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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

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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.
how do i find the partial derivative?
bec i think i got to dp/dt part but d^2/ dx^2 i dont know how to do it

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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.
For this particular function, you'll have your work cut out for you. Plenty of chain rule and product rule.
I actually just need to code this in mathematica, not figure it all out. lol
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:*+D*t%5D%29++*+Exp%5B-x%5E2%2F+4+D*t%5D%2Ct%5D And here's the 2nd-order partial with respect to x:*+D*t%5D%29++*+Exp%5B-x%5E2%2F+4+D*t%5D%2Cx%5D%2Cx%5D

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