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Hints: Follow the steps i) Find $y_h$ for the homogenous equation $y'''-y=0$ by assuming $y=e^{mx}$. You should have three roots which gives the three solutions of the homo. eq and write $y_h$ as $$y_h=c_1 y_1+ c_2 y_2 + c_3 y_3$$ ii) Find a particular solution for $y'''-y=t$ by assuming $y_p=a+b t$ iii) write down the general the general solution ...

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You are right, and the given formula for $\dot z$ is incorrect. Let's check on a simple example, inviscid Burgers: $u_t+uu_x=0$. Here $H=up$, hence $\dot x=D_pH=u$. Along the characteristic $u$ is constant, thus $\dot z\equiv 0$. This is in agreement with your formula: $$\dot z = (D_p H) p-H=up-up=0$$ but not with the formula for $\dot z$ in the quoted ...

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The base of a triangle is the vertical dimension of the picture. The apex is the eye. You are not given the height of the eye, but I believe it is intended to be $5.5$ feet, which makes the triangle isosceles. The altitude is $10$ feet and decreasing.

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You can begin by dividing by $r^2$, then multiply by $\dot r$. You can now integrate the equation once to get a first order equation. In physical terms, this really boils down to the conservation of energy.

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Definition of Laplace transform is: $$F(s) = \int_0^\infty f(s) e^{-st}\, dt$$ Substitute the function and because integral is linear operation, it can be split into two integrals: $F(s) = \int_0^\infty \left(\begin{cases} t, & 0 \leq t < 3\\ 1, & 3 \leq t <\infty \end{cases} \right)e^{-st}\, dt = \int_0^3 t e^{-st}\, dt+\int_3^\infty ... 0 Call$f_1(t)=t$and$f_2(t)=\mathrm{e}^{-t}$. The point that you missed is that the convolution theorem states that$L_s(f_1)\,L_s(f_2)$is the Laplace transform of $$f(t)=\int_0^t f_1(u)\,f_2(t-u)\,\mathrm{d}u=\int_0^t\;u\;\mathrm{e}^{-(t-u)}\,\mathrm{d}u=\mathrm{e}^{-t}-1+t.$$ Take the Laplace transform of$f$and you find $$... 1 Lets say we have a spring with an external force. We solve the second order ODE and arrive at:$$x(t) = \dfrac{e^{-4 t}}{9} [3 \cos(4 \sqrt{3} t) - 11 \sqrt{3}) \sin(4 \sqrt{3} t)] + 2 \sin(8 t) \\ y(t) = 2 \sin(8t)$$A plot of these functions is: You can see that the e^{-4t} becomes negligible when t is large. The steady state solution is shown in ... 0 Your eigenvectors are missing from your plot. Note that you already have them there in your solution. No need to scratch around for equations. Plot (2,1) and a constant multiple of, it say (-2,-1) and draw the line through the points. Actually only the first point is needed since the eigenvector passes through (9,0). Do the same for eigenvector represented ... 0 Since cross-product is bilinear, the correct formula is$$\frac{d}{dt}(\vec{\dot r}\times \vec{x})= \vec{\ddot r}\times \vec{x} + \vec{\dot r}\times \vec{\dot x}$$(given by Hoseyn Heydari). Tony Piccolo noticed that the formula from which the question originated had a typo. The correct form is$$D=\sum_\alpha m_\alpha[\mathbf r_\alpha\times\mathbf ... 2 Hints: You found the eigenvalues and eigenvectors correctly. The solution is given as: $$x(t) = \dfrac{1}{3} c_1 e^{-6 t} (e^{3 t}+2)- \dfrac{2}{3} c_2 e^{-6 t} (e^{3 t}-1)$$ $$y(t) = \dfrac{1}{3} c_2 e^{-6 t} (2 e^{3 t}+1)-\dfrac{1}{3} c_1 e^{-6 t} (e^{3 t}-1)$$ Now, using all of the information below, direction fields, solutions, eigenvalues, ... 3 Hints outlined. To find the critical points, we want to find where we simultaneously have$x'$and$y'$equal to zero. We find three critical points$(x, y) = (-1, 0), (0,0), (1, 0)$. We can now find the Jacobian matrix and analyze each critical point for it's type at those three critical points (see Phase Portrait below). We can draw the phase portrait ... 0 A different yet very simple and easy derivation of Euler's method is to consider the following$y'=f(x,y)$(I hope you don't mind me using$x$and$y$, the diagram for the graphical interpretation is in terms of$x$and$y$, sorry.) with$y(x_0)=y_0$If you imagine the graph of$y=y(x)$(the solution) then$y'=f(x,y)$is the slope of the tangent to the ... 1 The unique solution of this IVP is$\sin t$,$t\in(-\pi/2,\pi/2)$, and I doubt that it can be obtained by series expansion. Note though, that left of$-\pi/2$and right of$\pi/2$uniqueness is violated! 0 Not sure if this works, I wanted to leave it as a comment but it is too long. We have $$\left(\sum_{n = 0}^{\infty}{(n+1) a_{n+1} x^n }\right)^2 = 1 - (\sum_{n = 0}^{\infty}{a_nx^n})^2$$ Pick pencil and paper. Treat the sums as "finite". Ask yourself: what is the coefficient of$x^0$in the left-hand side and in the right-had side? They must be equal. ... 2 I believe you are on the right track. Starting with $$\frac{1}{x} dx = -A e^{y/B} dy$$ you have the equation separated. Now you can integrate each side to get $$\ln(x)=-ABe^{y/B} + C,$$ where$C$is an integration constant to be determined with the boundary condition, e.g.$y(x=0)=y_{0}$. With some algebra you can write the solution for$y$explicitly ... 2 Hint: $$\frac{d(x^2)}{dy} = 2x\frac{dx}{dy}.$$ And in general form, if you have an equation$x' = f(x)g(t)$with initial data$x(t_0) = x_0$, then the method of separation of variables writes $$\int_{x_0}^{x(t)}\frac{du}{f(u)}=\int_{t_0}^{t}g(s)ds.$$ After that you study the existence of these integrals, find them, and try to solve the result with respect ... 1 Note that$\cos(\phi-\pi)=-\cos(\phi), \cos(\phi - \pi/2)=\sin(\phi)$so$A_3=A_c+D\theta \cos(\phi)$which gives$A_1+A_3=2A_c=A_2+A_4$If these are not satisfied your equations are contradictory. You can then write $$A_3-A_1=2D\theta \cos(\phi)\\A_4-A_2=2D\theta \sin(\phi)\\ \phi=\arctan\left(\frac {A_4-A_2}{A_3-A_1}\right)\\ \theta=\frac ... 1 We have \displaystyle L(e^{at})=\frac1{s-a} We know, \displaystyle L(e^{at}f(t))=F(s-a) where F(s)=L(f(t)) So as \displaystyle L(t^n)=\frac{n!}{s^{n+1}}, for integer n, \displaystyle L(e^{at} t^n)=\frac{n!}{(s-a)^{n+1}} So, L(t^n)\cdot L(e^{at})\ne L(e^{at} t^n) 2 Laplace transformation of the convolution (not the product) of two functions equals the product of Laplace transformation of each function. 4 The equation is separable, so you can write \frac{dy}{y^2}=dx Therefore \frac{-1}{y}=x+c and y=\frac{-1}{x+c} 1 You can solve it just by writing$$\frac{dy}{y^2}=dx$$and then integrating both sides. 3 Hint: Use Separation of Variables. We end up with:$$\int \dfrac{1}{y^2}~ dy = \int dx$$3 Hint: This is a first-order nonlinear ordinary differential equation, let$$y = v t \rightarrow y' = v + v' t$$This is also Bernoulli's equation. 2 This is separable: It can be rearranged to yield$$\cos t dt = -\frac{e^y \sin y}{1 + e^y} dy$$Integrating the right side is possible, but not so very nice. 1 For 0<t<1, define the integrated matrix I(t)x= \int_0^t S(w)x\,dw. Note that {1\over t}I(t) x\to x as t\to 0, so for t small enough, I(t) is invertible. Now fix such a t. Your semigroup is differentiable from the right on I(t)(\mathbb{R}^n)=\mathbb{R}^n since direct calculations give$${S(s)(I(t)x)-I(t)x\over s}\to (S(t)-I)x,$$as ... 0 You Have written that the DE is in the form p'(x)=p''(x)+(2\pi*\frac fc)^2p(x)=0 but in "Problem 1:Phonetics" it is stated as p''(x)+(2\pi fc)2p(x)=0? which I think is an error, since after running through the problem the it must be p''(x)+(\frac{2\pi f}{c})^2p(x)=0 since the sine and all trig functions are Transcendental Functions they must have a ... 0 Consider t=1 in what you want to prove, S(1)=e^A[2]. You can see from your definition that S(1)= S(m (1/m))= S(1/m + ... + 1/m)= S(1/m)...S(1/m)= S(1/m)^m, so e^A=S(1/m)^m and therefore S(1/m)=e^{A{1/m}}. Therefore, S(n/m)=e^{(n/m)A} by similar argument for all 0\le n/m \le 1 [1] - Note that that case n=0 requires S(0)=I; all ... 0 This may not be true; I will put my first impression and think about it. On a small neighborhood of the identity matrix we have a well-defined logarithm. So, taking L(t) = \log S(t) we have the equation L(s+t) = L(s) + L(t). This is Cauchy's functional equation in each matrix entry. The interesting thing is that a bad solution of Cauchy's equation is ... 1 This is an interesting question. Actually this algebraic method is known as "factorization", at least this is what Schrodinger called it. This is how he solved the Hydrogen atom initially, he did not use his wave equation (which is very strange), he used factorization methods. This algebraic/ factorization approach is now called "supersymmetric" quantum ... 4 Hint: Separation of variables. Write:$$\int \dfrac{1}{\cos^2(2y)}~ dy = \int \cos^2(x) ~ dx$$0 Let y(x)=\sum_{i=0}^{\infty}a_i x^i. Find y' and y'' by differentiating y(x).$$y'(x)=\sum_{i=0}^{\infty}ia_i x^{i-1}y''(x)=\sum_{i=0}^{\infty}i(i-1)a_i x^{i-2}$$Substitute all these in the differential equation and try to find a recurrence relation for a_i. Hope you can continue. You can find an example of the power series method ... 1 This is a first order linear ODE. We solve it via integrating factor method. Re-writing your ODE by dividing through by dx we obtain$$ \cos x\cdot y'(x)=\sin x (\cos x-2y), \ \to y'(x)=\tan x(\cos x-2y). $$Note, I divided out by \cos x on both sides which is how we got tangent. We now use distributive property on the right hand side and simplify ... 1 Analyticity and harmonicity are local properties. When proving them, you focus on a small neighborhood of a generic point z_0 in the disk. It helps to make this neighborhood small enough so that it stays away from the boundary. For example, let it be the disk of radius (1-|z_0|)/2 with center z, denoted N below. By the reverse triangle inequality, ... -1 first five terms using Picard iterations with y (0) = 0$$\left\{t \text{Sin}[1],\text{Csc}[1] \left(-\text{SinIntegral}[1]+\text{SinIntegral}\left[e^{t \text{Sin}[1]}\right]\right),\int_0^t \text{Sin}\left[e^{\text{Csc}[1] \left(-\text{SinIntegral}[1]+\text{SinIntegral}\left[e^{s \text{Sin}[1]}\right]\right)}\right] \, ds,t \text{Sin}\left[e^{\int_0^s ... 1 A hint: If$f$were given explicitly you would solve this differential equation by the method of "separation of variables". Look at the formulas you get in this way. 0 A particular solution of the difference equation $$s_n=\alpha s_{n-1}+\beta s_{n-2}+a+bn+cn^2\qquad(n\geq2)$$ can be found with the "Ansatz" $$s_n=A+Bn+Cn^2,\tag{1}$$ and solving for$A$,$B$,$C$by comparing coefficients. If$\alpha+\beta\ne1$, i.e.,if the associated homogeneous problem $$s_n=\alpha s_{n-1}+\beta s_{n-2}\tag{2}$$ has no constant ... 1 This IVP enjoys existence and uniqueness of solutions. Also, its solution is global (i.e., defined in the whole of$\mathbb R$) as the flux function$f(y)=\sin(\mathrm{e}^y)$is bounded. The solutions$\varphi=\varphi(x)$of IVP of scalar autonomous ODEs (i.e., of the form$y'=f(y)$,$\,y(0)=y_0\$) with smooth flux have the following properties: a. If ...

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Separate variables by rearranging terms and then integrating; examine the domain, the set of reals on which the function is defined, and assign the longest interval on which the function is defined to the maximal interval for existence for solutions.

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