MHB Positive Roots of ODE Solution: K>1/4 has Infinite Zeros

AI Thread Summary
The differential equation y'' + (k/x^2)y = 0 exhibits different behaviors based on the value of k. If k < 1/4, the solution has no positive roots, while if k = 1/4, there can be at most one positive root. For k > 1/4, the solution includes sinusoidal components, leading to an infinite number of positive roots. The transformation into a Cauchy-Euler equation and the analysis of the characteristic roots reveal these outcomes. Thus, the presence of infinite positive zeros is confirmed for k values greater than 1/4.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Show that this differential equation has infinite positive zeros?


Show that every nontrivial solution for y''+ (k/x^2) y=0, has an infinite number of positive zeroes if K>1/4 and only a finite number if K<1/4 and if K=1/4.

I have posted a link there to this thread so the OP can see my work.
 
Mathematics news on Phys.org
Re: apu's question at Yahoo! Anwers regarding the number of positive roots of solution to ODE

Hello apu,

We are given the 2nd order linear ODE:

$$y''+\frac{k}{x^2}y=0$$

Multiplying through by $x^2$, we obtain the Cauchy-Euler equation:

$$x^2y''+ky=0$$

Using the substitution $x=e^t$, we find:

$$x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dx^2}-\frac{dy}{dt}$$

Hence, the ODE is transformed into the linear homogenous ODE:

$$\frac{d^2y}{dx^2}-\frac{dy}{dt}+ky=0$$

The characteristic roots are:

$$r=\frac{1\pm\sqrt{1-4k}}{2}$$

We know the nature of the solution depends on the discriminant.

Case 1: The discriminant is positive.

$$1-4k>0$$

$$k<\frac{1}{4}$$

The solution is then:

$$y(t)=c_1e^{\frac{1+\sqrt{1-4k}}{2}t}+c_1e^{\frac{1-\sqrt{1-4k}}{2}t}=e^{\frac{1-\sqrt{1-4k}}{2}t}\left(c_1e^{\sqrt{1-4k}t}+c_2 \right)$$

We conclude that the solution has no positive roots.

Case 2: The discriminant is zero.

$$1-4k=0$$

$$k=\frac{1}{4}$$

In this case, because of the repeated characteristic root, the general solution is:

$$y(t)=c_1e^{\frac{t}{2}}+c_2te^{\frac{t}{2}}=e^{ \frac{t}{2}}\left(c_1+c_2t \right)$$

Here, we find that there can at most one positive real root.

Case 3: The discriminant is negative.

$$1-4k<0$$

$$\frac{1}{4}<k$$

The general solution is then given by:

$$y(t)=e^{\frac{t}{2}}\left(c_1\cos\left(\frac{\sqrt{1-4k}}{2}t \right)+c_2\sin\left(\frac{\sqrt{1-4k}}{2}t \right) \right)$$

The sinusoidal factor guarantees an infinite number of positive real roots.
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top