# Initial value problem $dy/dx=y^{1/3}$, $y(0)=0$ has one of the following solution

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.

• The answer depends on your definition of $y^{1/3}$ when $y<0$. – Christian Blatter Nov 28 '12 at 18:57
• @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 $y<0$. – Federico Dec 5 '18 at 16:37

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.

• I see why this happened. This is because the function is ill defined on the negative axis. So hope the question asker takes note. – Gautam Shenoy Nov 29 '12 at 6:26
• i know the uniqueness criteria for first order differential equations and from that criteria i know that the above problem doesn't have a unique solution. so can we say either unique solution or infinitely many solutions? – Mittal G Feb 16 '16 at 4:03
• @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 $y<0$. – Federico Dec 5 '18 at 16:39

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.