# Tag Info

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If $x\ge 1$, then $|t-x|=x-t$ and $f(x) = \int_0^1t(x-t)dt = x/2-1/3$. If $x\le 0$, then $|t-x|=t-x$ and $f(x)=\int_0^1t(t-x)dt = 1/3-x/2$. If $x\in (0,1)$, then we split the integral in two: $$f(x)=\int_0^1t|t-x|dt =\int_0^xt|t-x|dt+\int_x^1t|t-x|dt$$ $$=\int_0^xt(x-t)dt+\int_x^1t(t-x)dt = x^3/2-x^3/3+(1-x^3)/3-x(1-x^2)/2$$$$=1/3-x/2+x^3/3.$$ Can you ...

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Hint If $x\in [0,1]$ then $$f(x)=\int_0^x(x-t)tdt+\int_x^1(t-x)tdt$$ If $x\ge1$ then $$f(x)=\int_0^1(x-t)tdt$$ If $x\le0$ then $$f(x)=\int_0^1(t-x)tdt$$ so calculate these integrals and you find that $f$ is a polynomial on every interval.

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First, you use $n$ in your post twice, in different meanings. First $n$ denotes the dimension of the phase space, and second, $n$ means the order of the terms you keep. I will use $d$ instead of $n$ in the former case. A lot is known about the case $d=2$, although not everything. Assume that we are given polynomial system on the plane $$\dot x=P(x,y),\\ ... 0 We can just factor it and solve two first order ordinary differential equations instead. Write$$y''+4y=y''+2iy'-2iy'+4y=(y'+2iy)'-2i(y'+2iy)=0.$$Substitute z=y'+2iy and get z'-2iz=0\implies z(t)=A e^{2it}. Substitute back:$$y'+2iy=Ae^{2it}\iff(e^{2it}y)'=Ae^{4it}$$We then get the desired solution$$y=Ae^{2it}+Be^{-2it}.$$Needless to say, the same ... 1 I am not sure about everything you wrote as it seems some information is missing, so you still need to work through some things. We are given the system:$$\tag 1 x' = (\epsilon x+2y)(z+1) \\ y' = (-x+\epsilon y)(z+1) \\ z' = -z^3$$For \epsilon = 0, (1) reduces to:$$\tag 2 x' = 2y(z+1) \\ y' = -x(z+1) \\ z' = -z^3$$As an aside, note that z is ... 0 You should review the rules for taking derivatives, specifically the chain rule. (Either that, or expand all squares in F, but this is more algebraically painful.) The basic idea is that the derivative of the square of some function g should be computed as (g^2)'=2 g g'. For example, the derivative of (3x-7y+5)^2 with respect to y is ... 2 You can write w\times\alpha as \Omega\alpha, where \Omega is an (antisymmetric) matrix. Then the problem reduces to a linear ODE. 2 Do you know how to solve general linear system? \dot x = A x? Convert w \times \alpha to  W \alpha where W is matrix associated with cross product. If you have a point rotating around a fixed axis, calculate its velocity. 1 In the first representation, we have a function$$ f:\Bbb R^{n+2}\to\Bbb R $$which is not the case for the second case. Such functions are far more easy to manipulate than general functionals such as the second equation suggests. 0 Let L = \lim_{x \to \infty} f(x)+\int_0^x f(t)dt. Let \phi(x) = e^x \int_0^x f(t)dt, note that \phi is differentiable, \phi(0) = 0 and \lim_{x \to \infty} e^{-x}\phi'(x) = L. Let \epsilon>0, then for some x_0 if x \ge x_0 we have |e^{-x}\phi'(x)-L| < \epsilon, and so |\phi'(x)-e^{x}L| < e^x \epsilon. Integrating over [x_0,x] ... 5 Let  L = \lim_{x\to\infty} \left( f(x) + \int_{0}^{x} f(t) \, dt \right)  denote the limit. We easily check that$$ f(x) + \int_{0}^{x} f(t) \, dt = \frac{\frac{d}{dx} \left( e^{x} \int_{0}^{x} f(t) \, dt \right) }{\frac{d}{dx} e^{x}}. $$Since e^{x} \to \infty as x \to \infty, it satisfies the condition of L'hospital's rule and hence$$ ...

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Hint: It is a linear equation $$\frac{1}{x'}=\frac{y}{(y+2)e^y - 2x} \\ x'=\frac{(y+2)e^y - 2x}{y}\\ x'+\frac{ 2}{y}x=(1+\frac{2 }{y})e^y$$ The factor is $$\mu=e^{\int\frac{2}{y}} =y^2$$ Then the solution is $$x=\frac{1}{y^2}[\int y^2(1+\frac{2 }{y})e^ydy+c]$$

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$x\equiv y \pmod{z}$ means that $z$ divides the difference $y-x$. If you know $x,y$, then $z$ can be any divisor of their difference.

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The equation $$x = y \bmod z$$ means nothing more nor less than $$x-y = kz$$ for some integer $k$. Your question originally specified $x=5, z=7$, so I will use those as examples. Plugging in, we get $$5-y = 7k$$ for some integer $k$; rearranging we have $$y = 5-7k$$ for some integer $k$. This is all the information that is available, so the solution for ...

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Hint Look at the quantity $x^2 + y^2$ as a function of time. Incidentally: a system itself is not "asymptotically stable". What is stable is a given fixed point of the system. In your case I am assuming you are talking about the fixed point $\vec{x}_0 = (0,0)$.

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$y"(x)+y(x)-y^{-3}=0$ is an autonomous ODE of second order. Let $y'(x)=f(y)$ then $y''(x)=f'(y)y'(x)=f'(y)f(y)$ $f'(y)f(y)+y-y^{-3}=0$ is a first order ODE $f^2=-y^2-y^{-2}+C$ $y'(x)=f(y)=\sqrt{-y^2-y^{-2}+C}$ $x(y)=\int{(-y^2-y^{-2}+C)^{-1/2}}dy$

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We are given: $$B = \begin{bmatrix}a-3 & 5 \\ -2 & a+3\end{bmatrix}$$ To find the characteristic polynomial and eigenvalues, we set up and solve: $$|B - \lambda~I| = 0 \implies \begin{vmatrix}a-3 -\lambda & 5 \\ -2 & a+3 -\lambda\end{vmatrix} = 0$$ This yields: $$(a-3 - \lambda)(a+3-\lambda) + 10 = 0 \implies \lambda^2 -2 a \lambda + a^2 ... 2 The matrix exponential is given by:$$\tag 1 e^{At} = \sum_{k=0}^{n-1} \alpha_k A^k$$where the \alpha_i's are determined from the set of equations given by the eigenvalues of A, as:$$\tag 2 e^{\lambda_i t} = \sum_{k=0}^{n-1} \alpha_k \lambda_i^k$$We are given:$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & ...

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Yes, it is possible, try: $$x'= x^{1/2}, x(0) = 1$$ Using separation of variables, we find: $$x(t) = \dfrac{1}{4}(t+c_1)^2$$ Solving for the IC, we have two solutions: $$x(t) = \dfrac{1}{4}(t + 2)^2, ~~x(t) = \dfrac{1}{4}(t -2)^2$$ One can come up with more examples too: $x' = x^{3/2}$ $x' = |x|$ $\ldots$

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The system of differential equations $$x' = 2x + y^3$$ and $$y' = -y$$ As answered by Robert Lewis, the second equation does not make any problem and its solution is $y = y_0e^{-t}$. This first result can be inserted in the first equation, the solution of which being easily obtained using the method of variation of parameters. Its solution is given by $$x= ... 0 I was able to fix the system! I changed one of my parameters to include a negative that way my system became$$ \frac{ds}{d\tau}=-3s^3-3(\lambda-s^3)e  \epsilon\frac{de}{d\tau}=s^3-(K+s^3)e $$Then my time-scale decomposition gave me a stable steady state at  e = \frac{s^3}{K+s^3}  and I was able to finish the decomposition, which showed that s ... 2 The eigenvalues are:$$\lambda_{1,2} = -1 ~ \pm ~ 4i$$To find the eigenvectors, we solve:$$[A - \lambda I]v_i = 0$$For the first eigenvalue, we have:$$\begin{bmatrix}-1 -(-1 + 4i) & 4\\-4 & -1 -(-1 + 4i)\end{bmatrix}v_1 = 0 \implies \begin{bmatrix}- 4i & 4\\-4 & - 4i\end{bmatrix}v_1 = 0$$Finding the RREF (R_2 = R_2 + iR_1, R_1 = ... 1 Since y = y_0e^{-t}, there are no periodic solutions. If a solution had period T > 0, then y_0^{-(t + T)} = y_0 e^{-t}, clearly impossible unless y_0 = 0. If y_0 = 0, x = x_0 e^{2t} and a similar argument applies. Hope this helps. Cheerio, and as always, Fiat Lux!!! 1$$\frac{d(x^{''}(t))}{d(x^{'}(t))}=\frac{d\cos(tx(t))}{dx^{'}(t)}=\frac{d\cos(tx(t))}{d(tx(t))}\frac{d(tx(t))}{dx^{'}(t)}=\boxed{-\sin(tx(t))\left(t\frac{dx(t)}{dx^{'}(t)}+x(t)\frac{dt}{dx^{'}(t)}\right)}$$0 There is a very nice proof of the existence of a maximal torus in a compact Lie group involving Lefschetz fixed point theorem, given in the third chapter of "J.F.Adams, Lectures on Lie groups, 1983". 0 Note Edited In: I must apologize, but in my haste I mis-read the equations as 0 = A f(x,y) + B \dfrac{\partial^2 f(x,y)}{\partial x^2} + C \dfrac{\partial^2 f(x,y)}{\partial y^2} + D \dfrac{\partial^2 g(x,y)}{\partial x \partial y}, \tag{1} 0 = A g(x,y) + B \dfrac{\partial^2 g(x,y)}{\partial x^2} + C \dfrac{\partial^2 g(x,y)}{\partial y^2} + D ... 1 For existence and uniqueness you are asked to check the assumptions of the Picard-Lindelöf theorem. It is not necessary to repeat the proof of it, which uses the Banach fixed point theorem. Replace e^c by a multiplicative constant C. Which tells that along the way you have missed a sign discussion, or earlier the absolute value signs in ... 1 Yes, you are correct, the family of solutions of a homogenous linear differential equation forms a vector space, a subspace of the space of twice continuously differentiable functions. The set of solutions of an inhomgenous ODE is then an affine subspace, a shifted version of the homogenous solution set. The Wronski matrix and their properties tells you ... 2 For the homogeneous, we have:$$m^2-14 m+65 = 0 \implies m_{1,2} = 7~ \pm~ 4i$$This gives:$$y_h(x) = e^{7 x} ( c_1 \sin 4 x+c_2 \cos 4 x)$$For the particular, choose:$$y_p(x) = a$$Subbing back into ODE, we find 65 a = 13 \implies a = \dfrac{1}{5}. Thus, we have:$$y(x) = y_h(x) + y_p(x) = e^{7 x} ( c_1 \sin 4 x+c_2 \cos 4 x) + \dfrac{1}{5}$$... 0 Hint: Try to use separation of variables Separate and then integrate$$\int\frac{y}{1-4y^2}dy=\int\frac{1}{x}dx$$1 With some manipulation, we have$$(y^2 - xy)dx + x^2dy = 0(xy - y^2)dx = x^2dy\begin{align}\frac{dy}{dx} &= \frac{xy - y^2}{x^2}\\ &= \frac{y}{x} - \left(\frac{y}{x}\right)^2\end{align}$$Now, let v = \frac{y}{x} \implies y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx} (product rule). Then,$$v + x\frac{dv}{dx} = v - v^2$$... 0 Just as in NasuSama's answer, let the differential equation as$$\dfrac{dv(t)}{dt} + \dfrac{\alpha}{m}v(t) = -g$$and integrate for an rhs equal to zero. The solution is$$v(t)=c_1 e^{-\frac{\alpha t}{m}}$$Now apply the variation of parameters considering that c_1 is a function of t and replace in the original differential equation. After ... 0 Assuming that the first mass-spring system is pinned with respect to the world frame, and the second mass-spring system is pinned to the first, you have$$ f_1(t) = m_1 x_1'' + c_1 x_1' + k_1 x_1 $$and$$ f_2(t) = m_2 x_2'' + c_2 x_2' + k_2 x_2 $$separately. 0 This is more of a physics question, but if you have two masses connected by a spring then you need to write express Newton's 2nd law for each mass in some appropriate reference frame. In the end it would be a system of two equations. One gives you the motion of the first mass, while the other gives you the motion of the second mass. 0 Normally for coupled systems you have two position variables, one for each mass. So you have$$m_1x''+c_1x'+k_1x=f_1(t)\\ m_2y''+c_2y'+k_2y=f_2(t)$$These are still not coupled. You need a term that is usually k_3(x-y) added to the first and subtracted from the second to represent the coupling. You can write this as a single matrix equation where the ... 5 Your difficulty stems from the use of the letter y for two different purposes: (a) as coordinate variable in the (x,y)-plane, and (b) as variable for (unknown) functions x\mapsto y(x) whose graphs are lying in the (x,y)-plane. When dealing with ODEs for the first time we are given a function f:\ (x,y)\mapsto f(x,y) defined in some region \Omega ... 3 The notation \dfrac{\mathrm dy(x)}{\mathrm dx} is short for \dfrac{\mathrm dy}{\mathrm dx}(x) or y'(x), if you prefer. In this context, the equality \dfrac{\mathrm{d}y(x)}{\mathrm{d}x} = f(x,y) should be read as \dfrac{\mathrm dy}{\mathrm dx}(x)=f(x,y(x)). As for your last example, you got the wrong idea, y is a function whose domain is a ... 0 \dot{x}(t+x^2) = x \frac{dx}{dt}(t+x^2) = x tdx+x^2{dx}= xdt tdx-xdt=-x^2{dt} \frac{xdt-tdx}{x^2}=dx d(\frac{t}{x})=dx \int d(\frac{t}{x})=\int dx \frac{t}{x}=x+C t=x^2+Cx we can write this in x^2+Cx-t=0 form.Then we can solve this as quadratic equation.So D=C^2+4t then x=\frac{-C\pm \sqrt{C^2+4t}}{2}=-\frac{C}{2} \pm ... 3 As you asked for methods I decided to post an alternative. Here we use the trick of eliminating the inhomogeneity. First, consider only the equation for v(t)=\dot{x}(t),$$\dot{v}=\mu v^2-g.$$The inhomogeneity -g prevents us from simple integration of the ODE, so let's make it disappear by defining w(t)=v(t)-\gamma, which leads to$$\dot{w}=\mu ...

1

Use the fact $$\ddot{x} = \dot{x}\frac{d}{dx}\dot{x} = \frac{d}{dx}\frac{\dot{x}^{2}}{2}$$ then the equation becomes $$\frac{d}{dx}\frac{v^{2}}{2} = \mu v^{2} - g$$ Then you can solve directly to get $$v^{2}\mathrm{e}^{-2\mu x} = \int 2g \mathrm{e}^{-2\mu x} dx + C_{1} = -\frac{g}{\mu}\mathrm{e}^{-2\mu x} + C_{1}$$ if $\dot{x}(0)=x(0)=0$ then  ...

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