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## New answers tagged differential-equations

0

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 ...

0

\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}$$The equation should not become 1/x^2 when you cancel the factors for the removable discontinuity. 1 The canonical answer (and in fact only answer) that would be expected in a good differential equations course is the following picture: You can add a few more arrows in each reagion but that's it. Really one would never expected to "plot" a 2-dimensional vector field on \mathbb R^4, right? So we should also not really plot a 1-dimensional vector ... 0 A vector field is a vector valued function. The second image shows a vector field f : \mathbb{R}^2 \to \mathbb{R^2}, where v = (v_x, v_y) = f(t, y). The first image shows a scalar valued function f : \mathbb{R} \to \mathbb{R}, where v = f(y). If you consider the vectors of one dimensional vector spaces as vectors, which in the view of algebra ... 0 They are two different visual representations of the same thing. The vector field corresponding to a differential equation y'= f(t,y) assigns to each point (t,y) of the plane a vector [1, f(t,y)]. The second picture is more direct: it shows a region of the t-y plane. The little lines represent the directions of these vectors at a sample of points ... 0 Hint: I would prefer the following differential equation:$$\frac{dI(t)}{dt}=k\cdot (1,000,000-I(t))-600$$The more people have been infected the less new infections can happen. 0 I would suggest you use an Ansatz of the form $$u(x,y) = A x^2 + 2 B x y + C y^2 + D x + E y + F,$$ and try to determine possible values for A to F using the initial condition and the PDE. The method of characteristics would give you the same result, but you'll have to use the fully nonlinear version. 1 Hint: Since you have a final condition v(T,x)=x^2 instead of an initial condition, and you would like to solve the problem backwards, for the time interval t \in (0,T), it is very useful to reverse time, that is, to introduce \tau = T- t, such that the equation becomes v_{\tau} = \frac{1}{2} v_{xx},\qquad v(\tau=0,x) = x^2 ... 0$$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) $$Let : W_{-1}( -e^{-1-\lambda f(t) })=-y(t) \quad\to\quad -ye^{-y}=-e^{-1-\lambda f(t)}\quad\to\quad f(t)=\frac{1}{\lambda}\left( y-\ln(y)-1\right) f'(t)=\frac{1}{\lambda}\left( 1-\frac{1}{y}\right) y'(t) ... 0 The first moment doesn't capture anything about the "spread" of the distribution, so you can always take some of the probability above the mean and move it a bit further away and same with the mass below the mean. Or move a bit of the mass from below the mean to above the mean or vice versa to get the mean. So just knowing the mean doesn't tell you much for ... 0 Hint: Use variation of constants y=c(x)\cdot y_h where y_h is your homogenous solution and c(x) is the function you need to find. 1 Note that$$\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 ...

1

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$. That differential equation is, according to Maple, an Abel's equation of the second kind, class B, and does not seem to have a ... 0 If you perform the substitution z=t+x, you get$\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 $$tx+\frac12 (x^2-t^2) = constant=0+\frac12.$$ 4 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 $(2)$: $$\frac1{\lambda_{max}(P)}V(x)\le \Vert ... 4 No problem using elimination, but your system is incompatible. 7 The equivalence is really between these systems of equations$$ \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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