# Help - Seperation of variables problem, multiple solutions.

1. Mar 5, 2013

### girlphysics

Help -- Seperation of variables problem, multiple solutions.

1. The problem statement, all variables and given/known data

Suppose that dy/dx = √y and y(0) = 0. What is y(x)? There is more than one answer to this problem. You must list five correct solutions.

2. Relevant equations

Seperation of Variables/ integration

3. The attempt at a solution

I got the first solution to be y= x^2 /4 and c=0. I dont know how to get the four other solutions.

2. Mar 5, 2013

### CompuChip

If you restrict a function to a subset of its domain, it is technically a different function. Could this be what is meant? Because the given IVP seems unambiguous to me.

3. Mar 5, 2013

### I like Serena

Hey you!

There must be some mistake in the problem statement.
The solution y=x^2/4 is the only one!

4. Mar 5, 2013

### pasmith

y(x) = 0 is also a solution, since then $y' = \sqrt y = 0$ for all $x$.
This suggests something like
$$y(x) = \left\{\begin{array}{r@{\quad}l} \frac{(x - a)^2}{4}, & x < a \\ 0, & a \leq x \leq b \\ \frac{(x - b)^2}{4}, & x > b\end{array}\right.$$
for $a < 0 < b$.

5. Mar 5, 2013

### girlphysics

Hey! There is no mistake, my professor talked about it in class and said there are infinitely many solutions, and that he wants us to get the 3rd solution. The second solution someone posted below. He said it is difficult to get the third. any ideas?

6. Mar 5, 2013

### I like Serena

My mistake.
This turns out to be an interesting problem!
I did not realize this system had more than one solution.
I see now that they are caused because separation of variables leads to a system that is not defined for y=0, causing multiple solutions.

Anyway, I believe pasmith came up with the key to the solutions.

You can pick $y=0$ for $x<0$ and $y=x^2/4$ for $x\ge 0$.

Or you can pick $y=0$ for $x<0$, $y=0$ for $0 \le x < 2$, and $y=(x-2)^2/4$ for $x \ge 2$.
Or...

However, I believe his part of the solution for x<a is faulty, since the derivative becomes negative which does not match with $\sqrt y$.

Last edited: Mar 5, 2013
7. Mar 6, 2013

### I like Serena

If you're still interested, you can find more information on your problem here.

8. Mar 6, 2013

### girlphysics

Thank you so much! I finally understand. I really appreciate it.