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

- Mimi_x3

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

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- Mimi_x3

reduction of order

- Mimi_x3

@Kainui @Astrophysics

- lochana

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

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## More answers

- lochana

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

- lochana

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

- lochana

m will be 0 and -1/r

- lochana

@ganeshie8

- lochana

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

- lochana

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

- ganeshie8

interesting...
(dy/dx)^2 is not same as d^2y/dx^2
should that be a problem ?

- lochana

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

- lochana

R'' means second derivative I believe..

- lochana

do you?

- Mimi_x3

that solution no right

- Mimi_x3

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

- lochana

@Mimi_x3 I didn't get it:(

- ganeshie8

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

- ganeshie8

im sure it is a typo or something..

- lochana

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

- lochana

My answer is based on that.

- ganeshie8

i think that works only if the coefficients are constants
here r is the dependent variable right ?

- ikram002p

this is not homogeneous i suppose :-\

- ikram002p

well not separable variables

- lochana

yes. I agree with that.

- ganeshie8

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

- lochana

okay.

- Astrophysics

In reduction of order the r term disappears right

- ganeshie8

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

- Astrophysics

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

- ganeshie8

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

- ganeshie8

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

- Astrophysics

Ok I see

- ganeshie8

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

- ganeshie8

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...

- lochana

@ganeshie8 It is a nice solution. thanks for sharing

- ikram002p

was confused about this all the time :3

- lochana

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

- lochana

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

- lochana

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

- lochana

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

- lochana

R = constant. would that be correct?

- lochana

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

- ganeshie8

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 \(a\)

- ganeshie8

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

- lochana

I see

- ganeshie8

\(\color{red}{r^2}R'' + R'\color{red}{r}=0\)
the coefficients, \(\color{red}{r^2}\) and \(\color{red}{r}\) are variables here. so we need to try something else..

- lochana

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

- ganeshie8

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...

- IrishBoy123

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

- ganeshie8

that will do haha !

- lochana

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

- ganeshie8

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...

- lochana

mine was complex numbers

- Kainui

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 \(R=r^n\)

- Kainui

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.

- ganeshie8

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

- ganeshie8

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

- Kainui

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.

- ganeshie8

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

- Kainui

Yeah, I'm not disagreeing with you on that

- Kainui

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.

- ganeshie8

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 :)

- ganeshie8

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

- Kainui

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.

- ganeshie8

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

- ganeshie8

\(ar^2R''+brR'+cR=0\)
if the solution is of form \(r^n\), it is easy to guess that \(n=0\). i don't buy that method yet...

- IrishBoy123

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 \(R = R(r)\)

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