Linear ODE Solutions Without Initial Conditions and the Arbitrary Constant C

jdstokes
Messages
520
Reaction score
1
Suppose I already have a solution u to a first order ODE.

If I try to solve this ODE without initial conditions and I get another solution w, then it can be regarded as a function of an arbitrary constant: w=w(C).

Is it true to say that u = w(C) for some C? If so, how do I find such a C?
 
Physics news on Phys.org
That's a little vague. You use the boundary conditions to find C. In general, an ODE doesn't even have a unique solution unless you make some assumptions about the form of the ODE. Can you be more concrete?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top