Constants of integration depend on the preferred way of representing solutions. For example, in
u'' + u = 0,
we can represent the two homogeneous solutions either as sin x and cos x, or as eix and e-ix. A general solution is a linear combination of the two, and depending on your choice of representation, constants of integration can be real or complex.
The representation of the two linearly independent solutions is nearly completely arbitrary. You can also choose sin x + cos x as one of the solutions, and six - cos x as the other. Integration constants are determined only once you have chosen the representation of your two linearly independent solutions.
1.part of söderlind : general solution is a functionvectorspace ,you take the arbitrary constant from R , C, or any other Körper(german)/Legeme(dansk).
1. part of the calculation.
2.part. practice want a particulary solution(IVP) y=const1, y´ =const2,(inicial codition/values) constants again normaly R but C also possible (electrical problems or fluid mechanics?)
Chosen the base(2 lin. independent func) in 1.part
and the inicial condition/values fixed you solve and get the constants
if you start with pure R you will normaly stay there(you try find example!!). but starting i C you could easy come in R (conjugated numbers)
iF you have Boundary value problem(BVP) it gets complicated, often the donot have solutions. onother day
Hello,
considering your initial value problem : y' = 2\sqrt(y) and y(0)=-1< as in C -1 has two square roots, and as \sqrt(-1)\neq 0, you can integrate near this initial value. However the solution is complex.
In any case the constant of integration is in the space (a vector space, a field) where you search the solutions.
Best Regards,
Gabriel
Dear Colleague,
writing y^(1/2) = x+c and looking for such a solution where x and c would be real, then y^(1/2) should be real, and y should be positive as a square of a real valued function. If you have y(0) = -1 < 0, then this cannot happen.
Sincerely, Octav
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