# How did they get |x|=±x? How did they know y=0 is also a solution?

1. Mar 13, 2013

### InvalidID

Consider example 3 in the attachment. Better quality version: http://postimage.org/image/fjz61dbyr/

How did they get $|x|=±x$? If I remember correctly, I was told on physics forums that $|x|≠±x$. The curious thing about this one is that it doesn't even involve square roots!

How did they know y=0 is also a solution?

Last edited by a moderator: Mar 14, 2013
2. Mar 13, 2013

### Staff: Mentor

They didn't.

They have |y| = eCe(1/3)x3 = Ae(1/3)x3, where A is a positive constant.

This means that y = Ae(1/3)x3 or y = -Ae(1/3)x3, with A again being positive.
By noticing that y = 0 satisfies the differential equation.

3. Mar 13, 2013

### eumyang

If we are talking about functions and graphs, then the graph of y = |x| isn't the same as the graph of y = ±x. But this is something different. It's essentially an absolute value equation, where if $| x | = b$ then $x = \pm b$.

4. Mar 13, 2013

### InvalidID

I don't understand how you went from:

to:

without the use that $|x|=±x$ where x is anything (not the same x in the example).

$$y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } },\quad A>0\\ Set\quad y=0\\ 0=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }$$

But there is no solution?

5. Mar 13, 2013

### Staff: Mentor

Let's make it simple. Assuming that B is a positive number, the equation |y| = B is equivalent to y = B or y = -B.

That's all they did.

The original differential equation is y' = x2y.
Notice that if y = 0, then y' = 0, so substituting this function and its derivative into the differential equation yields a true statement. That means that y = 0 is a solution of the DE.

6. Mar 13, 2013

### xAxis

But that's not the equation. The equation is the first one: dy/dx = x^2y, y=0 satisfy the equation.

7. Mar 13, 2013

### null

So in other words, you're using the fact that $|y|=B$ is the same as writing $y=±B$, right?

Like this?

$$\frac { dy }{ dx } ={ x }^{ 2 }y\\ Set\quad y=0\\ \frac { dy }{ dx } =0\\ ∴\quad when\quad y=0,\quad y'=0$$

And then we substitute $y=0$ and $y'=0$ back into the differential equation to get a true statement? But we got those values from the DE, so obviously it will be true... Isn't this circular logic?

8. Mar 13, 2013

### Number Nine

What does it mean for a function to be a solution to a differential equation?

9. Mar 13, 2013

### null

Assuming that a differential equation works like a regular equation, a solution is one that consists of 3 values: one for x, one for y, and one for y' that satisfy the equation when plugged in.

10. Mar 13, 2013

### null

Okay, so I get that y=0 satisfies the DE.

But why is it implied that just because it satisfies the DE, that the solution also works for y=0? I thought the solution was restricted to not work for y=0?

11. Mar 14, 2013

### Staff: Mentor

Huh? The last part makes no sense. y = 0 is a solution of the DE, because it satisfies (i.e., makes a true statement of) the DE.
Read the problem a little more closely. There is no restriction on y. What they did was break things up into two cases:
Case 1: y ≠ 0, from which they obtained the exponentional function.
Case 2: y = 0.

12. Mar 14, 2013

### null

Opps. My mistake.

Here is what I should've said:

Why is it implied that just because y=0 satisfies the DE, that y=0 also satisfies $y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }$? In order to obtain $y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }$, we had to restrict y=0 (basically case 1 that you referred to).

13. Mar 14, 2013

### HallsofIvy

A is an arbitrary constant- taking any specific value for A gives a solution.

So what if A= 0?