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$$N'(x)=aN(x)+bN(x)^2$$ The change of variable $\quad N=\omega^{1/2}\quad$ is strange since it leads to an ODE more complicated than the original one. $N'=\frac{\omega'}{2\:\omega^{1/2}}$ $$\frac{\omega'}{2\:\omega^{1/2}} = a\:\omega^{1/2}+b\:\omega$$ $$\omega'= 2a\:\omega +b\: \omega^{3/2}$$ The transformed ODE is non-linear. So, what is the interest ...

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From Newton's Law of Cooling, $\frac{d T(t)}{d t} = -k(T - T_f)$ you'd expect the result for the temperature as a function of time, $T(t)$, to be of the form $T(t) = T_f + A e^{-k t}$ where $T_f$ is the temperature after a long time, so $T_f = -7^o C$ and $A$ and $k$ (with $k > 0$) are constants. (Notice that because of the $-ve$ sign, when $T < ... 0 -5.57 appears to be correct. (I have re-verified). (assuming this is a newton's law of cooling problem): My Solution: $$\frac{dT}{dt}=-k(T-T_{env})$$ $$dT=-k(T-T_{env})dt$$ $$\frac{1}{T-T_{env}}dT=-kdt$$ $$\int\frac{1}{T-T_{env}}dT=\int-kdt$$ $$\ln|T-T_{env}|=-kt+C$$ $$|T-T_{env}|=e^{-kt+C}$$ $$T-T_{env}=\pm e^ce^{-kt}$$ $$let\: H=\pm e^c$$ ... 1 Exactness is really a property of differential forms rather than differential equations. Even though $$M(x,y) \, dx + N(x,y) \, dy = 0$$ represents the same differential equation as $$f(x,y) \, M(x,y) \, dx + f(x,y) \, N(x,y) \, dy = 0$$ (if$f(x,y) \neq 0$), it is not true that if one of the differential forms $$M \, dx + N \, dy$$ and $$(fM) \, dx + ... 0 You know the eigenvalues are 1 and 2, thus consider g(x)=e^{-x}f(x) with$$ g'-g=e^{-x}(f'(x)-2f(x)),\\---\\ g''-g'=e^{-x}(f''(x)-3f'(x)+2f(x))=e^{-x}·kδ(x−a)=e^{-a}·kδ(x−a) $$This can be integrated once, and using the Heaviside function as the integral of delta,$$ g'-g=ke^{-a}·H(x-a)+C $$Substituting again h(x)=e^{-x}g(x) results in$$ ... 0 To convert the non-homogeneous differential equation to a homogeneous differential equation, simply remove the "non-homogeneous part"! That is, to convert the non-homogeneous differential equation y'(t)= M(t)y(t)+ h(t) just write it as y'(t)= M(t)y(t). After finding the general solution to that equation, you can get the general solution to the original ... 0 You are confusing your particular problem ($y=kx$) for the general orthogonal trajectory problem. The curves you are searching for must intersect each member of the given family of curves at a right angle. That means the key point is the slope of the tangent line at each point on each curve in your family. Therefore, for each$k$you are interested in ... 0 Just as, in solving a system of linear equations, one reduces to a single equation in one unknown, so in a system of linear differential equations, one reduces to a single equation, of higher order. The first of these equations says that$y''= \omega z'$. Differentiating again,$y'''= \omega z''$so that$z''= \frac{y'''}{\omega}$. Replace z'' in the ... 1 An IVP $$\dot x=f(x,y),\qquad\dot y=g(x,y), \quad x(t_0)=x_0,\quad y(t_0)=y_0$$ defines a motion in the plane as a process in time. The resulting curve may have vertical tangents at some points, or wind three times around the origin before going off to infinity. Sometimes you are not interested in the exact "timetable" but only in the resulting curve ... 8$y'=x-y^2$is Riccati ODE classically solved thanks to a change of function, in this case : $$y=\frac{f'}{f}$$$y'=\frac{f''}{f}-\frac{f'^2}{f^2}$$$\frac{f''}{f}-\frac{f'^2}{f^2}=x-\left(\frac{f'}{f}\right)^2$$ $$f''=x\:f$$ This is an Airy ODE. The solutions are the Airy functions$\text{Ai}(x)$and$\text{Bi}(x)$which derivatives are$\text{Ai}'(x)$and ... 0 Continuing with the solution of How to Solve the Coupled Differential Equations? The general solution of : $$\ddot q + \omega^2q=\omega^2\frac{E}{B}$$ is : $$q(t)=C_1\cos(\omega t)+C_2\sin(\omega t) +\frac{E}{B}$$ then $$y(t)=\frac{C_1}{\omega}\sin(\omega t)-\frac{C_2}{\omega}\cos(\omega t)+\frac{E}{B}t+C_3$$ Now I guess that you can find$z$2 The mistake is marked on the screen copy below : 1 The comments do not seem to lead to anything fruitful. I give you the first step, and then you confirm that you have the same in your solution, OK? Write the differential equation as $$x'-\frac{5}{y-4y^6}x=\frac{y^3}{1-4y^5}$$ Thus, an integrating factor is $$\exp\Bigl(\int-\frac{5}{y-4y^6}\,dy\Bigr).$$ I get it to be Using this, I get the final ... 1 By Gronwall's inequality,$f$is identically zero. 1 You need a basis of the generalized eigenspace. Therefore you need to make sure that the vectors are linearly independent. You have$v_1,v_2$and now you need to complete them to a basis of the generalized eigenspace by adding a vector from the eigenspace that is independent of the two. (It actually suffices that it is independent of$v_1$but I find it ... 0 Hint. Substitution$z=y^2$, and then you obtain the IVP$z'-z=5x-5,$with$z(0)=36$. 0 Linear dependence/independence is a feature that must be true at every point of your space. When you say that$y_i$are linearly independent on a set$Z$, that means there doesn't exist a fixed set of$a_i$(at least one which is nonzero), such that: $$a_1y_1(x)+\cdots+a_ny_n(x)=0, \forall x\in Z.$$ So suppose that$y_i$are linearly independent on$[b,c]$... 1 Following link given by Artem: $$x(t) = 0 \times 1_{t < K} + (t-K)^{3/2} \times 1_{t \ge K}$$ is a solution (check that$\lim_{t \to K} x'(t)$exists) passes through$(t_0,0)$for$K \ge t_0$1 here are some ideas i have. first we establish that $$\frac{dx}{dt} = -x + k + t^3 \tag 1$$ has the solution$x_k(t) =k+ ce^{-t}+ x_p(t)$where$x_p(t)$is a particular cubic polynomial solution of$(1).$the solution exists for all$t.$now, argue that the solution$x$to $$\frac{dx}{dt}=-x+ \sin x+t^3 \tag 2$$ lies between the solutions$x_{-1}(t)$and ... 1 I use book Applied Stochastic Control and Jump Diffusion (2nd edittion) written by Bernt Oksendal and Agnes Sulem. You can find more details therein. Let $$dX_{t}=a(t,X_{t})dt+b(t,X_{t})dB_{t}+\int_{\mathbb{R}}\gamma(t,X_{t},z)N(dz,dt)$$ then infinitesimal generator is of the form $$\mathcal{A}f(t,x) = \frac{\partial f}{\partial t}(t,x) + ... 0 If we look at the particle level, heat trasnfer is kind of brownian motion. See how heat transfer. It is not a suprise that both equations look like the same. 0 The trajectory is a subset of the phase space -- which often is \mathbb{R}^n, but, depending on the dynamical system, this can be a suitably chosen manifold. A solution to the dynamical system is a mapping \phi: \mathbb{R}_{>0} \to \mathbb{R}^n, \phi: t \mapsto \phi(t). The solution curve this mapping traces out in \mathbb{R}^n is called the orbit ... 0 The equation can be classified as being elliptic. The procedure to transform the equation to canonical form can be found here. I advise you to follow that procedure and, if you do not succeed in finding a fundamental solution based on the new canonical form of the equation, post a new question on this site. 0 Not an answer but, by a series approach can you establish that$$(x+a) - (x+a)^3/3! + ( x+a)^5/5! - ... =( x - x^3/3! + x^5/5! -...) \cdot ( 1 - a ^2/2! + a^4/4! - ... ) +(1- x^2/2! +x^4/4!- ... )\cdot ( a - a ^3/3! + a^5/5! - ....) ? $$0 Separation of variables u(x,y)=X(x)Y(y). y^2 X''(x)Y(y)+x^2 X(x)Y''(y)=0 \displaystyle \frac{Y''(y)}{y^{2} Y(y)}=-\frac{X''(x)}{x^{2} X(x)}=-\frac{k^{4}}{4} \displaystyle X(x)=a\, D_{-\frac{1}{2}}(kx)+b\, D_{-\frac{1}{2}}(ikx) \displaystyle Y(y)= c\, D_{-\frac{1}{2}} \left( \frac{1+i}{\sqrt{2}} ky \right)+ d\, D_{-\frac{1}{2}} \left( ... 1 \begin{equation*} \mathbf{x}^{\prime }=\mathbf{W\cdot x\Rightarrow x}(t)=\exp [\mathbf{W}% t]\cdot \mathbf{x}(0),\;\mathbf{W}=\left( \begin{array}{cc} 1 & 3 \\ 3 & 1 \end{array} \right) \end{equation*} We can diagonalise \mathbf{W}. Thus \begin{equation*} \mathbf{W}=\left( \begin{array}{cc} 1 & 3 \\ 3 & 1 \end{array} \right) ... 1 Here's a hint: You can use the Existence and Uniqueness theorem for linear differential equations to prove many different identities. If you need to prove an identity y_1(x)=y_2(x), first prove that both y_1(x) and y_2(x) are solutions to the provided differential equation. Then show that y_1(a)=y_2(a) and y_1'(a)=y_2'(a) for a value a by using ... 1 The differential equation is linear in \theta, so that means that if you find two independent solutions \theta_1 and \theta_2, then any linear combination of those two, c_1 \theta_1 + c_2 \theta_2 is again a solution to the differential equation. In general, you will need all the linearly independent solutions in order to be able to satisfy the ... 0 I'm afraid this is a type of so-called Functional Differential Equation, a notoriously difficult type of differential equations. The source of the difficulties is that only knowing an initial condition is not enough to `evolve' the function using the differential equations. To get a flavour of the difficulties, search for 'delay differential equations', ... 0 You can suppose that u\ne0. Then you have$$1=uu' = (\frac{1}{2}u^2)'$$Integrate to get$$\frac{1}{2}u^2 = x + c$$1 Note that$$\frac{d}{dx}u^2=2u\frac{du}{dx}=2$$and so u(x)^2=2x+c, etc (you must choose the intervals of definition so that u doesn't vanish, because of the initial equation). Incidentally, the method of separation of variables is not a "leap of faith". Simply, it can be applied to almost no equation. 0 Your system is more widely known as the Gray-Scott model; a nice overview and description of the Turing instability analysis can be found here. To find out more about the relationship between the real part of the eigenvalues of the Jacobian and its trace and determinant, see my answer to this question. I believe that, with these two ingredients, you'll be ... 0 You can make an informed guess for y(t), ignoring the initial condition for the moment. Hint: try to choose y such that the second term on the right hand side vanishes. Does that choice of y satisfy the ODE? What does that mean for the associated value of a? 0 This is the improved Euler or Heun's method. Let y(x) be an exact solution of y'=f(y), y(t_0)=y_0. Then y''=f'f, y'''=f''f^2+f'^2f and$$ f(y+hf(y))=f(y)+hf'(y)f(y)+\frac{h^2}2f''(y)f(y)^2+O(h^3)=y'+hy''+O(h^2) $$so that$$ y_1=y_0+\frac h2 (f(y_0)+f(y_0+hf(y_0))\\ =y_0+hy'_0+\frac{h^2}2y''_0+O(h^3)=y(t_0+h)+O(h^3) $$This gives the local ... 1 I don't think there is relevance between the number of solutions of the linear system Ax = b and that of the system of ODE x' = Ax. For the system of ODE x' = Ax, there must be n linearly independent solutions arouse from the eigenvalues of A, irrespective of the rank of A. For completeness, I would like to add that the only relation between ... 0 Although you mention explicit Euler method, the formula as posted is an implicit finite difference approximation that is equivalent to integrating with the trapezoidal rule to solve the differential equation$$y'(x) = f[x,y(x)],$$numerically over an interval [x_n,x_{n+1}]. Use the notation h = x_{n+1} - x_n, y_n = y(x_n) and y_{n+1} = y(x_{n+1}). ... 1 The differential equation being$$y'=\frac{(x^2-y^2)}{3xy}$$you can notice that it is perfectly homogeneous and, as David Quinn already commented, define y=x z which makes y'=z+x z'. Now replace$$z+x z'=\frac{(x^2-x^2 z^2)}{3x^2z}=\frac{(1-z^2)}{3z}$$Multiply each side by 3z$$3z^2+3xzz'=1-z^24z^2+3xzz'=1$$that is to say$$4z^2+\frac32 x ... 2 You are solving the so-called wave equation. After separating variables we obtain for the space-dependent part$X(x) = C\sin\alpha x + D\cos\alpha x$. The first boundary condition yields$X(0) = 0$, and results in$D=0$. The second boundary condition becomes$X(\pi) = 0$and leads to$C\sin\alpha\pi = 0$. If$B=0$, then the solution is trivial, i.e.$X ...

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The functions $f(x) = \cos \alpha x$ and $g(x) = \sin \alpha x$ are linearly independent if and only if their Wronskian is nonzero everywhere. But the Wronskian is $\alpha$ in this particular case.

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Your proof seems correct to me. Let me suggest a simpler argument (somewhat based on yours): Since $f$ is only continuous the solutions may not be unique but still, in order to start at $k=u(a)$ and reach some bigger value at a later time necessarily $u'(t)>0$ for some $t\in(a,b)$. But since $u(b)=k$ the solution needs to return and pass in particular ...

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Let $f(x)= A\sin(\alpha x)+B\cos(\alpha x)$. We will assume that $f=0$ and arrive at a contradiction. If $f=0$ for all $x$, then so is its derivative $f'$. Therefore, we have both $$A\sin(\alpha x)+B\cos(\alpha x)=0 \tag 1$$ and $$\alpha (A\cos (\alpha x)-B\sin(\alpha x))=0 \tag 2$$ for all $x$. If $\alpha \ne 0$, then $(1)$ and $(2)$ imply ...

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Here's a quick proof I made up: Say $A \cos \alpha x + B \sin \alpha x = 0$ with $A, B, \alpha \neq 0$. Then $\cos \alpha x = -(B/A) \sin \alpha x$. But $\cos \alpha x$ attains its maximum at $0$ while $-(B/A) \sin \alpha x$ attains its maximum at $\pi/(2\alpha)$, and not at $0$. This is a contradiction, so $A \cos \alpha x + B \sin \alpha x \neq 0$.

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That is because $\alpha$,$A$ and $B$ are constants, and when since $x$ can be arbitrary, the expression $Acos(\alpha x)+B sin(\alpha x)$ can not be zero for all $x$.

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If $Y$ is the solution of $L[y]=0$, then differentiating the equation once again, we will have that $Y'$ is the solution of $$ay''+by'+cy=0.$$ But I have shown for your previous question, that it implies $Y'\rightarrow0$ as $t\rightarrow\infty$. Similarly, $Y''\rightarrow0$ as $t\rightarrow\infty$. Now, $L[y]=0$ implies $$cy=d-ay''-by'$$ So, as $t ... 0 Hint: in this case your general solution has the form $$y_g =y_h+y_p$$ Where$y_h$is the solution of homogenous eq. (your previous problem) and$y_p$is the particular solution. 2 We need to prove that the solution$Y$of$L[y]=0$tends to$0$as$t$goes to infinity. Note that$ay''+by'+cy=0$has characteristic equation$ar^2+br+c=0$. Since$a,b,c$are positive numbers, the solutions$r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$have negative real part. So the solution is either$Y(t)=C_1e^{\alpha t}+C_2e^{\beta t}$(with$\alpha, ...

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As you have a "nice form" differential equation here, you can use Separation of Variables and integration to solve. Note that $$\frac{dP}{dt}=0.4873P^2$$ can be written as $$\frac{dP}{P^2}=0.4873dt$$ Now you can integrate both sides $$\int\frac{dP}{P^2}=\int0.4873dt$$ Thus $$-\frac{1}{P}+C_P=0.4873t+C_t$$ To finish, note that $C_P$ and $C_t$ are ...

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Most of the theory that is now taught actually grew out abstract semigroups, starting with the work of Oliver Heaviside in signal analysis. The semigroups didn't come later; they came first. Heaviside reasoned very abstractly that if you start with the state of a fixed linear circuit at $t=0$, say $\phi_0$, and you evolve $t$ seconds into the future to ...

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For nonlinear equations, you can't do "eigenfunction expansions" because there is no superposition principle: a linear combination of solutions is not likely to be a solution. So you can't build up a solution from simpler solutions. About the only hope, unless you can somehow transform the problem to a linear one, is that you can solve the whole problem in ...

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Your solution is wrong because It is not a solution of the differential equation. It's always a good idea to check your solution by plugging it in to the differential equation and seeing if the two sides are equal. Its derivative is not continuous at $t=\pi$.

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