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apu's question at Yahoo! Answers regarding the number of positive roots of solution to ODE
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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.
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.