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Show by direct substitution that

$$ P(x,t) = \frac{1}{\sqrt{4 \pi D t}} \exp \left( - \frac{x^2}{4Dt} \right) $$

is a solution to the equation

$$ \frac{\partial P}{\partial t} = D \frac{\partial^2 P}{\partial x^2}$$

(Please show the full working and reasoning including any annotations that would help a beginner to understand.)

What does it mean by direct substitution? Direct substitution of what into what? What common mathematical techniques am I expected to use here?

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You know how to take partial derivatives, no? Take the partial derivative of $P$ with respect to $t$; take the second partial derivative of $P$ with respect to $x$, and multiply by $D$. Then you need to show that they are identical... –  J. M. Nov 21 '11 at 13:51
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$\frac{\partial^2P}{\partial x^2}$ is easy to compute, so do so, multiply by $D$ and then verify that $\frac{\partial P}{\partial t}$, which takes a little more computation, gives the same result. –  Dilip Sarwate Nov 21 '11 at 13:53
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It means: Plug in, and check that you get an equality using that function. –  Arturo Magidin Nov 21 '11 at 14:01
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Show by direct substitution that $x={\color{red}2}$ is a solution to the equation $x^2=x+2$: ${\color{red}2}^2={\color{red}2}+2$. –  robjohn Nov 21 '11 at 19:39

1 Answer 1

up vote 2 down vote accepted

You know that $P$ is a certain function of $x$ and $t$.

So you can find $\dfrac{\partial^2 P}{\partial x^2}$ and $\dfrac{\partial P}{\partial t}$.

Take whatever you got when you found $\dfrac{\partial P}{\partial t}$ and put it in place of $\dfrac{\partial P}{\partial t}$ in the differential equation $\dfrac{\partial P}{\partial t} = D\dfrac{\partial^2 P}{\partial x^2}$. That is "direct substitution". Likewise, take whatever you got when you found $\dfrac{\partial^2 P}{\partial x^2}$ and put it in place of $\dfrac{\partial^2 P}{\partial x^2}$ in the differential equation. That is "direct substitution". Then check that the thing to the left of "$=$" actually is equal to the thing to the right. If they are equal, then the given function $P(x,t)$ is a solution of the differential equation; otherwise it's not.

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