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Determine $p(z)$ by solving $$ \frac{dp}{dz}=-\frac{1}{\lambda}p$$ Where $\lambda$ is a constant and find the particular solution that satisfies the initial condition $p(0)=P$ where $P$ is a constant. $$ \frac{dp}{dz}=-\frac{1}{\lambda}p$$ $ \int\frac{dp}{p}=\int-\frac{1}{\lambda}dz $, where $p$ does not $=0$. $p(z)\equiv 0$ for $p = 0$. $$ \ln|p|=-\frac{1}{\lambda}\int dz $$ $$ \ln|p|=-\frac{1}{\lambda}z +c,c\in\mathbb{R} $$ $$ |p|=e^{-\frac{1}{\lambda}z+c} = e^{-\frac{1}{\lambda}z}e^{c}=ke^{-\frac{1}{\lambda}z}, c\in\mathbb{R},k>0 (k:=e^c)$$ for $p > 0$: $$ p = ke^{-\frac{1}{\lambda}z}, k>0$$ for $p < 0$: $$ -p = ke^{-\frac{1}{\lambda}z}, k>0$$ $$ p = -ke^{-\frac{1}{\lambda}z}, k>0$$
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Mon, 23 Nov 2020 21:25 GMT