Do the initial values for second order differential equations need to include a value for the derivative?

Question

I've been learning about second order differential equations on my own, and one thing I've noticed is that whenever there's an IVP, one of the conditions will use the actual function (ex. $y(0)=2$), but the other condition will always involve the derivative of that function (ex. $y'(0)=3$). Is there any particular reason for this, or is it just convention? Or to put it another way, for any given second order differential equation, would the initial conditions $y(x_0)=y_0$ and $y(x_1)=y_1$ be just as useful as the initial conditions $y(x_0)=y_0$ and $y'(x_0)=y_0'$ for solving an IVP?

Answer 1

Yes, the second variant also exists, but is called a boundary value problem. There does not always exist a solution to a given BVP, especially if the differential equation is non-linear. In the other direction, there can also be multiple solutions to given boundary values.

In general, if you transform the equation(s) into a first-order system, then the dimension of it tells you how many "integration constants" or degrees of freedom in the general solution there are. In principle, these can be fixed by a system of general equations in the state space, where the number of equations is equal to the state dimension. Each equation can use value at a different time, or combine the state at several times. The more fancy that gets, the more difficult it is to find a numerical solution.