Help in solving a second-order linear differential equation

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SUMMARY

The discussion focuses on solving the second-order linear differential equation given by \(\frac{{d^2 y}}{{dx^2 }} + (Ax + B)y = 0\). A change of variable is suggested, where \(Ax + B = \lambda u\), leading to the transformed equation \(\frac{A^2}{\lambda^2} \frac{d^2 y}{d u^2} + \lambda u y = 0\). By setting \(\lambda = -A^{\frac{2}{3}}\), the equation simplifies to the Airy equation, \(\frac{d^2 y}{d u^2} - u y = 0\). The discussion concludes that there are three solutions for \(\lambda\), including one real and two complex values, as dictated by the condition \(\frac{\lambda^3}{A^2} = -1\).

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[tex] \frac{{d^2 y}}{{dx^2 }} + \left( {Ax + B} \right)y = 0[/tex]

I have tried lots of substitions, but a solution won't pop out. Can anyone help solve this?

Thanks.
 
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Make the change of variable

[tex]Ax + B = \lambda u[/tex]

(lambda is a constant) this will give you

[tex]\frac{A^2}{\lambda^2} \frac{d^2 y}{d u^2} + \lambda u y = 0[/tex]

so if you then set

[tex]\lambda = -A^{\frac{2}{3}}[/tex]

you then have

[tex]\frac{d^2 y}{d u^2} - u y = 0[/tex]

which is the Airy equation (in u). Have a look on Wikipedia or elsewhere on Airy functions and such - or just type in "Airy Equation".

edit: note you will actually get three different solutions as the condition for lambda is

[tex]\frac{\lambda^3}{A^2} = -1[/tex]

which means that there are three values of lambda that satisfy this (i.e. three distinct cube-roots) - one will be real-valued (already given) plus two complex ones.

Here is a link for the Airy function

http://mathworld.wolfram.com/AiryFunctions.html
 
Last edited:

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