I have prove that one solution of the problem dy/dx = 3y^(2/3), y(2) = 0 is from
ID: 1947065 • Letter: I
Question
I have prove that one solution of the problem dy/dx = 3y^(2/3), y(2) = 0 is from the following statement: Let y= g(x) be a solution to the IVP on (-INF,INF). Of course g(2) = 0. If g vanishes at some point x > 2, the let b be the largest of such points; otherwise, set b = 2. Similarly, if g vanishes at some point x < 2, the let a be the smallest(furthest to the left) of such points; otherwise, set a = 2. Here we allow b = INF and a = -INF. (Because g is a continuous function, it can be proved that there always exist such largest and smallest points.) Using the results of parts (a) and (b) prove that if both a and b are finite, then g has the following form:
(x-a)^3 if x is mayor or equal of a
g(x) = 0 if a < x minor or equal of b
(x-b)^3 if x > b
what is the form of g if b = INF? If a = -INF? If both b = INF and a = -INF?
Note: for the first two statements I have the following results:
For (b): the values of a and b are (-INF, -c) for a and (-c, INF) for b.
For (a):y=(x-c)^3 and the constant c is not in the interval (I).
Note: Also I have sent you the image of the hole problem. If I am doing something wrong please correct me.
Explanation / Answer
dy/dx = 3y^(2/3)
(1/y^2/3)dy = 3dx
(y^-2/3)dy = 3dx
Integrating both the sides,
3y^1/3 = 3x + C
For x = 2, y = 0
0 = 6 + C
C = -6
Thus,
3y^1/3 = 3x - 6
y^1/3 = x - 2
y = (x-2)^3
Hence, y = g(x) = (x-2)^3 be a solution to the IVP on (-,).
Of course g(2) = 0.