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

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?

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Mark44
Mentor
Consider example 3 in the attachment. Better quality version: http://postimage.org/image/fjz61dbyr/

How did they get ##|x|=±x##?
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.
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?
By noticing that y = 0 satisfies the differential equation.

eumyang
Homework Helper
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!
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 $x = \pm b$.

They didn't.
I don't understand how you went from:

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

This means that y = Ae(1/3)x3 or y = -Ae(1/3)x3, with A again being positive.
without the use that ##|x|=±x## where x is anything (not the same x in the example).

By noticing that y = 0 satisfies the differential equation.
$$y=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } },\quad A>0\\ Set\quad y=0\\ 0=A{ e }^{ \frac { { x }^{ 3 } }{ 3 } }$$

But there is no solution?

Mark44
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.

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

But there is no solution?
But that's not the equation. The equation is the first one: dy/dx = x^2y, y=0 satisfy the equation.

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

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.
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?

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

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

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?

Mark44
Mentor
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?
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.
I thought the solution was restricted to not work for y=0?
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.

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.
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).

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.

HallsofIvy
Homework Helper
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).
A is an arbitrary constant- taking any specific value for A gives a solution.

So what if A= 0?