# Tag Info

2

By the product rule $$\frac d{dx}\bigl(xp(x)\bigr) = xp'(x) + p(x)$$ so we get $$7xp'(x) = 4\bigl(p(x+1) - p(x)\bigr)$$ Obviously, constant $p$s are a solution, if $p$ were not constant, $n = \deg p \ge 1$, say, we have that the degree of $p(x+1) - p(x)$ is less than $n$. On the other hand, $\deg p' = n-1$, hence the degree of $xp'(x)$ is $n$. But that ...

0

The Mean Value Theorem says that the $y$-Lipschitz constant would be $$\sup_{\substack{x\in[0,a]\\y\in\mathbb{R}}}\left|\frac{\partial}{\partial y}f(x,y)\right|$$ So we compute \begin{align} \frac{\partial}{\partial y}f(x,y) &=-2x^4ye^{-xy^2}\\ &=-2x^{7/2}x^{1/2}ye^{-xy^2}\\ &=-2x^{7/2}ue^{-u^2}\\ \left|\frac{\partial}{\partial ... 1 Artem's answer is complete, though I thought I would give a few more details in case you want to learn a bit more about existence/uniqueness questions. The key part of what I have to say is the last paragraph, in which I introduce my favourite example of non-unique solutions. First of all, the existence of solutions to an initial value problem is (perhaps ... 2 It suffices to see that the strangle-looking integral is a particular solution. You probably came up with the following type of expression, or at least you should agree it is an obvious way to go:u_p(t)=\int_0^t\int_0^v\int_0^u e(s)dsdudv.\tag{1}$$Now this is equal to$$\int_0^te(x)\cdot{\rm Area}(\{(u,v):s\le u\le v\le ...

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Let $A(t)$ be the distances from the finish of $A$ at time $t$ from when A runs out of gas, and the initialy velocity of $A$ and $B$ be $v$. Then $B$ takes a time of $\frac{5}{v}$ to reach the finish. $A(0)=0 \ , \ A'(0)=-v$ $\frac{\mathrm{d}^2A}{\mathrm{d}t^2}=k(\frac{\mathrm{d}A}{\mathrm{d}t})^2\\$ Seperation of variables to find ...

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Better to find df/dy, then you will get either 2a^3 or 2a^4.

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I'd recommend you check out the Mathematics section here for free online video courses offered by some of the top institutions.

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1). Try Verhulst's book on nonlinear differential equations. While the focus of this book is nonlinear ODEs, it does have a chapter on linear equations. It is a nice middle between the formula-computation based ODE books, and something like Arnold's ODE book. 2). For a more comprehensive but still accessible book, try the one by L. Perko. This is a gem.

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I would recommend Differential Equations, Mechanics, and Computation You can read first pages to figure out whether it works for you. The book is 1) introductory, 2) rigorous, 3) with emphasis to physics.

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"Theory of Ordinary Differential equations" by Coddington. If you find the previous one advanced, there is "An Introduction to Ordinary Differential Equations" by Coddington too. The first one is very complete and have a lot of things that I have never seen in other books. The second one is very basic (there is almost no qualitative theory), but proves the ...

1

Try Birkhoff/Rota and Hirsch/Smale (I actually don't know the latest edition with Devaney).

2

To solve $$x'=Ax+b$$ use $$x=e^{At}x_0-A^{-1}b$$ where $e^{At}x_0$ is the solution to $x'=Ax$. Often times it is useful to look at the $1$-dimensional analog: $$x'=ax+b$$ where the solution is $$x=x_0e^{at}-b/a$$ I just put that solution into matrix form and checked it out to make sure it worked. Low tech math.

2

If there exists a vector $v$ such that $Av=b$ then consider the variable $y=x-v$. One finds $y'=Ay$. If there is no such $v$ then $A$ is not invertible. If $A$ is still diagonalizable, however, then the tricky part of $b$ is the bit lying in the kernel of the matrix. Thus write $b=b_0+c$ where $Ab_0=0$ but there is some $Av=c$ solution. Try computing ...

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The answer is that: If the imposed boundary condition leads to a well-posed weak formulation (solution exists and is unique), then no matter what this boundary condition is, we can not get a weak solution $u\in H^2_0$ in general. We simply can not find such non-trivial function satisfying: \left\{ \begin{aligned} -\Delta u &= f \quad \text{ in } U, \\ ... 2 Yes, there is a unique solution to this problem on all \mathbf R. The local existence and uniqueness follow from the standard theorems, since the right hand side is C^\infty(-\infty,\infty) in both variables. To show that the solution exists globally note that |\sin y|\leq 1 for any y, hence the solution to the problem with the initial conditions, ... 1 It seems to me that y'=-2+t-y should be y'=-2+x-y. Letting u=y-x+3, then the DE becomes u'=-u. $$Hence u=Ce^{-x}. So \lim_{x\to\infty}u=0 and hence the solution of y'=-2+x-y is asymptotical to y=x-3. 3 The equation is equivalent to$$\frac{dy}{\sin{y}} = dx \implies \int dy \, \csc{y} = x + C$$where C is a constant of integration. Performing the integration of the left, we get$$\log{\tan{\frac{y}{2}}} = x + C$$C may be determined from the initial condition:$$C = \log{\tan{\frac{Y}{2}}}-X$$Now we solve for y in terms of x. ... 3 If we accept that the vector field F = (x^3 -3xy^2)\vec{i} +(y^3-3x^2y)\vec{j} +2\vec{k} is conservative, so there is a function, say f such that, \nabla f=F. So$$f_x=x^3-3xy^2,~~f_y=y^3-3x^2y,~~f_z=z$$Assume that f_x=x^3-3xy^2, so$$f(x,y,z)=\frac{x^4}4-\frac{3}2x^2y^2+h(y,z)$$Then f_y=-3x^2y+h_y(y,z) but f_y=y^3-3x^2y, so we get ... 2 You can find the particular solution by doing an integration by parts. \frac12 \int_0^t (t-s)^2 e(s)ds = [0-0] -2* \frac12 \int_0^t (t-s) E(s)ds  where E(s) is a primitive of e(s) such that E(0)=0 Similarly we have : -\int_0^t (t-s) E(s)ds = - ( [0-0] - \int_0^t F(s)ds) where F(s)... And u(t)=\int_0^t F(s)ds is a solution of u'''(t)=e(t) ... 1 The particular solution$$u(t)=\frac{1}{2}\int^{1}_{0}(t-s)^2e(s)ds $$is the Green's function solution G(t,s) for this problem. 1 Again, as referenced above, I would Laplace transform. The LT of the RHS is not hard:$$\int_0^{\infty} dt \, f(t) e^{-s t} = \int_3^{\infty} dt \, t\, e^{-s t} = \frac{(3 s+1) e^{-3 s}}{s^2}$$The equation for the LT Y(s) is then$$Y(s) = \frac{(3 s+1) e^{-3 s}}{s^2 (s^2+4)}$$The ILT y(t) may be found by using the theory of residues on$$y(t) = ...

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A related problem. In fact, you need to solve these two differential equations $$\begin{cases} y'' + 4y = 0 &t < 3 \\ y'' + 4y = t & t > 3\end{cases} .$$

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Find two solutions for for $y'' + 4y = 0$ (Set $y(t) = e^{st}$ and solve for $s$, note that $s$ may be complex. If you don't want to use complex numbers, set $y(t)=e^{\sigma t}\sin(\omega t + \phi)$ instead). Combine those solutions to find a solution of the IVP $y_1'' + 4y_1 = 0$ with boundary conditions $y_1'(0) = 0$ and $y_1(0)=0$. Then find a solution ...

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I would use a Laplace transform: $$\scr{L}(y''+4y)=\scr{L}f(t)$$ $$s^2Y(s)+4Y(s)=F(s)$$ $$Y(s)=\frac{F(s)}{s^2+4}$$ $$y(t)=\scr{L}^{-1} \frac{f(s)}{s^2+4}$$ The Laplace transform of $f(t)$ is $\frac{1}{s^2}e^{-3s}$. $$y(t)=\scr{L}^{-1}\frac{1}{s^2(s^2+4)}e^{-3s}$$ Using partial fraction decomposition: ...

0

Write $L$ for a linear differential operator. For example, the equation $\ddot{y}+\omega^2y = f(t)$ can be written as $$Ly = f(t) \qquad \text{where}\qquad L\equiv \frac{\mathrm d%2}{\mathrm d t^2} + \omega^2$$ Eigenfunctions $y_n$ such that $Ly_n=\lambda_n y_n$ are interesting because if someone gives you the equation $Ly = \sum_n c_n y_n(t)$ you can solve ...

1

Zero is an eigenvalue. Let $x(t)=1$ to see this. Whether or not zero is a natural number depends on your definition of $\mathbb N$ which is not standardized. Either convention can be used.

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The expression for Laplacian operator in the $D$-dimensional spherical coordinates is $$\Delta=\frac{\partial^2}{\partial\rho^2}+\frac{D-1}{\rho}\frac{\partial}{\partial\rho}+\frac{1}{\rho^2}\Delta_{S^{D-1}},$$ where $\Delta_{S^{D-1}}$ is a differential operator called Laplace-Beltrami operator on the $(D-1)$-dimensional sphere $S^{D-1}$. It depends only on ...

1

If $S$ is speherically symmetric, say $S(x) = S(\rho)$ with $\rho = |x|$, we have \begin{align*} \Delta S &= \sum_i \partial_i^2S\\ &= \sum_i \partial_i \bigl(\partial_\rho S\cdot \partial_i\rho\bigr)\\ &= \sum_i \left(\partial_\rho^2 S \cdot (\partial_i \rho)^2 + \partial_\rho S \cdot \partial_i^2\rho\right) \end{align*} Now ...

2

Here is a solution by the method of characteristics. So, we assume that the equation is $$u_t+u_x=\cos^2 u,\quad u(x,0)=u_0(x).$$ Consider the characteristics defined by $$\frac{dx}{ds}=1,\quad x(0)=\tau,\\ \frac{dt}{ds}=1,\quad t(0)=0,\\ \frac{du}{ds}=\cos^2 u,\quad u(0)=u_0(\tau).$$ Obviously, from first two equations $$x=s+\tau,\quad t=s\implies ... 2 We have$$\dot \psi u= A \psi u$$Suppose that \psi(t_0)\neq0 for some t_0. Then by continuity, in some neighbourhood$$ A u = \frac{\dot\psi}\psi u$$(and hence \dot \psi/\psi is an eigenvalue of A assuming u\neq0.) But the left-hand side is a constant; therefore so is the other, and the result follows. 1 So u_1 being a solution gives that for all x$$ 0 = a_2(x) u_1''(x) + a_1(x) u_1'(x) + a_0(x) u_1(x) = a_1(x) + a_0(x)\cdot x $$hence we must have a_1(x) = -a_0(x)x for all x \in \mathbb R, u_2 being a solution gives$$ 0 = a_2(x) u_2''(x) - a_0(x)x u_2'(x) + a_0(x)u_2(x) = -a_2(x)\sin x - a_0(x)x\cos x + a_0(x)\sin(x) $$hence$$ -a_2(x)\sin x ...

0

Define $h(x)=-y(-x)$. Then $$h'(x)=y'(-x)=f(y(-x))=f(-h(x))=f(h(x)),$$ where we have used the eveness of $f$. Now, notice that $h(0)=-y(0)=0=y(0)$. As the solution should be unique you obtain that $y$ is odd.

3

Clearly, $\alpha<0<\beta$. Let $r=\min\{-\alpha,\beta\}$ so that we at least have a solution on $\left]-r,r\right[$. Then show: If $y$ is a solution on $]\alpha,\beta[$, then $x\mapsto-y(-x)$ is a solution on $\left]-\beta,-\alpha\right[$. By uniqueness, these coincide on $\left]-r,r\right[$. Then you can combine $y$ with $x\mapsto- y(-x)$ to find a ...

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I think that "Boundary Value Problems and Green's Functions" by Stakgold and Holst is a good text.

1

The determinant $f(t)=\mathrm{det}\,X$ of the fundamental matrix satisfies the equation $$(\ln f)'=(\ln\mathrm{det}\,X)'=\left(\mathrm{Tr}\ln X\right)'=\mathrm{Tr}\left(X'X^{-1}\right)=\mathrm{Tr}\,A,$$ so that $\displaystyle f(t)=\mathrm{const}\cdot\exp\int^t\mathrm{Tr}\,A(t)\,dt$. The statement for $X^{-1}$ follows from this relation and boundedness of ...

1

A possible approach is the following. Consider the Green function for the problem $$\ddot G(t) = \delta(t)$$ that in your case takes the simple form $G(t-t')=(t-t')\theta(t-t')$. Then, $$x(t)=-\int_0^t(t-t')\theta(t-t')A\sin(\omega x(t'-\tau))dt'+x_0+v_0t.$$ You can solve this equation iteratively, starting with $x(t)=x(0)+\dot x(0)t$, and will ...

1

$$e^{ix}=\cos(x)+i\sin(x)$$ Leading to: $$C_1e^{\alpha t} e^{i \beta t}+C_2e^{\alpha t} e^{-i \beta t}=e^{\alpha t} (C_1 \cos(\beta t)+iC_1 \sin(\beta t)+C_2 \cos(\beta t)-iC_2 \sin(\beta t))$$ $$=e^{\alpha t} ((C_1+C_2) \cos(\beta t)-(iC_2-iC_1) \sin(\beta t))$$ $$=e^{\alpha t} (A \cos(\beta t)-B \sin(\beta t))$$ Now, any $(A \cos(\beta t)-B \sin(\beta ... 0 It is 11 minutes because every minute the number doubles. So if half the computers are turned off at ten minutes and number of computers turned of doubles every minute it would be 1/2 *2 =2/2=1 in the next minute so at EXACTLY 11 minutes all computers would turn off. 1 Assuming you mean that$N^\prime(x) = G(x)N(x)$, it follows by linearity of the operator$A \mapsto A^\ast$that $$(N^\ast)^\prime(x) = (N^\prime(x))^\ast = (G(x)N(x))^\ast = N(x)^\ast G(x)^\ast.$$ Hence, by the noncommutative Leibniz rule, $$(N(x)^\ast\sigma N(x))^\prime = (N(x))^\ast \sigma N(x) + N(x)^\ast \sigma^\prime N(x) + N(x)^\ast \sigma ... 2 Let A(t) be the amount of substance A at time t, and let B(t) be the amount of substance B at time t. For suitable decay constants a and b, we have$$A'(t)=-aA(t),$$and$$B'(t)=-A'(t)-bB(t)=aA(t)-bB(t).$$The second equation comes from the fact that B atoms are "born" (through decay) at rate aA(t), and die at rate bB(t). How we handle ... 4 I would certainly like to mention six Painlevé equations. These are the only nonlinear 2nd order ODEs of the form w''=F(t,w,w') whose solutions cannot have movable critical points. Painlevé equations have a lot of applications in various areas of mathematics, including integrable models, random matrices, algebraic and differential geometry and ... 7 For the first order non linear OE, you certainly are aware of Clairaut's equation. I see you are searching for higher order, so considering the following:$$y^{'''}=(x-1)^2+y^2+y'-2\\\\ y(1)=1,~y'(1)=0,~y''(1)=2$$We can find a particular solution, of course by using series method and the undetermined coefficient method simultaneously. W e will see that ... 6 A very simple non-linear system to analyze is what I like to call the "Parachute Equation" which is essentially$$\ddot{x}+k\dot{x}^2-g=0 \tag{1}$$With initial conditions x(0)=0 and \dot{x}(0)=0. where \displaystyle k=\frac{\pi \rho C_d D^2}{8m} such that: m is the mass of the body and parachute, \rho is the density of the fluid in which the ... 3 You assume that there exist a_i such that$$y = a_0 +a_1x + a_2x^2 + \cdots.$$From here it is very easy to calculate power series expansions for y', xe^x, and 2y^2. Then you equate like powers of x on both sides of the equality to get a relation between the a_i, and get a sequence of fairly simple equations in the a_i. The initial condition ... 1 You made a mistake there somewhere: the value of X(0) makes no sense. Here's what I get for the LT:$$X(s) = \frac{\dot{x}_0 + (s+\delta) x_0}{s^2+\delta s+\omega_0^2} + \frac{\gamma s}{(s^2+\delta s+\omega_0^2)(s^2+\omega^2)} $$where x_0 = x(0) and \dot{x}_0 = \dot{x}(0). As you point out, the ILT is given by$$\frac{1}{i 2 \pi} \lim_{R \to ... 0 I think that you can't solve this equation in$H^2_0$. If a function$u$is in$H^2_0$then you have$u=0$and$\nabla u=0$at the boundary and these are too much boundary conditions for this equation. Let me think how one can prove the nonexistence. 2 Let${\bf y}=(x,y)$. Then the coupled system can be written as $${\bf y}'=A{\bf y}$$ The general solution is $${\bf y}=c_1e^{4t}(1,1)+c_2e^{-t}(-3,2)$$ assuming that you did the eigenvector/eigenvalue calculations correctly. 0 Substitute$u=\frac{y}{x^4}$.$u'=\frac{x^4y'-4yx^3}{x^8}=\frac{xy'-4y}{x^5}y'=x^4u'+4\frac{y}{x}x^5u'+4y-4y-x^4\sqrt{u}=0u'=\frac{\sqrt{u}}{x}2\sqrt{u}=\ln{x}+C\sqrt{y}=\frac{1}{2}(x^2\ln{x}+Cx^2)$1 You are confused with the terminology. Consider the initial value problem $$\tag 1 u'=A u, \quad u(0)= u_0.$$ The matrix analogue is $$\tag 2 \Psi'=A u, \quad \Psi(0)= \Psi_0,$$ where$\Psi: \mathbb{R} \rightarrow \mathbb{C}^{n \times n}$. There is a unique solution$\Psi$, given$\Psi_0$, the columns of being solutions of$ (1) $. ... 0 total$3$solutions 1.$y=0$is also a solution. 2.y=0 for$x \le 0$and$y=(2/3x)^{(3/2)}\$ (positive squ root) 3.similarly consider neg squ root.

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