
#1
Dec312, 04:55 PM

PF Gold
P: 3,173

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! 



#2
Dec312, 06:32 PM

Mentor
P: 16,562

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? 



#3
Dec312, 07:02 PM

PF Gold
P: 3,173

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. 



#4
Dec312, 08:33 PM

Mentor
P: 16,562

Second order DE[tex]A^\prime B + 2A B^\prime[/tex] The form of my B is also much simpler. 



#5
Dec312, 09:01 PM

PF Gold
P: 3,173

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. 



#6
Dec312, 09:42 PM

Mentor
P: 16,562

Anyway, with that choice of B, what does your equation become then? 


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