Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Do you know the chain rule? –  in_wolfram_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
1  
It is so called an Euler Equation. For details you may check math.dartmouth.edu/archive/m23s06/public_html/handouts/… –  беркай Jun 11 '13 at 9:33
    
Thanks. Nice reference! –  user78723 Jun 12 '13 at 3:46
add comment

2 Answers

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 $$y(x)=C_1 x^{\frac{1-a+\sqrt{(a-1)^2-4b}}{2}}+C_2 x^{\frac{1-a-\sqrt{(a-1)^2-4b}}{2}}.$$

share|improve this answer
add comment

Hint:

Let $u = \ln x$.

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

share|improve this answer
    
Good hint. Thank you! –  user78723 Jun 12 '13 at 3:44
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.