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It turns out the problem was in the calculation of the initial velocities which should be $$\dot{r}=\dot{x}\cos(\theta)+\dot{y}\cos(\theta) \quad \text{and} \quad\dot{\theta}=\frac{\dot{y}\cos(\theta)-\dot{x}\cos(\theta)}{r}$$ Then the the system above is correct and both formulations yields the same plot of the correct initial conditions in polar ...

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$$f'(x) = e^{x - 1} + 3x^2 + 12 x^{-4}$$ Note that $f(1) = 1 + 1 - 4 + 10 = 8$, so $f^{-1}(8) = 1$. So $$\left. \frac{df^{-1}}{dx} \right|_{x = 8} = \frac{1}{f'(f^{-1}(8))} = \frac{1}{f'(1)} = \frac{1}{16}.$$

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$$4 dx + 2 {\cos(y)\over \sin(y)} dy = 0$$ Note that the ODE is sperable. We can simpliy integrate it: $$4x+2\ln|\sin(y)|=c.$$ Apply IV (initial value): $$4\cdot 0+2\ln|\sin(\pi/2)|=c \implies c=0.$$ $$2\ln|\sin(y)|=-4x \implies \ln|\sin(y)|=-2x \implies |\sin(y)|=\exp(-2x).$$ As our IV was $y=\pi/2$, we can assume $y \in (0,\pi)$. This implies that ...

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$$\frac{dx}{dy}=-\frac{1}{2}\frac{\cos y}{\sin y}$$ $$x(\pi/2)=0$$ $$x=-\frac{1}{2}\int \frac{\cos y}{\sin y}dy$$ $$=-\frac{1}{2}\ln(\sin y)+C$$ $$0=-\frac{1}{2}\ln(\sin \frac{\pi}{2})+C=C$$ $$x=-\frac{1}{2}\ln(\sin y)$$ $$y=\sin^{-1}(e^{-2x})$$

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The total derivative of $f\left(x(t),y(t)\right)$ is : $$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ Considering now the same function but depending on a parameter $a$, say $f_a\left(x(t),y(t)\right)$ : If $a$ isn't function of $t$ and if $t$ is the only variable the total derivative is the same : $... -1 Hint: This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0503.pdf. Let$y=\left(\dfrac{dh}{dx}\right)^2$, Then$\dfrac{d^2y}{dh^2}=\pm2(h^{-2}-h^{-3})y^{-\frac{1}{2}}$0 Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2: Let$F(t,x,s)=2ts-\ln s-x$, Then$\dfrac{dt}{ds}=-\dfrac{\dfrac{\partial F}{\partial s}}{\dfrac{\partial F}{\partial t}+s\dfrac{\partial F}{\partial x}}=-\dfrac{2t-\dfrac{1}{s}}{2s+s(-1)}=\dfrac{\dfrac{1}{s}-2t}{s}=\dfrac{1}{s^2}-\dfrac{2t}{s}\dfrac{dt}{ds}+\dfrac{...

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The equation $$\frac{\mathrm{d}}{\mathrm{d}x} f(x) = f(x)$$ is a linear (thus Lipschitz continuous), first-order ordinary differential equation on $\mathbb{R}$. By the Picard-Lindelöf theorem, such an equation has a unique solution for any initial condition of the form $$f(0) = y_0$$ with $y_0 \in \mathbb{R}$. In particular, for the condition $$f(... 0 This proof uses "Little-o" notation. (read on wikipedia for it's definition and properties) let's start with y(t+\Delta t), and let's develop it:$$ y(t+\Delta t)=y(t)+y'(t)\Delta t+o(\Delta t) $$now let's tackle this, using the intermediate value theorem for definite integrals:$$ e^{-\int_{t}^{t+\Delta t}H(t')dt'}y(t)=e^{-H(t+\theta \Delta t)\Delta t}y(...

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No matter what continuous function $a$ is, there is small enough $\Delta t$ to make this a reasonable approximation. However, there are "arbitrarily bad" continuous functions that need arbitrarily small $\Delta t$ to make this reasonable. A class of situations where you can quantify this is when $a$ is continuously differentiable near $a$; in this case this ...

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Together with the general solution of the PDE, the conditions $u(0,t)=u(L,t)=0$ lead to : $$u(x,t)= \sum_{n=1}^{\infty}c_n \sin\left(\dfrac{n \pi x}{L}\right)\exp\left(-\left(\frac{n\pi}{L}\right)^2 \beta t \right)$$ The condition $u(x,0)=f(x)$ can be expressed in terms of Fourier series of various manner, depending on the chosen bounds. In the present case,...

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For linear equations, the technique of separation of variables is used to find all separated solutions of the form $X_1(x_1)X_2(x_2)\cdots X_n(x_n)$. You find them all if your equation can be separated. If you make a change of variables, then you will generally find a different set of separated solutions. For example, you might separate $X(x)Y(y)$, or ...

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Assume that $f(x)$ is a function such that $f'(x)=f(x)$ for all $x\in\Bbb{R}$. Consider the quotient $g(x)=f(x)/e^x$. We can differentiate $$g'(x)=\frac{f'(x)e^x-f(x)D e^x}{(e^x)^2}=\frac{f(x)e^x-f(x)e^x}{(e^x)^2}=0.$$ By the mean value theorem it follows that $g(x)$ is a constant. QED.

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This may not be an answer you are looking for, but its a nice one to consider. Consider $y=\cos(ix)-i\sin(ix)$. You may find that: $$\frac{dy}{dx}=-i\sin(ix)-i^2\cos(ix)=\cos(ix)-i\sin(ix)$$ Thus, $y'=y$ is satisfied. Since $y(0)=1$, $y'(0)=1$, $\dots$, then by Taylor's theorem, we have $e^x=\cos(ix)-i\sin(ix)$, or more commonly known as $$e^{ix}=\cos(... 8 Consider the equation y'=y. Our goal is to solve for the function y=f(x). Roughly speaking$$\frac{dy}{dx}=y \implies \frac{dy}{y}=dx \implies \int\frac{dy}{y}=\int dx \implies\ln(y)=x+C \implies y=e^{x+C}=Ae^x$$for some constant A 2 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 0 At least one solution of the above differential equation is $$\mathbf V(t) = e^{\mathbf A t} \mathbf V(0) e^{\mathbf A^\text{T} t}.$$ Since e^{\mathbf A^\text{T} t}= (e^{\mathbf A t})^\text{T}, this can be written as $$\mathbf V(t) = e^{\mathbf A t} \mathbf V(0) (e^{\mathbf A t})^\text{T}.$$ The ... 1 It vanishes except for n=0. In that situation$$a_0'(t) = F(t).$$saw that JJ beat me to it. 1$$\sum\limits_{n=0}^\infty \left[ a_n'(t)+a(t)\left( \frac{n\pi}{L} \right)^2c^2 \right]\cos\left(\frac{n\pi x}{L}\right)=F(t)\implies \begin{cases} a_n'(t)+a(t)\left( \frac{n\pi c}{L} \right)^2=0 \qquad n>0\\ a'_0(t)=F(t) \end{cases}\begin{cases} a_n(t)=A_n\exp \left(-\left( \frac{n\pi c}{L} \right)^2 t\right) \qquad n>0\\ a_0(t)=A_0+\...

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If I properly understand, you look for the solution of $$h'(x)=\frac k{(1+x^n)^{1/n}}$$ The solution exists but it involves the hypergeometric function $$h(x)= k x \, _2F_1\left(\frac{1}{n},\frac{1}{n};1+\frac{1}{n};-x^n\right)+C$$ The only simple forms are $$n=1 \implies h(x)=k \,\log (1+x)+C$$ $$n=2 \implies h(x)=k\, \sinh ^{-1}(x)+C$$ Don't be afraid ...

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Sketch of proof: The tangent to $\gamma$ meets the coordinate axes at points $$P_y(t):=\left(0,\gamma_y(t)-\frac{\gamma'_y(t)\gamma_x(t)}{\gamma'_x(t)}\right)\\P_x(t):=\left(\gamma_x(t)-\frac{\gamma'_x(t)\gamma_y(t)}{\gamma'_y(t)},0\right)$$ Now, assume $\gamma'_x(t)\ne 0$ and $\gamma'_y(t)\ne0$, impose that $2(\gamma_x(t),\gamma_y(t))=P_x(t)+P_y(t)$, ...

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The particular solution is: $$y(x) = \frac{-e^x + \log{(e^x + 1)}}{e^{2x}} + \frac{\log{(e^x + 1)}}{e^{x}}$$ We can separate out the terms as : $$y(x) = \frac{-e^x}{e^{2x}} + \frac{\log{(e^x + 1)}}{e^{2x}} + \frac{\log{(e^x + 1)}}{e^{x}}$$ Now, the first term (which is our "extra" term) is $$\frac{-e^x}{e^{2x}} = \frac{-(e^x)}{(e^x)^2} = \frac{-1}{e^x}... 4 Let's write \lambda_2=-1+i\sqrt{2}, \lambda_3=-1-i\sqrt{2}. We have$$A-\lambda_2 I=\begin{pmatrix} -2-i\sqrt{2} & 0 & 2 \\ 1 & -i\sqrt{2} & 0 \\ -2 & -1 & 1-i\sqrt{2} \end{pmatrix} Applying row operations, we obtain \begin{align*} A-\lambda_2 I&\to \begin{pmatrix} 1 & 0 & \frac{-2+i\sqrt{2}}{3} \\ 1 & -i\sqrt{... 1 If the quotient of two solutions is constant that means that one is a constant multiple of another, which means that one is linearly dependent on the other, which means they cannot form a basis of solutions (a basis is a linearly independent and spanning set of solutions). For instance, e^x and 2e^x are both solutions to your ODE, but clearly they are ... -1 I can understand this and you should feel lucky that you haven't touch on calculating derivative of an inverse function. if y=f(g(x)), you can write it in two equations.y=f(u)u=g(x)$$Then$$\frac{dy}{dx}=\frac{df(u)}{du}\frac{dg}{dx}$$After that, replace u with g(x) for the first term. 2 The general solution of a homogeneous linear ODE of order n consists of the linear combinations of a basis consisting of n linearly independent solutions. In the case n=2, for two solutions to be linearly independent is equivalent to neither being a constant multiple of the other, i.e. neither is identically 0 anThe theory says what you need is a ... 1 Not very rigourous, but the notation speaks for itself:$$\frac{\mathrm d\bigl(f(g(x))\bigr)}{\mathrm d\mkern1mu x}=\frac{\mathrm d\bigl(f(g(x))\bigr)}{\mathrm d(g(x))}\times\frac{\mathrm d(g(x))}{\mathrm d\mkern1mu x}.A concrete example: \begin{align*}\frac{\mathrm d\bigl(\sin\sqrt{x^2+1}\bigr)}{\mathrm d\mkern1mu x}&=\frac{\mathrm d\sin\sqrt{x^2+1}}... -3 The Taylor formula is:f(x+\varepsilon)=f(x)+\varepsilon f'(x)f\circ g(x+\varepsilon)=f\circ g(x)+\varepsilon (f\circ g)'(x)f\circ[g(x)+\varepsilon g'(x)]=f\circ g(x)+\varepsilon (f\circ g)'(x)$$As \varepsilon\rightarrow 0 so does \varepsilon g'(x) so we can apply Taylor again with \varepsilon g'(x) as the \varepsilon:$$f(g(x))+\...

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Oh, I hate the Laplace transform! I have yet to find a differential equation that cannot be solved more easily using simpler methods. Here, the differential equation is $y''+ y= x^2$. The associated homogeneous equation is $y''+ y= 0$. Its characteristic equation is $r^2+ 1= 0$ which has roots $r= \pm i$ so the general solution to the associated ...

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As for the Laplace solution you asked for, you can split the fraction like this: $$\frac 1{s^3(s^2+2)}=\frac As+\frac B{s^2}+\frac C{s^3}+\frac{Ds+E}{s^2+1}$$ $$=\frac{As^4+As^2+Bs^3+Bs+Cs^2+2C+Ds^4+Es^3}{s^3(s^2+1)}$$ $$=\frac{(A+D)s^4+(B+E)s^3+(A+C)s^2+Bs+C}{s^3(s^2+1)}$$ By identification, you find $B=0,E=0,C=1,A=-1,D=1$

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My question is how do I find this inverse Laplace transform of $\dfrac{1}{s^3(s^2+1)}$? Hint. If one wants to proceed on your route, by a partial fraction decomposition, one has $$\frac{1}{s^3(s^2+1)}=-\frac{1}{s}+\frac{1}{s^3}+\frac{s}{1+s^2}$$ giving $$\mathcal{L}^{-1}\left(\frac{1}{s^3(s^2+1)}\right)(t)=-1+\frac{t^2}2+\cos t$$ using standard ...

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Following this: http://eqworld.ipmnet.ru/en/solutions/ode/ode0310.pdf Let $y = f(x), w = \frac{x}{y}y', z=\frac{x^3}{y^3}$, then $$w'_x = \frac{1}{y}y'+\frac{x}{y}y''-\frac{x}{y^2}(y')^2,$$ $$w'_x = w_z'\frac{dz}{dx}=w'_z (3\frac{x^2}{y^3}-3\frac{x^3}{y^4}y'_x),$$ $$xw_x'=3w_z'(z-zw),$$ $$xw_x'=\frac{x}{y}y'+\frac{x^2}{y}y''-\frac{x^2}{y^2}(y')^2=w+z-w^2,$$...

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Your result is correct and the result of the book is also correct. Strange isn't it ? Not at all : just a matter of symbolism for two functions each one being ANY function. $$\frac{-2x^2-3xy+2y^2}{5} = (xy-2x^2)+\frac{2}{5}(y-2x)^2$$ $$e^{\frac{-2x^2-3xy+2y^2}{5}} = e^{\frac{2}{5}(y-2x)^2} e^{xy-2x^2}$$ $e^{\frac{2}{5}(y-2x)^2}$ is a function of $(y-2x)... 1 It seems the answer is only valid within inter$(-\pi, \pi)$, beyond that interval, y is negative and$\sqrt y$is not valid. 2 What you've done is correct, indeed, in general if we have$T(x) = \int_{a}^b f(x, t) \, \mathrm{d}t$then$T(1)$is given by$\int_a^b f(1, t) \, \mathrm{d}t$. 1 The point of separation of variables is not just to get some solution, but to get a general solution, which can be used to produce a solution for any initial condition. Consider the case of the initial value problem$u'(t) = u(t)$,$u(0)=1$. You solve this problem by first writing down the general solution$u(t) = C e^t$, and then plugging in the initial ... 0 The true Fabius function is no-where analytic. This implies a lack of an analytical function describing it. The linked page a has a discussion concerning a function that approaches the Fabius function when taken to infinity. (though it gets reasonably close at around n=20): Recursive Integration over Piecewise Polynomials: Closed form? 0 I think it's important to understand what an equation is in the first place. Heuristically, it's a statement about an unknown quantity that is true or false for various values of the quantity [yes, this is also true for inequalities and other relations, but bear with me.] For instance,$x^2 - 4x + 3 = 0$is an algebraic equation. The unknown is a number$...

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Putting $t = (1+z^n)^{1/n}$, we have: $$h(z) =C \int\frac{t^{n-2}}{t^n -1}dt$$ for $n=4$, this is: $$h(z) = C\int \frac{t^2}{t^4 - 1}dt = \frac{C}2 \left[ \int \frac{dt}{t^2 + 1} + \int \frac{dt}{t^2 -1} \right] = \frac{C}2 \left[\arctan t + \frac12 \ln \left| \frac{t-1}{t+1} \right| \right] \\ = \frac{C}2 \left[ \arctan\left( \sqrt[4]{1 + z^4} \right) + \... 1 @ Michal : your solution is correct. But it isn't the only way. It might be less disturbing if one can find a particular solution in order to change of function:$$u_x+u_y+u=e^{x+2y}$$The form of the right term suggests to look for a particular solution on the form Ce^{x+2y} which leads to C=\frac{1}{4} \quad\to\quad U=\frac{1}{4}e^{x+2y} Change of ... 1 y + sqrt(1 - x^2) = arcsin(x) sin(y + sqrt(1 - x^2)) = x cos(y + sqrt(1 - x^2)) * [dy/dx + 1/2sqrt(1-x^2) * -2x] = 1 cos(arcsin(x)) * [dy/dx - x/sqrt(1-x^2)] = 1 sqrt(1-x^2) * [dy/dx - x/sqrt(1-x^2)] = 1 dy/dx = 1/sqrt(1-x^2) + x/sqrt(1-x^2) dy/dx = (1+x)/sqrt(1-x^2) 2 I was too hasty before. Here's one solution (which likely could be simplified): You need in fact two different energy estimates. Multiply by u_t and integrating by parts gives you that for (as you noted)$$ E(t) = \int_B u_t^2 + |\nabla u|^2 ~\mathrm{d}x $$you have$$ \dot{E}(t) = - 2 \int_B a u_t^2 ~\mathrm{d}x $$Multiply by u and integrating by ... 0 I believe that some approximations are in order here. This seems like the kind of question where we assume damping is relatively small. Note that the response function isn't even symmetric about \omega_0. There may be some fancy way to define FWHM here, but if we just take the limit where \gamma is small, and the peak value occurs at about \omega_0, ... 1 You can use some constant k, but then one realizes that along the characteristics there are two constants or invariants k,c. If k is an arbitrary number and c is an arbitrary number then couldn't we express k as some function of c? 1 \newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \... 2 First, we should find the inverse Laplace transform of \frac{1}{(s+1)^4}. If you look at a table of Laplace transforms, you'll see that$$L(t^ne^{at}) = \frac{n!}{(s-a)^{n+1}}$$(this formula can be shown by induction & integration by parts). So we can see that$$L(\frac{t^3e^{-t}}{3!}) = \frac{1}{(s+1)^4}$$Most tables will also mention that$$L(f'(...

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We can take $x=0.$ Now every harmonic function $u$ on $\mathbb R^n$ has a unique expansion into harmonic homogeneous polynomials. That is, $$u(y) = \sum_{k=0}^{\infty}P_k(y),$$ where each $P_k$ is a homogeneous harmonic polynomial of degree $k.$ The convergence is uniform on compact subsets. These polynomials have the further property that if $j\ne k,$ ...

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$$(h-1)h'=2$$ is a separable differential equation: it can be simply re-written as $$\frac{d}{dt}\left(\frac{h^2}{2}-h\right) = 2$$ from which: $$(h-1)^2=(4t+C)$$ and $$h = 1\pm\sqrt{C+4t}$$ readily follow. Have also a look at the generating function for Catalan numbers.

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The easy way for this problem, as is the case for many pdf problems, is to work with CDF's instead. Here, since $f(x) = \frac{x-1}{2}$ on $(1,3)$, $$F(x) = \left\{ \array{0 & x\leq1\\ \frac{(x-1)^2}{4} & 1< x < 3 \\ 1 & x\geq 3 }\right.$$ And this needs to match the CDF of the uniform distribution ojn $(0,1)$ $$F(y) = y$$ So y= \frac{...

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