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

[MarkFL]

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.

 

[MarkFL]

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.