On the nature of certain things - Atheist Nexus2015-05-27T23:15:34Zhttp://www.atheistnexus.org/forum/topics/on-the-nature-of-certain?groupUrl=mathematics&commentId=2182797%3AComment%3A1945948&groupId=2182797%3AGroup%3A68303&feed=yes&xn_auth=noIf you're solving this differ…tag:www.atheistnexus.org,2012-05-08:2182797:Comment:19459482012-05-08T15:55:26.579ZRick Taylorhttp://www.atheistnexus.org/profile/RickTaylor
<p>If you're solving this differential equation using separation of variables, you would start by writing:</p>
<p></p>
<p>dy/y = dx</p>
<p></p>
<p>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…</p>
<p>If you're solving this differential equation using separation of variables, you would start by writing:</p>
<p></p>
<p>dy/y = dx</p>
<p></p>
<p>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.</p> Well integrate dy/dx = ky and…tag:www.atheistnexus.org,2009-08-12:2182797:Comment:4590302009-08-12T17:25:39.638ZAthianarchisthttp://www.atheistnexus.org/profile/Athianarchist
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:<br />
<br />
y = be^(kx), k > 0<br />
y = 0, k = 0<br />
y =…
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:<br />
<br />
y = be^(kx), k > 0<br />
y = 0, k = 0<br />
y = -be^(kx), k < 0<br />
<br />
instead of y