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The trick here is to separate places in a function (i.e., which argument is which) and the names of variables occupying those places. Suppose, for instance, that we define $$ f(x, y) = x^3y $$ and then ask "What's the derivative of $f(y, x)$ with respect to $x$?" Using the chain rule, one might be tempted to say that it's $$ \frac{\partial f} {\...


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The calculus argument is that you have $\frac{dx}{dk}\approx -\frac{x^2}{4}$ suggesting $\int -\frac1{x^2}\, dx \approx \int \frac14\, dk$ i.e. $\frac{1}{x}\approx \frac{k}{4}+B$ for some constant $B$, i.e. $x\approx\frac{1}{\frac{k}{4}+B}$. But here you have a discrete recursion so may need some adjustment. When $0 < P_0=c \le 1$ you have the bounds $$\...


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$$\dfrac{dp}{dr}r+p=0$$ Or more generally $$\dfrac{dp}{dr}F(r)+G(p)=0$$ Is a linear first order homogeneous ODE. As suggested in the comments can be solved by separation of variables or integrating factor.


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\begin{gather*} \frac{dy}{dx} =\frac{y^{3}}{2x^{3}} +\frac{3y}{2x}\\ \\ Let\ y=tx\\ \frac{dy}{dx} =t+x\frac{dt}{dx}\\ t+x\frac{dt}{dx} =\frac{t^{3}}{2} +\frac{3t}{2}\\ x\frac{dt}{dx} =\frac{t^{3} +t}{2}\\ \\ \int \frac{2dt}{t^{3} +t} =\int \frac{dx}{x}\\ \\ Now,\ \int \frac{2dt}{t^{3} +t} =\int \frac{2dt}{t\left( t^{2} +1\right)} =\int \frac{2tdt}{t^{2}\left(...


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The line print( np.dot( D, yl ).round( decimals=9 ) ) calculates the integrals of the Fourier transform at set of frequencies using the rectangle integration method: \begin{equation} F(k) = \int\limits^{\infty}_{-\infty}e^{-ikx}\left(\partial_x + x\right)e^{-\frac{x^2}{2}}dx=\left.e^{-ikx}e^{-\frac{x^2}{2}}\right\vert^{\infty}_{-\infty} +\int^{\infty}_{-\...


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Such recursions that look like Bernoulli equations can often be solved using the "Bernoulli trick". That is, consider the corresponding negative power of $P$, $$ P_k^{-1}=P_{k-1}^{-1}\left(1-\tfrac14P_{k-1}\right)^{-1} =P_{k-1}^{-1}+\tfrac14+\tfrac1{16}P_{k-1}+\tfrac1{64}P_{k-1}^2+...\ge P_{k-1}^{-1}+\tfrac14 $$ so that $$ P_k\ge \frac{k}4+P_0^{-1}\...


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Compare $$(x^2+6y^2)dx-4xydy=0~~~~(1)$$ with $$Mdx+Ndy=0 \implies \frac{\partial M}{\partial y}=12y,~~ \frac{\partial N}{dx}=-4y $$ It's integrating factor is $$I=\exp \left(\int \frac{12y +4y}{-4xy}\right)=x^{-4}$$ Multiplying (1) with $x^{-4}$,(1) becomes an exact ODE: $$(x^{-2}+6x^{-4} y^2) dx- 4x^{-3} ydy==0$$ It's solution is $$\int (x^{-2}+6x^{-4}y^2) ...


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We can check if the ODE is exact: $$\Delta=\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}=\frac{\partial}{\partial y}(x^2+6y^2)-\frac{\partial}{\partial x}(-4xy)=16y\ne0.$$ So it's not exact, but as $$\frac{\Delta}{N}=\frac{16y}{-4xy}=-\frac{4}{x}=f(x),$$ we can get an integrating factor with the form $$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\...


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Let us compute $f'(u)$ for $u > 0$: $$ f'(u) = -e^{\alpha u}\log(u)-\alpha u e^{\alpha u}\log(u) - e^{\alpha u} = -e^{\alpha u}(\log(u)+\alpha u \log(u) +1) $$ This is continuous on $(0,+\infty)$, so it is locally bounded on $(0,+\infty)$. Hence, $f$ is locally Lipschitz on $(0,+\infty)$. However, $$ \lim_{u \to 0⁺} f'(u) = +\infty, $$ hence $f$ is not ...


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