, you get \(R = A \ln r + B\) in either case. maybe i still don't understand the question....but i am assuming that \(R = R(r)\)