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

In summary: Hence, we have shown that if k<\frac{1}{4}, there is a finite number of positive real roots, whereas if k>\frac{1}{4}, there are infinitely many positive real roots.In summary, the differential equation y''+\frac{k}{x^2}y=0 has an infinite number of positive roots if k>\frac{1}{4}, and a finite number of positive roots if k<\frac{1}{4}.
  • #1
MarkFL
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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.
 
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  • #2
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:

\(\displaystyle y''+\frac{k}{x^2}y=0\)

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

\(\displaystyle x^2y''+ky=0\)

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

\(\displaystyle x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dx^2}-\frac{dy}{dt}\)

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

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

The characteristic roots are:

\(\displaystyle r=\frac{1\pm\sqrt{1-4k}}{2}\)

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

Case 1: The discriminant is positive.

\(\displaystyle 1-4k>0\)

\(\displaystyle k<\frac{1}{4}\)

The solution is then:

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

We conclude that the solution has no positive roots.

Case 2: The discriminant is zero.

\(\displaystyle 1-4k=0\)

\(\displaystyle k=\frac{1}{4}\)

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

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

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

Case 3: The discriminant is negative.

\(\displaystyle 1-4k<0\)

\(\displaystyle \frac{1}{4}<k\)

The general solution is then given by:

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

The sinusoidal factor guarantees an infinite number of positive real roots.
 

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

What is an ODE solution?

An ODE solution refers to the solution of an ordinary differential equation (ODE). This is a mathematical equation that describes the relationship between a function and its derivatives. ODE solutions are used to model a variety of physical and biological phenomena.

What are positive roots?

Positive roots refer to the values of a variable that satisfy a given equation when substituted into it. In the context of ODE solutions, positive roots are the values that make the solution of an ODE greater than zero.

Why is K>1/4 important in relation to ODE solutions?

K>1/4 is an important value in relation to ODE solutions because it determines the number of positive roots the solution can have. When K is greater than 1/4, the solution will have an infinite number of positive roots. This is known as an oscillating solution.

How can K>1/4 have infinite zeros in an ODE solution?

This occurs because when K is greater than 1/4, the ODE has an oscillating solution. This means that the solution will cross the x-axis an infinite number of times, resulting in an infinite number of positive roots.

What is the significance of an ODE solution with infinite zeros?

An ODE solution with infinite zeros is significant because it represents a system that oscillates indefinitely without ever reaching a stable equilibrium. This type of behavior is commonly observed in physical and biological systems, making it an important concept in the study of ODEs.

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