# Tag Info

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By means of a translation, we can consider $x_0=0$. Thus, we can write the systems as: $$\dot x= A_1 x+g_1(x), \; \dot x= A_2 x+g_2(x),$$ where the spectrum of $A_1$ and the spectrum of $A_2$ have all negative real part. (For asymptotic stability). The functions $g_i(x)$ are the higher order terms, $g_i(0)=0$. Furthermore, we ...

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You know from the eigenvalues of the Jacobian that one eigenvalue is $0$ and the other is $-1$. That is, you have a one dimensional centre manifold and a one dimensional stable manifold. Since both manifolds intersect transversally, every nearby solution follow such invariant manifolds. Therefore, locally, every solution away from a centre manifold follows ...

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The system is $y'=Ay+u$, where $A= \begin{bmatrix} 2 & 1 \\ -5 & 0 \end{bmatrix}$, and $u = \begin{bmatrix} 0 \\ 5 \end{bmatrix}$. The critical points are where $y'=0$, this gives $y^* = -A^{-1} u = \begin{bmatrix} 1 \\ -2 \end{bmatrix}$. The linearised system can be written as $(\upsilon+y^*)' = \upsilon' = A(\upsilon+y^*) +u = A \upsilon$. Since ...

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This is Gronwall Inequality, but let's use Euler's method: Maybe we can discretize $$u'(t) \approx \frac{u(t+\Delta t) - u(t) }{\Delta t} \leq f(t) u(t)$$ then get: $$u(t+\Delta t) \leq u(t) \left( 1 + f(t) \Delta t \right)$$ By induction and using $1 + x \approx e^x$ for $x << 1$: $$u(t+ n\Delta t) \leq u(t) \prod_{k=1}^n \left( 1 + f(t ... 2 Set g(t)=u(t)e^{-\int_a^tf(s)ds} g'(t)=u'(t)e^{-\int_a^tf(s)ds}-u(t)f(t)e^{-\int_a^tf(s)ds}\leq 0, hence g is a decreasing function then g(t)\leq g(a) . 2 This is the well known Gronwall's inequality . A very rigorous proof can be found from Tao's book, nonlinear dispersive equations, Chapter 1. You should be able to derive it yourself as well. Note a naive separation of variables would not work, as f's sign is undetermined. 0 Car 1 just has to make sure that its speed at any point is the same as Car 2's speed at the same point. If you like, Car 1 executes a video replay of Car 2, delayed by d_0/v_0 seconds. 0 Hint: Let h(x) be the difference of both sides. What are h'(x) and h(\pi/2)? 1 Yeah, that is the Two-Body problem. I know that there are closed solutions up to \mathbf{q}\in \mathbb{R}^3. A complete and detailed solution to this problem can be found in the chapter 2 of H.D. Curts, Orbital Mechanics for Engineer Student. There the complete solution is given in polar form. The principal step for the solution of that problem was found ... 3 It is called the Kepler Problem. You can find a detailed analysis in its Wikipedia page. 3 The solution of the differential equation includes the step \ln |y|=2\ln |x| +C which does not include zero in the domain. Any steps after that would have to provide for that exception as well. 1 This is a very quick thing, I have not done it properly. Solved it backwards numerically with T=1, k=1/2 in excel (!!) with a step size of 0.0001, with a simple Euler recursion. Will be accurate enough to the left of x=1/2, but then again that doesn't show much... That is more or less what it will look like though. 0 You can always solve your problem for t \in I_i (the three different intervals in which your domain is divided) and then couple the solution guessing what the values of the constants of integration should be. Indeed, we have:$$ y(t) = \left\{ \begin{array}{ll} y_0 e^{-t} & 0 \leq t < 3 \\ c_1 e^{-t} + t^2 - 2t+2 & 3 \leq t < 4 \\ c_2 e^{-t} ...

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No, that will not hold in general. For example, suppose $p(s) = 1$ is the constant function. Then the left hand side is $$\int_0^L \left(\int_s^L dt\right) ds = \int_0^L (L-s) ds = \left.(Ls - \frac{1}{2}s^2)\right\vert_{s=0}^{s=L} = \frac{1}{2}L^2$$ but the right hand side is $$\int_0^L ds = L.$$

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Let $x(t) = y(t^2)$. Then, $x'(t) = 2ty'(t^2)$ and $x''(t) = 2y'(t^2)+4t^2y''(t^2)$. $tx''(t)-x'(t)-4t^3x(t) = 0$ $t[2y'(t^2)+4t^2y''(t^2)]-[2ty'(t^2)]-4t^3[y(t^2)] = 0$ $4t^3[y''(t^2)-y(t^2)] = 0$ $y'' - y = 0$ Can you solve this ODE?

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I don't see how you would solve the first equation in its own; the system is coupled. Picard iteration for first-order systems works the same as for first-order equations; you just integrate a vector-valued function. In your case, writing $z = (x,y)$, we have $$z(t) = (0,1)+\int_0^t f(z(s))\,ds \qquad \text{where } \ f(x,y) = (y,-\sin x)$$ So, starting ...

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The roof is leaking water with a constant speed of 1 litre per hour. The leak is: $1$ a constant rate of water added to the level A bucket is placed under the hole to catch the water. The water is evaporating with a speed proportional to the amount of water in the bucket. The evaporation is: $-c\; y(t)$ for some constant of proportionality. So: ...

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Let $y(t)$ be the amount of water in the bucket in liters, and let $t$ be the time in hours. The rate (liters/hour) at which water in the bucket is increasing due to the leak is a constant $1$ liters/hour. The rate at which the water volume is decreasing is proportional to the liters $y(t)$, with a proportionality constant of 0.2. So $$... 1 These equations are called equations which demand an integrating multiplier. They are solved by searching integrating multiplier and transforming to total DE (exact, as you call). There is no general form of these equations, but, if you want, you can come up with any cumbersome nonlinear function F(x,y) and a multiplier (this is only one of many possible) ... 1 The center subspace is the line y=0. The stable subspace is the line x=0. For every point (x_0,y_0), the solution passing through (x_0,y_0) is such that, for every t in the interval of definition around 0,$$x(t)=\frac{x_0}{1-x_0t},\qquad y(t)=y_0\mathrm e^{-t}.$$Right half-plane: If y_0\ne0, x_0\gt0, then x(t)\to+\infty and y(t)\to ... 1  \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} }  Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE:$$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$Assume F \in \mathcal{C}^1_{x_i} and compute the partial derivative of eq. (1) with ... 0 I do not think the reverse implication is true. As a counterexample, take p(x)=1, L=3/2, \alpha_1=\beta_2=2, \alpha_2=\beta_1=1. Then the problem at \lambda=0 becomes$$ v''=0,\;\;2v(0)-v'(0)=0,\;\;v(3/2)-2v'(3/2)=0. $$The equation v''=0 gives v=ax+b. Then the boundary conditions imply a=2b. Thus, we have the eigenfunction$$ v=2x+1, $$for ... 0 We find singular solution of single parameter straight line family of Clairaut's type. The powerful C discriminant (and also P discriminant ) are so useful, as also mentioned by achilles hui. The required singular solution or envelope is the eliminant between F(x,y,u) = 0 and its partial derivative with respect to parameter u.  x/u + y/v = 1, u v = 4 ... 1 Not a formal proof by any means, but I set up the system and ran it using Runge-Kutta , step size , .001 . The drawing shows t from t = 0 ... 40\pi , with initial position (0,1) . A second run using a variety of initial positions shows that the ellipse is a limit cycle. Points interior to the ellipse spiral out to the ellipse, points exterior to it ... 1 Because of the special structure of your equations, you could eliminate i1 to deal with 2 unknowns instead of 3. However, here I would pursue the general case. Notice that the equations could be re-written as$$\begin{array}{l}{i_1}' = {i_2}' + {i_3}'\\{i_2}' = 50\sin t - 6{i_{1,{\rm{H}}}} - 5{i_{2,{\rm{H}}}}\\{i_3}' = 50\sin t - ...

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$$F(x)=\left ( \int_0^x t e^t dt \right )^6$$ $$F'(x)=6 \left (\int_0^x te^t dt \right )^5 \cdot xe^x$$

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Let $f(x) = \int_0^x te^tdt$, so that $F(x) = [f(x)]^6$. According to the chain rule, $F'(x) = 6[f(x)]^5 \times f'(x)$, so you just have to work out what $f'(x)$ is, and through direct application of the Fundamental Theorem of Calculus we can say that $f'(x)=xe^x$. Alternatively, the integral does succumb to integration by parts, so there is another way to ...

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Hint: Just like in the comment. Write $F(x)$ as $$F(x) = g(x)^6$$ where $$g(x)= \int_{0}^{x}t^{t}dt$$ and then differentiate and you will need the fundamental theorem of calculus.

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Eric: You are correct that one needs $n\ge 0$ for the proof to work. Indeed, as this plot suggests, there are no zeroes on $x<0$.

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Let $z=\frac yx$. Then we have $$\dfrac{d^nz}{dx^n}=\dfrac{d^n(yx^{-1})}{dx^n}=\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}\dfrac{d^{n-k}(x^{-1})}{dx^{n-k}}=$$ $$\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}(-1)^{n-k}(n-k)!x^{-1-n+k}=$$ $$\sum_{k=0}^n(-1)^{n-k}\dfrac{n!}{k!}x^{-1-n+k}\dfrac{d^ky}{dx^k}$$ So we have ...

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They are the same up to arbitrary integration constant. HINT: $$\log a - c = \log a - \log e^c = \log \frac{a}{e^c}.$$

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To briefly outline one way of solving this (by induction), assume that $y = x^i$ are solutions for $i=1,...,n$ (where $n$ is the order of the particular ODE), and then note that in the $(n+1)$th case the above solutions are already covered as the $y^{(n+1)}$ terms disappear, and $y=x^{n+1}$ is a solution as after substituting this in, you're left with what ...

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Use induction. Suppose the property holds for $$A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n$$ Prove that it works for $n+1$ terms by writing the expression, use the fact that $\dfrac{\mathrm d^{n+1}y}{\mathrm dx^{n+1}}f = \dfrac{\mathrm d^{n}y}{\mathrm dx^{n}}\left(\dfrac{\mathrm dy}{\mathrm dx}f\right)$ and then show that your new equation is equivalent ...

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As explained in the comments, differentiating exponentials of matrices such as $\bar P(s,t)$ is slightly more complicated than in the scalar case since, quite generally, $$\left.\frac{\mathrm d}{\mathrm dt}\mathrm e^{A+tB}\right|_{t=0}=\sum_{i\geqslant0}\sum_{k\geqslant0}\frac{A^iBA^k}{(i+k+1)!},$$ which is neither $B\mathrm e^A$ not $\mathrm e^AB$ in ...

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Your differential equation is a complicated one. I give some hints on $I=]0,+\infty[$ (the study has to be done also on $]-\infty,0[$). A) First note that on $I$, your equation is $\displaystyle (x^{\prime}(t)+1)^2=\frac{x(t)+t}{t}$. We see that $x_0(t)=-t$ is a solution. For any solution, we must have $x(t)+t\geq 0$. Now suppose that on $I$, we have ...

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If you write equation in this form it is more clear: $2x(1+y)dy+(y+y^{2})dx=0$ $\frac{(1+y)}{y(1+y)}dy+\frac{1}{2x}dx=0$ $lny+\frac{1}{2}lnx=C$ $y=\frac{A}{\sqrt{x}}$ $y=-1$ could be a particular solution.

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EDIT: $x=t(x')^2+2tx'$ $p=x'$ $x=tp^2+2tp$ We differentiate in respect to $t$: $p=p^2+t2pp'+2p+2tp' \Rightarrow p'(2tp+2t)=(p-p^2-2p) \Rightarrow p'(p+1)2t=-(p^2+p) \Rightarrow p'(p+1)2t=-p(p+1) \Rightarrow p'(p+1)2t+p(p+1)=0 \Rightarrow (p+1)(2tp'+p)=0 \\ \Rightarrow p+1=0 \text{ or } 2tp'+p=0 \\ \Rightarrow p=-1 \text{ or } p'=-\frac{1}{2t}p \\ ... 0 Interestingly, For anyone else who stumbles upon this question, the answer is rather simply. Take a look at, The Cooperative Principle of Communication The Maxim of QUALITY states one should not mention what is false. Which is the answer to the question. 2 $$\frac{y'}{y+y^4}=x^4\tag{1}$$ and since $$\int\frac{dz}{z+z^4}=\frac{1}{3}\log\frac{z^3}{1+z^3}\tag{2}$$ by partial fraction decomposition, by integrating both terms in$(1)we get: $$\log\left(1+\frac{1}{y^3}\right)=-\frac{3}{5}x^5+C,\tag{3}$$ hence: $$1+\frac{1}{y^3}= K e^{-\frac{3}{5}x^5},$$ $$y= \frac{1}{\sqrt[3]{K ... 1 Solve the separable equation:$$y'=x^4y^4+x^4y$$Siimplify:$$y'=x^4(y^4+y)$$divide by y+y^4$$\frac{\frac{dy}{dx}}{y+y^4}dx=x^4dx$$Integrate both sides wrt x:$$\int\frac{\frac{dy}{dx}}{y+y^4}dx=\int x^4dx$$Evaluate the integral:$$-\frac13\ln(y^3+1)+\ln y=\frac{x^5}5+\text{constant}Solve for y:[Spoiler:Answer] 6 You may want to take all the y terms on one side as follows: \begin{align*} \frac{1}{y(1+y^3)}\frac{dy}{dx} & = x^4\\ \int \frac{1}{y(1+y^3)} \, dy & = \int x^4 \, dx\\ \end{align*} The integral on the left can be dealt with partial fractions. Hopefully you can take it from here. 2 Here's an answer from group theory. Use a Lie groupG(x,y)=(\lambda X,\lambda^\beta Y)\lambda_o=1$$Since x=\lambda X, y=\lambda^\beta Y, and y'=\frac{dy}{dx}=\frac{\lambda^\beta Y}{\lambda X}=\lambda^{\beta -1}\frac{dY}{dX}=\lambda^{\beta -1}Y', we substitute these into the original ODE and we get$$ \lambda^{2\beta} Y^2+2\lambda^\beta Y ... 1 You solve the homogeneous problem: $$y''+4y=0$$ You will find:y_h(x)=c_1 \cos(2x)+ c_2 \sin(2x)$So,for the non-homogeneous problem: $$y_n(x)=Ax+B$$ Replace at$y''+4y=8x$and find the constants$A$and$B$. The general solution is$y(x)=y_h(x)+y_n(x)$1 Follow the steps: 1) find the solutions of the homogeneous ode$y_1,y_2$$$y''+4y=0$$ 2) find a particular solution$y_p$by assuming $$y_p=Ax+B$$ then plug in back in the ode to find the constants$A$and$B$. See table 3) construct the general solution $$y = c_1y_1 + c_2y_2 + y_p$$ 0 Write the equation as $$\frac{u'}{u^k}-\frac{u'}{u^{k+2}}=1.$$ Integrate to get $$\frac{1}{(1-k)u^{k-1}}+\frac{1}{(k+1)u^{k+1}}=x-x_0,$$ where$x_0\$ is a free constant. This relation can be rewritten as $$u^{k+1}(x_0-x)(1-k^2)+u^2(k+1)+1-k=0.$$

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Hint; $$y(y + 1) + 2xy'(y + 1) = 0 \iff (y + 1)( y + 2xy' ) = 0$$ Another Hint: Linear Equations.

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