Mimi_x3
  • Mimi_x3
r^2R'' + R'r = 0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Mimi_x3
  • Mimi_x3
reduction of order
Mimi_x3
  • Mimi_x3
@Kainui @Astrophysics
lochana
  • lochana
how about we if could treat it as a quadratic equation?

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lochana
  • lochana
\(m = \frac{dy}{dx}\\ m^2 = \frac{d^2y}{dx^2} \)
lochana
  • lochana
so \( r^2m^2 + mr = 0\)
lochana
  • lochana
m will be 0 and -1/r
lochana
  • lochana
@ganeshie8
lochana
  • lochana
R = Ae^(0) + Be^(-1) ?
lochana
  • lochana
so general form is \(y = Ae^{m1x} + Be^{m2x}\) where \(\Delta > 0\)
ganeshie8
  • ganeshie8
interesting... (dy/dx)^2 is not same as d^2y/dx^2 should that be a problem ?
lochana
  • lochana
no. it says R''. not (R')^2 right?
lochana
  • lochana
R'' means second derivative I believe..
lochana
  • lochana
do you?
Mimi_x3
  • Mimi_x3
that solution no right
Mimi_x3
  • Mimi_x3
it's R(r) = E + Fln(r)
lochana
  • lochana
@Mimi_x3 I didn't get it:(
ganeshie8
  • ganeshie8
hey i was refering to ur second reply from top...
ganeshie8
  • ganeshie8
im sure it is a typo or something..
lochana
  • lochana
well I just read this article. http://www.intmath.com/differential-equations/7-2nd-order-de-homogeneous.php
lochana
  • lochana
My answer is based on that.
ganeshie8
  • ganeshie8
i think that works only if the coefficients are constants here r is the dependent variable right ?
ikram002p
  • ikram002p
this is not homogeneous i suppose :-\
ikram002p
  • ikram002p
well not separable variables
lochana
  • lochana
yes. I agree with that.
ganeshie8
  • 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
  • lochana
okay.
Astrophysics
  • Astrophysics
In reduction of order the r term disappears right
ganeshie8
  • ganeshie8
why, im letting \(R' = t(r)\) so the de changes from (R'', R', R, r) variables to (t', t, r)
Astrophysics
  • 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
  • ganeshie8
\(y=vx\) substitution works nicely for first order homogeneous equations
ganeshie8
  • ganeshie8
once we reduce the order, we can try any of those first oder de tricks
Astrophysics
  • Astrophysics
Ok I see
ganeshie8
  • ganeshie8
but the de we got after reducing the order is much simpler than that... it is separable !
ganeshie8
  • 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
  • lochana
@ganeshie8 It is a nice solution. thanks for sharing
ikram002p
  • ikram002p
was confused about this all the time :3
lochana
  • lochana
can't we tell by looking at it, R is going to be in Ar^n form.
lochana
  • lochana
because if \(R = Ar^n \\ R' = Anr^{n-1}\\ R'' = An(n-1)r^{n-2}\)
lochana
  • lochana
\(r^n\) is cancelled out after substitution. rest is \[n(n-1) + n = 0\]
lochana
  • lochana
\(n^2 = 0\) that means y is not a variable.
lochana
  • lochana
R = constant. would that be correct?
lochana
  • lochana
and this leads to my previous answer. It was also a constant
ganeshie8
  • 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
  • ganeshie8
that method doesn't work in our case because we don't have constant coefficeints...
lochana
  • lochana
I see
ganeshie8
  • 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
  • lochana
okay. I have no background in non-homogeneous DE. So I believe what you said is right.:
ganeshie8
  • 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
  • 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
  • ganeshie8
that will do haha !
lochana
  • lochana
yes. he is amazing professor. I like him too:)
ganeshie8
  • 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
  • lochana
mine was complex numbers
Kainui
  • 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
  • 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
  • ganeshie8
how do we know the solutions are of form \(r^n\) ?
ganeshie8
  • ganeshie8
we need two independent solutions somehow, but there is no guarantee that we get them using the method of constant coefficents..
Kainui
  • 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
  • ganeshie8
\(n(n-1) + n = 0\) gives me \(n=0\) so one independent solution is \(R = c\). i reach dead end
Kainui
  • Kainui
Yeah, I'm not disagreeing with you on that
Kainui
  • 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
  • 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
  • ganeshie8
constant coefficents case is different, we have euler to backup :)
Kainui
  • 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
  • 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
  • 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
  • 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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