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using derivatives verify that:
p(x,t)= 1/( Sqrt[ 4 Pi Dt]) * Exp[x^2/ 4 Dt]
is a solution of hte diffusion equation:
dp/dt= D d^2p/dx^2
 11 months ago
 11 months ago
using derivatives verify that: p(x,t)= 1/( Sqrt[ 4 Pi Dt]) * Exp[x^2/ 4 Dt] is a solution of hte diffusion equation: dp/dt= D d^2p/dx^2
 11 months ago
 11 months ago

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SithsAndGigglesBest ResponseYou've already chosen the best response.0
I'm sure you mean the partial derivatives where you wrote "dp/dt": \[\frac{\partial p}{\partial t}=D\frac{\partial^2p}{\partial x^2}\] Given \(p(x,t)=\dfrac{1}{\sqrt{4\pi Dt}}\exp\left({\dfrac{x^2}{4}Dt}\right)\), all you have to do is find the partial derivatives \(\dfrac{\partial p}{\partial t}\) and \(\dfrac{\partial^2 p}{\partial x^2}\) and plug them into the equation.
 11 months ago

liliyBest ResponseYou've already chosen the best response.0
how do i find the partial derivative?
 11 months ago

liliyBest ResponseYou've already chosen the best response.0
bec i think i got to dp/dt part but d^2/ dx^2 i dont know how to do it
 11 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
It's a lot like taking a regular derivative. The partial derivative of a function \(f(x,t)\) with respect to \(t\), or \(\dfrac{\partial f}{\partial t}\), is obtained by differentiating the function with respect to \(t\), but fixing \(x\) as a constant. As an example: Suppose \(f(x,t)=xt\). Then \(\dfrac{\partial f}{\partial x}=t\) and \(\dfrac{\partial f}{\partial t}=x\), since \(t\) is considered constant in the first and \(x\) is considered to be constant in the second.
 11 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
For this particular function, you'll have your work cut out for you. Plenty of chain rule and product rule.
 11 months ago

liliyBest ResponseYou've already chosen the best response.0
I actually just need to code this in mathematica, not figure it all out. lol
 11 months ago

SithsAndGigglesBest ResponseYou've already chosen the best response.0
In that case, just use the differentiation operator. In Mathematica, you'd write \(\dfrac{\partial p}{\partial x}=\) D[p(x,t),t] \(\dfrac{\partial^2 p}{\partial x^2}=\) D[D[p(x,t), x],x] where p(x,t) is the given function, obviously. Here's the partial with respect to t: http://www.wolframalpha.com/input/?i=D%5B1%2F%28+Sqrt%5B+4+Pi*+D*t%5D%29++*+Exp%5Bx%5E2%2F+4+D*t%5D%2Ct%5D And here's the 2ndorder partial with respect to x: http://www.wolframalpha.com/input/?i=D%5BD%5B1%2F%28+Sqrt%5B+4+Pi*+D*t%5D%29++*+Exp%5Bx%5E2%2F+4+D*t%5D%2Cx%5D%2Cx%5D
 11 months ago
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