### Author Topic: Question 1 from TT2, 2015S  (Read 1182 times)

#### Tristan Fraser

• Full Member
•   • Posts: 30
• Karma: 11 ##### Question 1 from TT2, 2015S
« on: March 21, 2018, 08:17:23 PM »
Quote
Consider the eigenvalue problem
$$x^2 X″+2xX′+\lambda X=0,\ \ x \ \ \epsilon (\frac{2}{3},\frac{3}{2}), \ \ X′(\frac{2}{3})=0; \ \ \ X′(32)=0 \ \ \ \ \ (0)$$

Assume $\lambda \geq 0$. Find all the eigenvalues and the corresponding eigenfunctions.

Hint: as (1) is Euler equation, look   for elementary  solutions  in the form $x^m$).

I wrote the same trick and got the same characteristic of

$$k(k-1) + 2k + \lambda = 0 \ \ \ \ \ (1)$$

What I do not understand is why, and how we were able to make the substitution of $t = ln (\frac{3x}{2})$ to arrive at   $$\ddot{X} + \dot{X} + \lambda X = 0 \ \ \ \ \ (2)$$

Since my solution instead relied on examining the cases of $\lambda \geq 0$, but even after plugging in $x = \frac{3}{2} e^{t}$ I do not see how we would get to the above eigenvalue problem (2).

#### Jingxuan Zhang

• Elder Member
•     • Posts: 106
• Karma: 20 ##### Re: Question 1 from TT2, 2015S
« Reply #1 on: March 21, 2018, 09:55:01 PM »
Tristan this is the trick used in 244 to solve Euler's equation:
$$\partial_t=x\partial_x;\,\partial^2_t=\partial_x+x^2\partial^2_x,\, x=e^t.$$

#### Victor Ivrii ##### Re: Question 1 from TT2, 2015S
« Reply #2 on: March 22, 2018, 01:27:15 AM »
Actually it is a trick, used to explain, in what form one should look for a solution. After this is understood, you need to do it directly, without reductions