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As a general principle, I'd say that mathematical modeling was the translation of intuition (derived from observation, experiment, and experience) into mathematical formalism. The goal is not so much to "solve the problem", that is frequently unrealistic. Rather the goal is to get a good basis for making testable predictions (which will either tend to ...


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The solution given by wdacda treats the equation as a Bernoulli equation. It can also be solved as an equation in separeted variables: $$ \frac{dm}{m^{0.75}\Bigl[1- \Bigl(\frac{m}{M}\Bigr)^{0.25}\Bigr]}=a\,dt. $$ Integrating between $0$ and $t$ $$ \int_{m_0}^{m}\frac{dz}{z^{0.75}\Bigl[1- \Bigl(\frac{z}{M}\Bigr)^{0.25}\Bigr]}=a\,t. $$ The integral is easily ...


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Hint Since you have $A\to B\to C$, I suppose that the kinetic model is $$\frac{dA}{dt}=-\alpha A$$ $$\frac{dB}{dt}=\alpha A-\beta B$$ $$\frac{dC}{dt}=\beta B$$ (remember that $B$ appears and disappears). Integrating the first equation is simple. Using the initial condition, you obtained $$A=M e^{-\alpha t}$$ which is perfectly correct. So, the second ...


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One possible way to model car traffic is to use min-plus or max-plus semirings. See for instance these slides: ´┐╝Fundamental Traffic Diagrams : A Maxplus Point of View, where you can find more references at the end. See also Bart De Schutter's Traffic and transportation networks publications, with 253 entries.


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This is a simple factoring. Omitting the factor $P$ at the end (which stays the same) $$ m*(F-c*P)-n = m*F -m*c*P -n = (m*F-n) - m*c*P = (m*F-n)*\left[1 - \dfrac{m*c*P}{m*F-n}\right] $$ What happened here is something like this: $$ x - a = x \left(1 - \dfrac{a}{x}\right) $$


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$$\frac{dP}{dt} = \left(m( F - cP) - n \right) P = \left(mF - mcP - n \right) P = \left( mF- n - mcP \right) P = \left((mF - n) \left( 1 - \frac{mcP}{mF - n} \right) \right) P = (mF - n)\left( 1 - \frac{mc}{mF - n} P \right) P$$


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We cannot find exact solution in a closed form, but I hope it will be useful. So, we have the system $$ \left\{\begin{aligned} \dot x &= \alpha x - \beta xy=x(\alpha - \beta y),\\ \dot y &= -\gamma y + \delta xy = y(-\gamma + \delta x) \end{aligned}\right. $$ From first equation $$ \frac{\dot x}{x} = \alpha - \beta y\Longrightarrow \beta y = \alpha ...


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This definitely sounds like a variation on Kirkman's schoolgirl problem; the original problem had $N = 15$, $n_{1}=\dots=n_{5}=3$, and it was desired that, for seven days in succession, no two girls were "seated at the same table" twice. For general $N = 3k+3$ (all table sizes $=3$, successive $(1/2)(N-1)$ days), solutions are given by Steiner triple ...


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Linear algebra Functional analysis Convex analysis / optimization Statistics Morphological analysis etc., etc. You can take a tour on Gabriel Peyre's Numerical tours, and see how these branches are beautifully inter-woven.


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Google these terms: Wavelets Curvelets Mumford-Shah Deblurring Deconvolution


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if you want to work in digital framework, i think topics like compressed sensing is state-of-art which can be supported by optimization problems.


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\begin{align*} \frac{\mathrm{d}P}{\mathrm{d}t} &= (m*(F-c*P)-n)*P \\ &= (m*F - m*c*P - n)*P \\ &= ((m*F - n) - m*c*P)*P \\ &= (m*F - n) * \left[\left(1 - \frac{m*c}{m*F - n}\right)*P\right]*P. \end{align*} The expression was most likely rewritten to account for a carrying capacity. May I ask what book you're reading?


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There are some physical limits which are often hidden in the usual presentations. The "physically reasonable" boundary condition is the heat flux boundary condition: there is a thermal conductivity coefficient $k$ for the boundary, and the system is in thermal contact with a temperature $u_{ext}$ at each point of the boundary. Then the boundary condition ...


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Separation of variables is fine. Note $$\begin{align} \dfrac{dy}{dx} = \dfrac{y( \gamma x-\delta)}{x(\alpha - cy)} &\implies \displaystyle\int^y\left(\dfrac{\alpha}{y'} - c \right)dy' = \displaystyle\int^x \left(\gamma -\dfrac{\delta}{x'}\right)dx' \\ &\implies \alpha\log y-cy=\gamma x-\delta \log x+\log K \\ &\implies \exp[\log x^\delta ...


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The solution of the linear initial value problem $$ 4 y'(t) = 1 - y(t), \quad y(0) = y_0, $$ is $$ y(t) = 1+(y_0 - 1)e^{-t/4}. $$ Substituting $y(t) = x(t)^{1/4}$ we get that the solution of the initial value problem $$ x'(t) = x(t)^{3/4} \bigl(1 - x(t)^{1/4}\bigr), \quad x(0) = x_0 \gt 0 $$ is $$ x(t) = \bigl(1+(x_0^{1/4} - 1)e^{-t/4}\bigr)^4. $$ ...


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First of all, I really like the question, I asked myself a lot of times how to explain and translate mathematical ideas into some "real world" applications. I am also currently working in an interdisciplinary field and the first thing one has to realize is be especially aware of the slang you are usually using and drop it Mathematical language has ...


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I don't know how much detail you need, but if you only need to draw a very rough phase-space diagram you can do this using only some very basic physical reasoning. If $a=0$ then the friction force vanishes and consequently energy (kinetic plus potential energy of the spring) is conserved. The equation system in this case can be written $$\frac{dE}{dt} = ...



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