<p>mine is y’‘/kq + y’/k +y’/q + y = C at the moment. I came up with it myself last week :D.</p>
<p>(k/q are different constants of proportionality. C is an amount).</p>
<p>mine is y’‘/kq + y’/k +y’/q + y = C at the moment. I came up with it myself last week :D.</p>
<p>(k/q are different constants of proportionality. C is an amount).</p>
<p>The one you can solve.</p>
<p>Do you want us to name another one or tell you how smart you are?</p>
<p>wait does anybody think diff eqs have a particular beautifulness to them :p?</p>
<p>I want you to name another, of course! :)</p>
<p>Well then the answer is obvious. y’=k.</p>
<p>x’=0</p>
<p>slope field pl0x</p>
<p>You realize that your equation is just y’‘/kq + [(k+q)/kq]y’ + y = C
which is just the standard second order linear equation?</p>
<p>
I rate him (her? it would make sense due to her lack of math ability) as a 2 out of 10, and this isn’t golf so that’s bad.</p>
<p>Euler’s Identity.</p>
<p>It’s so awesome that it trumps any old differential equation. :P</p>
<p>x’x=xx’ </p>
<p>1char</p>
<p>Is this for real? I wish I was smarter so I could actually answer this question. But, more power to you…</p>
<p>
</p>
<p>Well I suppose it’s not that uncommon an equation. But I think the case of it I came up with (where the constants are what they are) has interesting applications, potential to model certain behaviors.</p>
<p>I actually came up with it trying when I was trying to understand the basis of [Hubbert</a> peak theory - Wikipedia, the free encyclopedia](<a href=“http://en.wikipedia.org/wiki/Hubbert_peak_theory]Hubbert”>Hubbert peak theory - Wikipedia). The theory posits that the rate of production of oil follows a bell shaped curve over time. Interestingly, the solution to y’ of my equation (which is supposed to be the rate of the commodities’ production) gives a similar type of curve.</p>
<p>jk man</p>
<p>…</p>
<p>… :).</p>
<p>This topic is hilarious…but in all seriousness…</p>
<p>(1-x^2)y’’ - 2xy’ + n(n+1)y = 0</p>
<p>:) Legendre’s equation.</p>
<p>dx/dy is the only part of calculus I understand, so… yeah.</p>
<p>^You probably mean dy/dx</p>
<p>Hahahahaha. This thread…</p>