I'm an AL student , and i have been recently introduced to the concept of Homogeneous Second Order Linear Differential Equations.I could understand how the Characteristic Equation forms and why it was necessary but in the Case of Repeated Real Roots , our teacher explained that the solution – ce^(rx) was not the most general solution because second order differential equations are supposed to always have "2 degrees of freedom".I was trying to figure out Why it is the case that second order differential equations should always have 2 degrees of freedom , but I cant seem to find a reasonable answer.Any answers would be much appreciated
Why is it always necessary that a second order differential equation have 2 degrees of freedom
calculusordinary differential equations
Best Answer
A 2nd order DE specifies the second derivative of the solution. This leaves the value at some time $t_0$ and the first derivative at that same time unspecified. Those are your two degrees of freedom. One example might be a ball on a spring. Wherever and however it starts, the equation of motion tells you that the ball will follow a sinusoid motion. But it doesn't tell you where the initial location of the ball has to be, or the initial kick you can give it in its motion.
ADDED: This is not a complete explanation, but might give you an idea of how it all works. Let's say we have a DE
$\displaystyle \frac{d^2x}{dt^2} = f(t)$
and let's say we find out that it is satisfied by some $x_0(t)$. Now, what if we randomly pick two numbers $a$ and $b$ and add the following to $x_0(t)$ to make $x_1(t)$:
$\displaystyle x_1(t) = x_0(t) +a +bt $
Is $x_1(t)$ also a solution of the equation, or not? Yes, it is, because
$\displaystyle \frac{d^2x_1}{dt^2} = \frac{d^2x_0}{dt^2} = f(t)$
That is, the second derivative "killed" both the $a$ and $bt$ terms. Note that this is going to happen regardless of what values you choose for $a$ and $b$. So you have two "degrees of freedom" in your solution.