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How to solve (2x+1)^2 y'' -2 (2x+1) y' +4 y = 0?

Sciencolic (sciencolic)

I love Physics, Mathematics and Science.
Japan

Ordinary differential equation	288
Mathematics	597

19 Aug 10

I would like to know how to solve the following differential equation (you may refer to this post for the original question):

$\f[(2x+1)^2 y'' -2 (2x+1) y' +4 y = 0\f]$

Thank you very much.

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Ross Tang (ross_tang)

Degree in Physics and Mathematics, Master in Physics
香港

Ordinary differential equation	672
Mathematics	809

19 Aug 10

First of all, notice that it is an Euler equation, and you can first transform it into a standard form.

Let $\f$z = 2x+1\f$$ ,

$\f$\displaystyle \begin{align*} \frac{d y }{d x}&=\frac{d y}{d z}\frac{d z}{d x} \\ &= 2 \frac{d y}{d z} \end{align*} \f$$

And:

$\f$\displaystyle \frac{d^2 y }{d x^2}&=4 \frac{d^2 y}{d z^2}\f$$

Now your equation become:

$\f$\displaystyle z^2 \frac{d^2 y}{dz^2} - z \frac{d y}{dz} + y = 0\f$$

This is a standard Euler equation, which can be solved by the substituion:

$\f$\displaystyle z = e^t \Rightarrow t = \ln z\f$$

$\f$\displaystyle \begin{align*} \frac{d y}{d z}&=\frac{d y}{d t}\frac{d t}{d z} \\ &= \frac{1}{z}\frac{d y}{d t} \end{align*}\f$$

$\f$\displaystyle \begin{align*} \frac{d^2 y}{d z^2}&=\frac{d}{d z}\left(\frac{1}{z}\frac{d y}{d t}\right)\\ &=\frac{1}{z}\frac{d}{d z} \frac{d y}{d t}-\frac{1}{z^2}\frac{d y}{d t}\\ &=\frac{1}{z^2} \frac{d^2 y}{d t^2}-\frac{1}{z^2}\frac{d y}{d t} \end{align*}\f$$