8
$\begingroup$

I came across the following problem which says:

The initial value problem $y'=y^{1/3}$, $y(0)=0$ has:

(a) a unique solution,

(b) exactly two solutions,

(c)exactly three solutions,

(d)no solution.

The solution of the problem is given by: $2x=3y^{2/3}$. But I could not come to a conclusion. Clearly, (d) can not be true. But i am not sure about the other options. It will be helpful if someone throws light on it. Thanks in advance.

  • 1
    @ChristianBlatter No, it doesn't. The answer is "infinitely many solutions" regardless of what happens for y<0. Moreover, the only sensible interpretation in this context would be $y^{1/3}=-|y|^{1/3}$ for y<0.2018-12-05

2 Answers 2

13

The answer is "none of the above". There are infinitely many solutions.

Pick any $\alpha > 0$ and define $f_\alpha (x) = 0$ for $x \le \alpha$ and $f(x) = (2/3)^{3/2} (x-\alpha)^{3/2}$ for $x > \alpha$. All these functions are solutions.

  • 1
    @GautamShenoy I don't know what do you mean by "the function is ill defined on the negative axis", but it has nothing to do with the problem. The answer is "infinitely many solutions" regardless of what happens for y<0. Moreover, the only sensible interpretation in this context would be $y^{1/3}=-|y|^{1/3}$ for y<0.2018-12-05
-6

total $3$ solutions

1.$y=0$ is also a solution.

2.y=0 for $x \le 0$ and $y=(2/3x)^{(3/2)}$ (positive squ root)

3.similarly consider neg squ root.