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# On the nature of certain things

dy/dx = ky

How does one handle y = 0 in this situation? The problem I'm having is with the fact that when I integrate, there's no defined value of c that allows "y = 0" to fit, but simply plugging y = 0 into the equation to start with works.

Should things be coming out to
y = be^(kx), y > 0
y = 0, y = 0
y = -be^(kx), y < 0
? Or leave out the y = 0 bit? (really, I suppose, this comes down to being a question of how to deal with absolute values.)

Views: 9

### Replies to This Discussion

Well integrate dy/dx = ky and you'll get y = be^(kx) like you said, but this is of course an exponential and y --> 0 as x --> -inf so technically y can never really "equal" zero in this situation. And when you set the derivative equal to a constant you of course won't be able to integrate it properly at all, it becomes dy/dx = c, ==>; y = cx, and you won't get the exponential at all. But I think what you meant to say in the middle part was:

y = be^(kx), k > 0
y = 0, k = 0
y = -be^(kx), k < 0

If you're solving this differential equation using separation of variables, you would start by writing:

dy/y = dx

Notice when you do this, you are dividing both sides of the equation by y. That means you are assuming that y is not 0 (because division by 0 is not defined). Therefore, the logic of your derivation is not valid when y=0, and you must test that as a separate case. In general, whenever a derivation involves dividing both sides of an equality by a quantity that may be zero, you must test what happens when the quantity is zero as a separate case.

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