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Consider the differential equation

$x^2y'' + a\,x\,y' + b\,y = 0 \text{ where } y = y(x) \text{ and } a,b \in R$

Using the change of variable $u = \ln(x)$, how can I transform the differential equation in the form of?

$Z'' + \alpha Z'+ \beta Z = 0 \text{ where } Z = Z(u)$

And what are the values ​​of $\alpha \text{ and }\beta$ as a function of a and b?

Thanks in advance

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Do you know the chain rule? – in_wolframAlpha_we_trust Jun 11 '13 at 9:12
@in_wolfram_we_trust Yep, but I don't know how to use de chain rule to solve this. – user78723 Jun 11 '13 at 9:21
It is so called an Euler Equation. For details you may check… – беркай Jun 11 '13 at 9:33
Thanks. Nice reference! – user78723 Jun 12 '13 at 3:46

2 Answers 2

up vote 0 down vote accepted

For $u=\ln{x}$ we get: $$y'_x=y'_uu'_x=y'_u\frac{1}{x},$$ $$y''_{xx}=(y'_x)'_x=y'_u(-\frac{1}{x^2})+y''_{uu}\frac{1}{x^2}.$$ The differential equation becomes $y''-y'+ay'+by=0$, where $y=y(u)$. Values $\alpha$, $\beta$ are $$\alpha=a-1$$ and $$\beta=b.$$ Finally, the roots of characteristic equation $\lambda^2+(a-1)\lambda+b=0$ are $$\lambda_1=\frac{1-a+\sqrt{(a-1)^2-4b}}{2},$$ $$\lambda_2=\frac{1-a-\sqrt{(a-1)^2-4b}}{2},$$ so solution to the differential equation is $$y(u)=C_1 e^{\lambda_1 u}+C_2 e^{\lambda_2 u},$$ or $$y(x)=C_1 x^{\frac{1-a+\sqrt{(a-1)^2-4b}}{2}}+C_2 x^{\frac{1-a-\sqrt{(a-1)^2-4b}}{2}}.$$

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Let $u = \ln x$.

Then by chain rule we have $$ \begin{array}{lll} \frac{\partial y}{\partial x} & = & \frac{\partial y}{\partial u} \frac{\partial u}{\partial x}\\ & = & \frac{\partial y}{\partial u} \frac{1}{x}.\end{array} $$

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Good hint. Thank you! – user78723 Jun 12 '13 at 3:44

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