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Sugestions to solve this equation 
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#1
Nov1312, 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 


#2
Nov1312, 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?).



#3
Nov1312, 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



#4
Nov1312, 07:34 PM

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Sugestions to solve this equation
You can obtain a lot of solutions by setting b(x) = Ax^{n} and solving for y. That won't give you a general solution though.



#5
Nov1412, 04:26 PM

P: 291

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 


#6
Nov1412, 10:16 PM

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#7
Nov1512, 01:53 PM

P: 291

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... 


#8
Nov1512, 03:02 PM

P: 350

That suggests the substitution
[itex] t =∫(a/b(x)) dx [/itex] which works wonders :). 


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