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Second order DE

by fluidistic
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fluidistic
#1
Dec3-12, 04:55 PM
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1. The problem statement, all variables and given/known data
Find the general solution to ##A(x)y''+A'(x)y'+\frac{y}{A(x)}=0## where A(x) is a known function and y(x) is the unknown one.
Hint:Eliminate the term that contains the first derivative.


2. Relevant equations
Not sure.


3. The attempt at a solution
So I don't really know how to tackle it. I guess that the hint suggests a change of variable that would get rid of the y' term. So I tried z=y'A, z=Ay and z=y/A. All failed to express the ODE as a function of z and its derivative(s) and A and its derivative(s).
I am therefore stuck. Any other hint will be welcome!
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micromass
#2
Dec3-12, 06:32 PM
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What if you try the following substition [itex]y=Bz[/itex], where B is a known function that is yet to be decided.

What do you get if you perform this substitution?

What should we take for B if we want the z' term to vanish?
fluidistic
#3
Dec3-12, 07:02 PM
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Quote Quote by micromass View Post
What if you try the following substition [itex]y=Bz[/itex], where B is a known function that is yet to be decided.

What do you get if you perform this substitution?
Good idea! ##x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0##.

What should we take for B if we want the z' term to vanish?
##B=Ce^{-x}/A##. I checked out and indeed the terms in front of ##z'## vanishes.
I'm left to solve ##z''A'Ce^{-x} \left ( \frac{1}{A}+ \frac{A'}{A^2} \right ) + z \left ( \frac{A'}{A} + \frac{A'^2}{A^2}- \frac{Ce^{-x}}{A^2} \right ) =0##. That really does not look beautiful/easy to me.

micromass
#4
Dec3-12, 08:33 PM
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Second order DE

Quote Quote by fluidistic View Post
Good idea! ##x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0##.
I get something different here. I get that the term of z' is
[tex]A^\prime B + 2A B^\prime[/tex]

The form of my B is also much simpler.
fluidistic
#5
Dec3-12, 09:01 PM
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Quote Quote by micromass View Post
I get something different here. I get that the term of z' is
[tex]A^\prime B + 2A B^\prime[/tex]

The form of my B is also much simpler.
I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so ##B=kA^{-1/2}## where k is a constant.

Edit: I might have rushed through the math but I reach that ##z''+z \left ( 1-\frac{2}{A'} \right ) =0##. Which has different solutions depending on the sign of the term in front of z. I guess I made another mistakes.
micromass
#6
Dec3-12, 09:42 PM
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Quote Quote by fluidistic View Post
I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so ##B=kA^{-1/2}## where k is a constant.
OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

Anyway, with that choice of B, what does your equation become then?
fluidistic
#7
Dec3-12, 09:54 PM
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Quote Quote by micromass View Post
OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

Anyway, with that choice of B, what does your equation become then?
I now get ##B'=-\frac{1}{2}A^{-3/2}##, ##B''=\frac{3}{4}A^{-5/2}A'^2-\frac{A^{-3/2}A''}{2}##.
The equation becomes ##z''AB+z(AB''+A'B'+B/A)=0##. Now I have to replace the B and its derivatives in that.
Edit: I've just done it and it's not beautiful so far.


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