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Sugestions to solve this equation

by pedro_ani
Tags: equation, solve, sugestions
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pedro_ani
#1
Nov13-12, 03:42 PM
P: 2
Any sugestions on how to find the solutions to this equation?


[itex]y'' +\frac{b'}{b} y' - \frac{a^2}{b^2}y=0[/itex]

where [itex]a[/itex] is a constant
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Number Nine
#2
Nov13-12, 05:02 PM
P: 772
I assume that b is a constant as well. That's a pretty standard homogeneous linear equation; just use the standard methods (can you write out its characteristic equation?).
pedro_ani
#3
Nov13-12, 05:07 PM
P: 2
No, [itex]b[/itex] is a function, if it were a constant its derivative would be zero and the term with [itex]y'[/itex] would disappear

haruspex
#4
Nov13-12, 07:34 PM
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Sugestions to solve this equation

You can obtain a lot of solutions by setting b(x) = Axn and solving for y. That won't give you a general solution though.
bigfooted
#5
Nov14-12, 04:26 PM
P: 288
rewrite

[itex]by''+b'y'-\frac{a^2}{b}y=0[/itex]

to the standard form:

[itex]z''=-c^2(t)z[/itex],

with

[itex]c(t)=-\frac{2a + b'(t)}{2b}[/itex]

the general solution of the transformed equation is then

[itex]z=Asin(cx)+Bcos(cx)[/itex]

Then get the solution of y by transforming back:
[itex]y=z e^{\int{\frac{b'}{2b}dx}} = z\frac{b'}{2b}[/itex]

[itex]y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}[/itex]

That's of course, assuming a,b are such that all steps are valid
haruspex
#6
Nov14-12, 10:16 PM
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Quote Quote by bigfooted View Post
[itex]y=(A\sin(cx)+B\cos(cx))\frac{b'}{2b}[/itex]
Looks to me that if you substitute that in the original equation there'll be an unbalanced b''' term.
bigfooted
#7
Nov15-12, 01:53 PM
P: 288
Oops, you're right, I was thinking too simple! The transformation to standard form does not lead to an expression that can be easily solved by a ricatti equation.

Actually, maple gave the following:

[itex]y=A\sinh(\int -\frac{a}{b(x)}dx) + B\cosh(\int -\frac{a}{b(x)}dx)[/itex]

hope this helps a bit...
Vargo
#8
Nov15-12, 03:02 PM
P: 350
That suggests the substitution

[itex] t =∫(a/b(x)) dx [/itex]

which works wonders :).


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