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Solve Bessel's equation through certain substitutions 
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#1
Oct509, 06:55 PM

P: 5

1. The problem statement, all variables and given/known data, relevant equation
[tex] x^{2}y'' + xy' + (4x^{4}\frac{1}{4})y = 0[/tex] 2. The attempt at a solution I tried substituting z = x^{2} From this I have [tex]\frac{dy}{dx} = 2x \frac{dy}{dz}[/tex] and [tex]\frac{d}{dx}(2x\frac{dy}{dz}) = 2xy'' + 2y'[/tex] Then the original equation becomes: [tex]2z^{3/2}y'' + 4zy' + (4z^{2}\frac{1}{4})y = 0[/tex] where derivatives of y are now with respect to the new variable z. This does not look like a Bessel equation and I'm not sure how to make it look like one. Did I use the wrong substitution? I know how to solve once it's in the correct form, but could someone help me get it there please? 


#2
Oct509, 07:45 PM

P: 110

try [tex]z^{2}=4x^{4}[/tex]



#3
Oct509, 08:49 PM

P: 5

If I use z^{2} = 4x^{4} I get:
z = 2x^{2} and dz/dx = z' = 4x [tex] (\frac{dy}{dx}) = \frac{dy}{dz} \frac{dz}{dx} = 4x \frac{dy}{dz}[/tex] Then [tex]\frac{d}{dx}(4x\frac{dy}{dz})=4\frac{dy}{dz}+\frac{d}{dz}(\frac{dy}{dz} )\frac{dz}{dx}=4\frac{dy}{dz}+4x\frac{d^{2}y}{dz^{2}}[/tex] Plugging in: x^{2}(4y' + 4xy'') + x(4xy') + (4x^{4}  1/4)y = 0 if z^{2} = 4x^{4} (z/2)^{3/2} y'' + zy' + (z^{2}/4  1/16)y = 0 so it still doesn't work. The x^{3} term messes the whole thing up.... 


#4
Oct609, 06:14 PM

P: 5

Solve Bessel's equation through certain substitutions
Never mind, I got it. Thanks for your help!



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