Exploring Initial Conditions for ODE Solution

[littleHilbert]

Hi, World! Nice place here! My first post in this forum.

I've got a short question for a start.
If we wish to evaluate the constants for the general solution
$x(t)=C_1e^{-{\lambda_1}t}+C_2e^{-{\lambda_2}t}$
of this ODE:
$\ddot{x}+2{\gamma}\dot{x}+{{{\omega}_0}^2}x=0$
we can choose the initial conditions: $x(0)=x_0,\dot{x}(0)=v_0$
I cannot see at a glance why we can't choose an initial condition of acceleration and try to calculate the constants using this value. Why do we choose $x_0,v_0$ and not for example $x_0,a_0$ with $a_0={{\lambda_1}^2}C_1+{{\lambda_2}^2}C_2$?

 

[littleHilbert]

Hey, Guys...why silence?
Did I ask nonsense? :uhh:
I don't think it's nonsense. :grumpy:
In the meantime I came across some info on oscillations in Feynman's lectures.
It says we cannot specify acceleration with which the motion started because it is determined by the spring, once we specify $x_0$. But isn't the velocity also dependent on the properties of the spring then?

 

[kyle8921]

Hello and welcome to the forum! It's great to see people exploring and asking questions about initial conditions for ODE solutions.

To answer your question, the choice of initial conditions for an ODE solution is not arbitrary. It is based on the structure of the equation and the physical interpretation of the variables. In this particular equation, x(t) represents the position of a particle at time t, and \dot{x}(t) represents its velocity. Therefore, it makes sense to choose initial conditions that correspond to the position and velocity of the particle at time t=0.

Choosing x_0 and v_0 as initial conditions allows us to uniquely determine the values of C_1 and C_2 in the general solution, and thus fully define the solution for all values of t. If we were to choose a_0 instead, it would not provide enough information to determine the values of C_1 and C_2, and the solution would not be fully defined.

Additionally, the equation \ddot{x}+2{\gamma}\dot{x}+{{{\omega}_0}^2}x=0 is a second-order differential equation, meaning it involves the second derivative of x. This means that the initial conditions must also be specified for the first and second derivatives of x, which is why we choose x_0 and v_0 as initial conditions.

I hope this helps clarify the reasoning behind the choice of initial conditions for ODE solutions. Keep exploring and asking questions!