# Tag Info

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

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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: 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 $$P_k\ge \frac{k}4+P_0^{-1}\... 1 Compare$$(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 $$\mu(x)=\exp\left(\int f(x)\mathrm{d}x\... 1 Let us compute f'(u) for u > 0:$$ 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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