New answers tagged

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Why don't simply use the definition of the convolution of two distributions, $$(T\star S) (\varphi) = T_x (S_y (\varphi(x+y)))$$ Hence, take $S = \delta_1-\delta_2$ and you got $S_y(\varphi(x+y))=\varphi(x+1)-\varphi(x+2)$. Hence \begin{align*} (te^{2t} \star (\delta_1-\delta_2))(\varphi)& = ...


1

Substitute $y=z+x$ and the equation becomes $z'+1=1+xz-x^3z^2$ or $z'=xz-x^3z^2$. Substitute $z=e^{\frac{x^2}{2}}w$ and this becomes $e^{\frac{x^2}{2}}w'=-x^3e^{x^2}w^2$ or $-\frac{w'}{w^2}=x^3e^{\frac{x^2}{2}}$. We can now integrate to get $\frac{1}{w}=x^2e^{\frac{x^2}{2}}-2e^{\frac{x^2}{2}}+C$, so ...


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\begin{eqnarray} q(t) &=& y_{n-1} {(t-t_{n})(t-t_{n+1}) \over (t_{n-1}-t_{n})(t_{n-1}-t_{n+1}) } + y_{n} {(t-t_{n-1})(t-t_{n+1}) \over (t_{n}-t_{n-1})(t_{n}-t_{n+1}) } + y_{n+1} {(t-t_{n-1})(t-t_{n}) \over (t_{n+1}-t_{n-1})(t_{n+1}-t_{n}) }\\ &=&{1 \over 2 h^2} (y_{n-1} (t-t_{n})(t-t_{n+1}) - 2y_{n} (t-t_{n-1})(t-t_{n+1}) + y_{n+1} ...


1

We have, for $x > 0,$ $$ \varphi(2x) - \varphi(x) \rightarrow 0. $$ The Mean Value Theorem says $$ \frac{ \varphi(2x) - \varphi(x)}{2x - x} = \varphi'(\xi) $$ with $$ x < \xi < 2x. $$ So $$ \varphi(2x) - \varphi(x) = x \varphi'(\xi) . $$ Since $ \xi < 2x$ and negative second derivative, $ \varphi'(\xi) > \varphi'(2x), $ and $$ ...


2

$$ \varphi(x) = 1 - \frac{1}{(1+x)^n} $$ Your limit is $n,$ bigger than the $1$ you allow. Next I will try for $\infty$ This works, $$ \varphi(x) = 1 - e^{-x} $$


1

Your equation is this: $$\frac{x-2}{x^2-x-2}$$ Factoring the denominator: $$=\frac{x-2}{(x-2)(x+1)}$$ Cancelling like factors: $$=\frac{1}{x+1}$$ The equation should not become $1/x^2$ when you cancel the factors for the removable discontinuity.


1

The canonical answer (and in fact only answer) that would be expected in a good differential equations course is the following picture: You can add a few more arrows in each reagion but that's it. Really one would never expected to "plot" a $2$-dimensional vector field on $\mathbb R^4$, right? So we should also not really plot a $1$-dimensional vector ...


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A vector field is a vector valued function. The second image shows a vector field $f : \mathbb{R}^2 \to \mathbb{R^2}$, where $v = (v_x, v_y) = f(t, y)$. The first image shows a scalar valued function $f : \mathbb{R} \to \mathbb{R}$, where $v = f(y)$. If you consider the vectors of one dimensional vector spaces as vectors, which in the view of algebra ...


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They are two different visual representations of the same thing. The vector field corresponding to a differential equation $y'= f(t,y)$ assigns to each point $(t,y)$ of the plane a vector $[1, f(t,y)]$. The second picture is more direct: it shows a region of the $t-y$ plane. The little lines represent the directions of these vectors at a sample of points ...


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Hint: I would prefer the following differential equation: $$\frac{dI(t)}{dt}=k\cdot (1,000,000-I(t))-600$$ The more people have been infected the less new infections can happen.


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I would suggest you use an Ansatz of the form \begin{equation} u(x,y) = A x^2 + 2 B x y + C y^2 + D x + E y + F, \end{equation} and try to determine possible values for $A$ to $F$ using the initial condition and the PDE. The method of characteristics would give you the same result, but you'll have to use the fully nonlinear version.


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Hint: Since you have a final condition $v(T,x)=x^2$ instead of an initial condition, and you would like to solve the problem backwards, for the time interval $t \in (0,T)$, it is very useful to reverse time, that is, to introduce $\tau = T- t$, such that the equation becomes \begin{equation} v_{\tau} = \frac{1}{2} v_{xx},\qquad v(\tau=0,x) = x^2 ...


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$$f'(t) = g'(t)\left(f(t)\left(\frac{a}{g(t)} -1 \right)-\frac{a}{g(t) \lambda}\left( 1+ W_{-1}(-e^{-1-\lambda f(t)} \right) \right) $$ Let : $W_{-1}( -e^{-1-\lambda f(t) })=-y(t) \quad\to\quad -ye^{-y}=-e^{-1-\lambda f(t)}\quad\to\quad f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right)$ $f'(t)=\frac{1}{\lambda}\left( 1-\frac{1}{y}\right) y'(t)$ ...


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The first moment doesn't capture anything about the "spread" of the distribution, so you can always take some of the probability above the mean and move it a bit further away and same with the mass below the mean. Or move a bit of the mass from below the mean to above the mean or vice versa to get the mean. So just knowing the mean doesn't tell you much for ...


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Hint: Use variation of constants $y=c(x)\cdot y_h$ where $y_h$ is your homogenous solution and $c(x)$ is the function you need to find.


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Note that $$\begin{align} G(\vec r_1,\vec r_2)-G(\vec r_2,\vec r_1)&=\int_V \left(G(\vec r,\vec r_2)\nabla \cdot(p(\vec r)\nabla G(\vec r,\vec r_1))-G(\vec r,\vec r_1)\nabla \cdot(p(\vec r)\nabla G(\vec r,\vec r_2))\right)\,dV\\\\ &=\oint_S p(\vec r)\left(G(\vec r,\vec r_2)\nabla G(\vec r,\vec r_1)-G(\vec r,\vec r_1)\nabla G(\vec r,\vec ...


1

There is no simple answer to the first question. Finding the general solution in a closed form of a differential equation cannot always be done (consider $y'=\sin(t^2)$). Rather there are usually steps to finding general solutions of certain classes of differential equations. This differential equation is a linear differential equation of the form ...


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Subtract $t^2y(t)$ from both sides : $\frac{dy(t)}{dt} - t^2y(t) = e^t $ Let $ m(t) = e^{\int -t^2y(t)dt} = e^{-\frac{t^3}{3}} $ Then multiply both sides by $m(t)$ and substitute : $ -e^{-\frac{t^3}{3}}t^2 = \frac{d}{dt}(e^{-\frac{t^3}{3}}) $ Apply the reverse product rule to the left hand side, then integrate and you will get : $ \int ...


1

There is a constant solution: $y(t) = 1$, $x(t) = 0$. Moreover, this is a stable equilibrium if $A D - B C > 0$. In general you can reduce this system to a single first-order equation for $y$ as a function of $x$. That differential equation is, according to Maple, an Abel's equation of the second kind, class B, and does not seem to have a ...


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If you perform the substitution z=t+x, you get $\dot x=\dot z -1$ ,so, after a few computations, you can find that $\dot z= \frac{2t}{z}$, so the new equation is with separable variables and, since $z_0=0+1=1> 0$, you have $z^2=z_0^2+2t^2$,$z_0=1$ => $z=\sqrt{1+2t^2}$ => $x(t)=z(t)-t=\sqrt{1+2t^2}-t$ $\forall t\in \Bbb R$ ...


1

the general method is to change variable $x = vt$ as suggested by quinn. but for this problem, you can try this: $$0=(t+x)\, dx - (t-x)\, dt = t \, dx + x \ dt + x \, dx - t \ dt = d\left(tx+\frac12 (x^2-t^2)\right) $$ that is $$tx+\frac12 (x^2-t^2) = constant=0+\frac12.$$


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Hint...the method is fairly standard: substitute $x=vt$ You get a separable variable differential equation in $v$ and $t$


0

I can't find the exact solution but I will show the inequality: $1<f(x)<2,\forall x\in (1,+\infty).$ Since $f$ is strictly increasing in the domain of interest, we have $f(x) > 1,\; \;\forall x > 1, $ which justifies the lower bound of $f$. Consequently, $f'(x) = 1/(x^2+f^2(x)) < 1/(x^2+1),\quad\forall x > 1.$ Now we set $g(x) = \arctan x ...


0

Changing variables $y = x+t$ seems to reduce this down to a very manageable $$ y \frac{dy}{dt} - 2y = -2t $$ Can you show this is indeed the case and solve the simpler ODE?


1

What the argument shows is that a differentiable function $y(x)$ can satisfy both equations only at a severely restricted set $S$ of $x$ values: with the possible exception of $x=1/2$, every point in $S$ is isolated. In particular there is no open interval of $x$'s where both equations can hold. Conversely, for any $S \subset \mathbb{R}$ whose set of ...


1

Given that the population has a logistic form, at long times (large t), the population reaches a steady state. Thus $dP/dt = 0$ as t approaches infinity. Solving for P yields $\pm 112$


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Hint: This is a autonomous differential equation. With $p=0$ and $p=\pm \sqrt{224\cdot 56}=\pm112$ as trivial constant solutions. Draw an approximate plot (look up how to draw solutions for autonomous differential equations) of the solutions and see what happens for different initial values for $t \to \infty$.


1

First note that $y=0$ and $y=\beta$ are trivial solutions. The given equation is a Riccati type differential eqaution. If you have a solution $y_0$ to the equation you can construct the general solution using this Ansatz: $$y=y_0+\frac{1}{u}=\beta+\frac{1}{u}$$ Plugging this into the equation will result in $$u'=(2\beta-\alpha)u+\frac{\alpha}{\beta}$$ ...


2

I can't seem to get the Mathematica solution. Just chipping away at it, $$2y\frac{dy}{dx}=\frac{(y^2-1)}{x^3}(x^2y^2+y^2-1)$$ Let $s=y^2$. Then $$\frac{ds}{dx}=\frac{(s-1)}{x^3}(x^2s+s-1)$$ Let $s-1=t$. Then $$\frac{dt}{dx}=\frac t{x^3}(x^2(t+1)+t)$$ Let $t=\frac1u$. Then ...


1

$\frac{dh}{dt}=-\frac{c}{h^\frac{3}{2}} => h^\frac{3}{2}dh = -c dt => \frac{2}{5}h^\frac{5}{2} = -ct + const => h(t)=H-\frac{5}{2}(ct)^\frac{2}{5}$. If you know that for $t=1$ (hr) $h(1) = \frac{1}{2} H$, you can find c, and, setting $H=\frac{5}{2}(ct)^\frac{2}{5}$, find $t$. Hope this helps.


1

Assume that $f$ is continuous and that $\phi$, $g$ are differentiable. Suppose $$H(x) = \int_0^x f(s) \, ds.$$ The fundamental theorem of calculus tells you that $H'(x) = f(x)$. Suppose that $$M(x) = \int_0^{\phi(x)} f(s) \, ds.$$ Then $M(x) = H(\phi(x))$ so that the chain rule tells you $$M'(x) = H'(\phi(x)) \phi'(x) = f(\phi(x)) \phi'(x).$$ The ...


0

Subtract $ \frac {y(x)}{1-x^2} $ from both sides : $ \frac {dy(x)}{dx} + \frac {y(x)}{x^2-1} = - \frac {-x^3 + x}{x^2-1}$ Let $ m(x) = e^{\int \frac {1}{x^2-1} dx } = \frac {\sqrt{1-x}}{\sqrt{x+1}} $ Then multiplay both sides by m(x) and substitute : $ \frac {\sqrt{1-x}}{\sqrt{x+1}(x^2-1)} = \frac {d}{dx} \frac {\sqrt{-x+1}}{\sqrt{x+1}} $. Apply the ...


1

Mathematica gives the solution as $$ y(x) = - \sqrt {\frac{ 2 c x -1 } { 2 c x + x^2 -1 } } $$ If you use the sub $z = y^2$,you'll find the equation transform to $$ x^3 d z + ( 1-z) ( x^2 z + z -1) dx =0 $$ Then it looks like you should be able to find an integrating factor in $z$.


2

Substitute $y'=z$. Let us look at the general case where the power of the first derivative is $p$. $$Az'+Bz^p+C=0$$ Assuming $A\neq0$: $$z'=-\frac{B}{A}z^p-\frac{C}{A}$$ This is a Bernoulli type differential equation. You can try to solve it using the Ansatz: $$z=w^{\frac{1}{1-p}}$$


1

Hint:First take $tanx$ common and then try Linear differential equation of first order $\frac{dy}{dx} +P(x)y = Q(x)$ So here $P(x)=2cotx$ and $Q(x)=x\frac{cosx}{sin^2x}$ now just use the standard solution of differential equation $ye^{\int P(x)dx}$ = $\int Q(x) e^{\int P(x)dx}$ $dx$ So you get something like this:$e^{\int P(x)dx}=sin^2x$ $ysin^2x$ = ...


1

These type of problems arise in adaptive control where the controller uses estimations (the $\alpha_i$ in your case) of the unknown system parameters ($a_i$ in the example). The convergence analysis follows from the so called Barbalat lemma which states the following: If $f$ is uniformly continuous and the integral $\int_0^{\infty}{f(s)ds}$ exists and is ...


0

Hint: $y(x) = u(x)v(x)$ then $y'(x) = u'v + v'u $


1

Since $P$ and $Q$ are both positive definite, for $V(x) = x^TPx$ we have $$\lambda_{min}(Q)\Vert x\Vert^2\le x^TQx\le \lambda_{max}(Q)\Vert x\Vert^2\tag{1}$$ $$\lambda_{min}(P)\Vert x\Vert^2\le V(x)\le \lambda_{max}(P)\Vert x\Vert^2\tag{2}$$ From $(1)$: $$-x^TQx\le-\lambda_{min}(Q)\Vert x\Vert^2$$ and from $(2)$: $$\frac1{\lambda_{max}(P)}V(x)\le \Vert ...


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The equivalence is really between these systems of equations $$ \left\vert \begin{matrix} y' + y = 3x \\ y' - y = x \end{matrix} \right\vert \iff \left\vert \begin{matrix} y' + y = 3x \\ y = x \end{matrix} \right\vert \iff \left\vert \begin{matrix} y' - y = x \\ y = x \end{matrix} \right\vert $$ so your error is that you dropped one of the original ...


1

Given ODE is $y'-f(t)y=0$ So it is linear ODE with initial condition so by uniqueness theorem has unique solution on $\mathbb{R}$.


2

I might be mistaken, but if your function $y$ is a function of $\mathbb{R}$ to $\mathbb{R}$, the solutoin of the equation is $$y(x)=\exp^{\int^{x}_{0}1+f^2(t)dt}$$ and this is greater than $\exp(x)$. You can use comparison as well, saying that $$z_1'\leqslant (1+C)z_1$$ where $C\geqslant f^2$ and $$z_2'\geqslant z_2$$ and compare $y$ with the two solutions ...


3

Dividing both sides by y(x) and integrating gives you : $\int \frac{\frac{dy(x)}{dx}}{y(x)} dx\ = \int (f^2(x) + 1)dx $ Solving that gets you : $ ln(y(x)) = \int (f^2(x) + 1) dx + c_1 $ Finally, the solution of your ODE is : $ y(x) = c_1e^{\int(f^2(x) +1)dx} $ You then, can work out the solution to your question from this.


1

Solutions are indeed unique, but for an arbitrary $f$ the interval may not be the whole $\mathbb R$. From this observation, it follows that the only possibilities would be $2$ and $4$. In order to find which is the right one, you need to find explicitly some $f$ for which the solution is not defined in $\mathbb R$, or to show that no such $f$ exists.


0

I don't know why I didn't see this earlier, forgot $\alpha b$ was a zero of $J_n$: $$\int_0^bxJ_n^2(\alpha x)dx=\frac{b^2}{2}\left(\frac{n}{\alpha b}J_n(\alpha b)-J_{n+1}(\alpha b)\right)^2+(\alpha^2b^2-n^2)(J_n(\alpha b))$$ $$=\frac{b^2}{2}([0]-J_{n+1}(\alpha b))^2+\frac{(\alpha^2b^2-n^2)[0]}{2\alpha^2}$$ $$=\frac{b^2}{2}J_{n+1}^2(\alpha b)$$


1

What you are referring to is called the index of a contour or the winding number of a contour around some point $z_0$. Intuitively, it is the number of times a contour goes around this particular point. There are various ways of defining it all of which can be find on the associated Wikipedia page: https://en.m.wikipedia.org/wiki/Winding_number


1

A few more digits $$ 1.6798002778544903357 $$ (where the last digit shown has rounded up). The Inverse Symbolic Calculator doesn't recognize it, so you're probably out of luck.


1

Well, my take on this is that the ellipses might as well be defined by $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ With $a=1$ so whether it ends up being the semi-major or semi-minor axis, the axis parallel to the $x$-axis will be of length $2$. The free parameter of the family will be $b$, so we solve for $b$, $$\frac{y^2}{1-\frac{x^2}{a^2}}=b^2$$ Differentiate ...


2

Maybe this will help you to better understand what is going on: http://mathworld.wolfram.com/GeometricSeries.html The two results are actually equivalent for $|x|<1$ since the Picard iteration is converging to a geometric series, which is $\frac{1}{1-x}$ for $|x|<1$.


1

How to solve $\tan(y)$ to $y$? I can't solve this. Hint. You may use $$ \arctan (\tan (y))=y,\quad y \in \left(-\frac{\pi}2,\frac{\pi}2\right). $$ Edit. There is a mistake in your steps above, you rather have $$ \int \frac1{\tan (y)}\:dy=\int \frac{\cos (y)}{\sin (y)}\:dy=\ln \left|\:\sin (y)\:\right| $$ giving $$ \ln \left|\:\sin (y)\:\right|=\ln ...



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