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1

I wouldn't say that it always does. But, if a numerical method can't even handle a linear problem, then it's probably going to have trouble on a more significant problem. Generally, the different types of stability (zero stability, absolute stability, etc.) are more so a means of crossing methods off the list, so to speak, or to show what context the method ...


0

This actually still is a first-order DE. The order of a DE has absolutely nothing to do with anything happening to the functions, and everything to do with the orders of the derivatives that appear in the DE's. In your case, the first DE is separable: \begin{align*} A^{-1.5}\,dA&=-k_1\,dt \\ \frac{A^{-0.5}}{-0.5}&=-k_1t+C \\ \frac{1}{\sqrt{A}}&=0....


0

You can write $$\frac{dA}{A^{3/2}}=-k_1dt$$ and integrate. An we get by integrating $$\frac{1}{\sqrt{A(t)}}-\frac{1}{2}k_1t-C=0$$ Solve this equation for $$A(t)$$ and plug this into the second equation. It is $$A(t)=\frac{1}{(\frac{1}{2}k_1t+C)^2}$$ and also Maple can find an answer.


0

Assume that $x$ is only a function of $W(t)$, so $x(W(t))$. Differentiating this yields \begin{align} \frac{dx}{dt} &= \frac{\partial x}{\partial W} \frac{dW}{dt} = \frac{\partial x}{\partial W} i\,u(t)\,W(t), \\ &= u(t)\,x(W(t)). \end{align} This can be simplified to $$ \frac{\partial x}{\partial W} i\,W(t) = x(W(t)). $$ This can be solved ...


1

Why, many. $f(x)={x^2\over2}$ will do. If you want to exclude the "trick" answers where $f'(x)=x$ or $f(x)=f'(x)$, go with $f(x)={x^3\over9}$. Other examples are still plenty, I believe. Upd. The other examples seem less numerous than I initially believed, but anyway, $f(x)=\dfrac{x^n}{n^{n-1}}$ for any constant $n$ is good.


3

For the real-valued function $f(x)=e^x$, $f'(x)=e^x$.


0

I do not see any particular reason why you would want to transform your differential equation. Why not solve it in the given formulation? If $u(t)$ is a known input you can solve the first equation $$\dfrac{dW}{W} = i u(t) dt \implies \ln |W|=\int_{t_0}^{t}iu(\tau)d \tau$$ $$\implies |W(t)| = \exp\left[ \int_{t_0}^{t}iu(\tau)d \tau\right]$$ $$\implies W(t) ...


2

The characteristic polynomial of this differential equation has roots $a,b,c$ and thus is $$(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc,$$ corresponding to the differential equation $$ y'''=(a+b+c)y''-(ab+bc+ca)y'+abcy. $$


0

$$y′′+y′−6y=αe^{rt}\implies (D^2+D-6)y=αe^{rt}$$ Complementary finction (C.F.) is $\quad Ae^{2t}+Be^{-3t}\quad$where $A,B$ are arbitrary constants. Now particular integral (P.I.)$=\frac{1}{D^2+D-6} αe^{rt}$ $=\frac{αe^{rt}}{r^2+r-6}$ Consider a differential equation of the form $f(D)y=X$ If $X=e^{ax}$, then $1.$ P.I.$\quad = \frac{1}{f(D)}e^{...


2

HINT Solve the homogeneous equation, assuming a solution of the form $y=e^{at}$, which yields $a^2+a-6=0$. Pick a particular solution of the RHS, which should be $Ae^{rt}$ if $r$ does not solve the above equation, otherwise, $Ate^{rt}$, and plug in to find $A$. Combine the solutions, by adding the homogeneous and inhomogeneous together.


2

Hint: If $r\ne2$ and $r\neq-3$ then let $y=Ae^{rt}$ and find $A$ by substuting in the DE, otherwise let $y=Ate^{rt}$.


1

As suggested by metamorphy in the comments above, we set $y = z^2$, in which case $y' = 2 \, z \, z'$, and this DE becomes linear: $$ (1+x^2) z' - x \, z \;=\; 2\sqrt{1+x^2}\; \arctan(x) $$ Next, we define $z = \sqrt{1+x^2}\; w$, so that $w = \sqrt{y\, /\, (1+x^2)}$. Then $$ z' \;=\; \frac{x}{\sqrt{1+x^2}}\, w \;+\; \sqrt{1+x^2}\, w'\, , $$ and the DE ...


2

Let $y = z^2$. We have: $$ 2(x^2 + 1) z z' - 2x z^2 = 4 z \sqrt{x^2+1} \arctan(x)$$ $$ z' - \frac{xz}{x^2 + 1} = 2 \frac{1}{\sqrt{x^2+1}}\arctan(x)$$ Homogeneous equation: $$z' = \frac{xz}{x^2+1} $$ Therefore, $z = c \sqrt{x^2+1}$ Variation of constant: $$c' \sqrt{x^2+1} + c \frac{x}{\sqrt{x^2+1}} - c \frac{x}{\sqrt{x^2+1}} = 2 \frac{1}{\sqrt{x^2+1}}...


0

Maple provides a solution in implicit form... $$ \ln \left( \sqrt {xy \left( x \right) }+x \right) +3\,\ln \left( \sqrt {xy \left( x \right) }-x \right) -2\,{\frac {x}{\sqrt {xy \left( x \right) }-x}}-2\,\ln \left( x \right) - C_1=0 $$


4

Write your equation in the form $$y'-\sqrt{\frac{y^2}{xy}}-1+\sqrt{\frac{x^2}{xy}}=0$$ and now substitute $$u=\frac{y}{x}$$


0

The periodic solution can be represented as $y=c_1ϕ_1+c_2ϕ_2$. It follows that \begin{alignat}2 y(T)&=c_1ϕ_1(T)+c_2ϕ_2(T)&&=y(0)=c_1,\\ y'(T)&=c_1ϕ_1'(T)+c_2ϕ_2'(T)&&=y'(0)=c_2. \end{alignat} This looks like an eigenvector equation and leads to a determinant condition $$ (ϕ_1(T)-1)(ϕ_2'(T)-1)-ϕ_1'(T)ϕ_2(T)=0. $$ Then apply the ...


1

Integrate $\frac{y'}{y^2}\geq 2$ from time $t=t_0$ to $t=t_1$, $t_0<t_1$ gives $$ \int_{t_0}^{t_1}\frac{y'}{y^2}\,\mathrm{d}t \geq \int_{t_0}^{t_1}2\,\mathrm{d}t $$ and $$ LHS=\int_{y(t_0)}^{y(t_1)}\frac{\mathrm{d}y}{y^2}=\frac1{y(t_0)}-\frac1{y(t_1)}. $$


0

From $\varphi=\dot\theta,$ we have \begin{align*} \ddot{r}&=r\varphi^2\\ r\dot\varphi&=\dot{r}\varphi \\ \varphi&=Cr \quad\text{solve second equation for } \varphi \\ \ddot{r}&=C^2r^3 \\ \ddot{r} \dot{r}&=C^2r^3\dot{r} \\ \frac{\dot{r}^2}{2}&=\frac{C^2r^4}{4}+B. \end{align*} The first result follows from solving for $\dot{r},$ ...


0

You are correct that $\ddot{r} \dot{r} = r \dddot{r}/3$. But it's not quite enough to show that a solution of $\dot{r} = \sqrt{A r^4 + B}$ satisfies this equation, you'd have to show that every solution of $\ddot{r} \dot{r} = r \dddot{r}/3$ satisfies $\dot{r} = \sqrt{ A r^4 + B}$ for some constants $A$ and $B$. And if $A$ and $B$ are supposed to be real, ...


5

The symbol $\delta$ is actually defined by the formal property that for any test function $\phi$, i.e., a smooth function $\phi : \mathbb{R} \to \mathbb{R}$ that vanishes outside some finite interval $[a,b]$, $$ \int_{-\infty}^\infty\delta(x)\,\phi(x)\,\mathrm{d}x = \phi(0). $$ Thus, from a rigorous mathematical standpoint, the Dirac delta is actually the ...


0

Essentially, the Delta distribution makes you evaluate the wave function $\psi (x)$ and the potential at $x=0$, in this case. The Delta distribution is indeed somewhat of an abuse of notation, but it is usually interpreted as meaning the evaluation of a function at the point the makes the argument of the Delta function equal 0. For instance, $\delta (x-a)\...


1

Hint: I think you need to assume more than that, e.g., the Wronskian $W(\phi_1,\dots,\phi_n)$ does not vanish on $I$. Then expand the ratio of Wronskians $$ \frac{W(\phi_1,\phi_2,\dots,\phi_n,y)(t)}{W(\phi_1,\phi_2,\dots,\phi_n)(t)} = \frac{\begin{vmatrix} \phi_1(t)& \phi_2(t) & \dots & \phi_n(t) & y(t)\\ \phi_1'(t) & \phi_2'(t) & \...


1

We know that if $W(x)$ be the Wronskian of the differential equation $y''+p(x)y'+q(x)y=0$, then $$W(x)=Ae^{\int p(x) dx}\qquad \text{where $A$ is constant.}$$ Now if $W(x)=constant = c$(say), then $$Ae^{\int p(x) dx}=c$$ $$\implies e^{\int p(x) dx}=d$$ $$\implies {\int p(x) dx}= \log d=k \text{(say)$\qquad$ where $k$ is constant}$$ Differentiating both side ...


1

You can directly use Abel's identity to show that if the Wronskian of any two solutions of the differential equation $y''+p(x)y'+q(x)y = 0$ (on an interval $I$) is constant, then $p(x) = 0$. (If $\int_{x_0}^{x} p(t) dt$ is constant $\forall$ $x \in I$, then $p(t) = 0 $) The answer to your last question can be found here.


2

It is ok. When you transform $x\to y$ your old equilibrium $(1,0)$ becomes $(0,0)$ for the $y$-system. You only have to analyze that point for the new system. You cannot avoid by translation having several equilibria.


2

The characteristic equation for this ODE is $$\eta^{2}+3\eta+\left(2+\lambda\right)=0$$ with roots $$\eta=\dfrac{-3\pm\sqrt{1-4\lambda}}{2}=\dfrac{-3\pm i\sqrt{4\lambda-1}}{2}$$ Note that I wrote this with an imaginary part because your solution must be oscillatory if you want to satisfy the boundary conditions. From the boundary condition at $x=0$ you'...


4

$\frac {dy} {y+y^{2}}=2tdt$. So $\int \frac 1 {y+y^{2}}dy=t^{2}+C$. This can be written as $\int [\frac 1 y -\frac 1 {1+y}]dy=t^{2}+C$ or $\log \frac y {1+y}=t^{2}+C$. Hence $y=\frac {e^{t^{2}+C}}{1-e^{t^{2}+C}}$.


2

Hint: You have $$ \frac{1}{y(y+1)}\,y' = 2t$$


2

Assuming $y\neq 0,-1$. It is a separable equation $$ \frac{\mathrm{d}y}{y+y^2}=2t\,\mathrm{d}t. $$


1

This system is just a collection of 2 uncoupled ODEs, and so can be solved separately, just as we would a simple 1st order ODE.


1

These are two unrelated DE's and you have to solve them independently. The answer is $x(t)=ce^{-t}$ and $y=de^{-2t}$ where $c$ and $d$ are constants.


0

$$u''+2u'+u=e^{-t}\implies (D^2+2D+1)u=e^{-t}$$where $D\equiv\frac{d}{dt}$ The roots of the trial solution be $m=-1,-1$ So the complementary function (C.F.) is $\quad (A+Bt)e^{-t}\quad$ where $A,B$ are arbitrary constants. Particular integral (P.I.)$\quad=\frac{1}{D^2+2D+1}e^{-t}$ $=\frac{1}{(D+1)^2}e^{-t}$ $=\frac{t^2}{2}e^{-t}$ So the general ...


0

$$\ y'' + y = \sin(x) + \cos(2x)\implies (D^2+1)y=\sin(x) + \cos(2x)$$where $D\equiv\frac{d}{dx}$ Complementary function (C.F.) $\quad = A\cos x+ B \sin x\qquad$where $A,B$ are arbitrary constants. Particular Integral (P.I.)$\quad = \frac{1}{D^2+1} (\sin(x) + \cos(2x))$ $=\frac{x}{2D}\sin x-\frac{1}{3}\cos(2x)$ $=-\frac{x}{2}\cos x-\frac{1}{3}\cos(2x)$ ...


2

Note that in general we solve initial value problems. The reason for adding the homogenous solution to the particular solution is to find the general solution and be able to find solutions which satisfy any given initial conditions. A particular solution satisfies a particular initial condition which is not necessarily what we are looking for.


0

Any eigenfunction must be a non-zero constant multiple of the solution of $$ y''+\lambda y =0,\;\; y(0)=1,\; y'(0)=0, $$ which has unique solution $$ y(x) = \cos(\sqrt{\lambda}x). $$ Then, in order to satisfy $y'(2)=0$, $\lambda$ must satisfy $$ \sin(2\sqrt{\lambda})\sqrt{\lambda}=0. $$ $\lambda=0$ is a such a solution. The ...


1

Your error is in the form of the particular solution. Homogeneous equation: $$y''+y=0$$ ...has solution $y=C_1\sin x + C_2\cos x$ So your particular solution should be: $$y_p=Ax\sin x+Bx\sin x +C \sin(2x)+D\cos(2x)$$


2

For the particular solution you need $$ Ax \sin x +Bx \cos x +C\sin(2x)+D\cos(2x)$$ The cconstants are found by plugging this solution in your equation and identifying the coefficients on both sides.


1

Denote with $v_i=x_i'$, $i=1,2$ Then your equation becomes $2v_1'=-4x_1+3x_2$ and $\frac{9}{4}v_2'=-\frac{27}{4}x_2+3x_1$ Therefore your four equations are $$ 2v_1'=-4x_1+3x_2$$ $$ \frac{9}{4}v_2'=-\frac{27}{4}x_2+3x_1$$ $$x_1'=v_1$$ $$x_2'=v_2$$ The coefficient matrix is $$\begin{bmatrix} -2 & \frac{3}{2} & 0 & 0 \\ \frac{4}{3} & -3 &...


1

You need a left and right solution, the left satisfying the left boundary condition, the right the right. With some linear equations solving these should turn out as $$ y_1(x)=A\cosh(x)~~\text{ and }~~y_2(x)=B\sinh(\pi-x) $$ These then get combined into a solution to $$ y''-y=\delta(x-s) $$ meaning that $$ y_1(s^-)=y_2(s^+)~~\text{ and }~~y_1'(s^-)+1=y_2'(s^...


1

You seem to know that you multiply the $e^{3x}$ by $x$ because $e^{3x}$ is part of the homogeneous solution. You have to do the same thing with $1 = e^{0x}$. Part of the forcing function is $-18x$. The derivatives of $x$ are $x$ and $1$, and $1$ is the other homogeneous solution. So your form for that part of the particular solution gets multiplied by $x$...


0

Since $y_h=C_1$ is a solution to the homogeneous part, you need to use $y_p=\alpha x e^{3x}+Ax^2+Bx$ instead of $xe^{3x}+Ax+B$ for the particular solution.


1

A picture is worth a thousand words (taken from Wikipedia) This is of course just an illustration, and in your case the actual plot is different. The general solution to the ODE $\dot{x}=Ax$ is given by $$x\left(t\right)=C_{1}v_{1}e^{\lambda_{1}t}+C_{2}v_{2}e^{\lambda_{2}t}$$ where $\lambda_{1,2}$ are the eigenvalues of $A$ associated with the ...


1

Since $$\left(g^{\mu\nu}-\frac{k^\mu k^\nu}{k^2-m^2}\right)\left(g_{\nu\rho}+Ck_\nu k_\rho\right)=\delta^\mu_\rho+k^\mu k_\rho\left(C\left(1-\frac{k^2}{k^2-m^2}\right)-\frac{1}{k^2-m^2}\right)=\delta^\mu_\rho$$simplifies to $\delta^\mu_\rho$ if $C=-\frac{1}{m^2}$,$$D(k)=-\frac{i}{k^2-m^2}\left(g_{\nu\rho}-\frac{1}{m^2}k_\nu k_\rho\right).$$


0

$\frac{x^{5/2}}{c - 2*\sqrt{x}} \quad$ is solution of $\quad\frac{dy}{dx}=\frac{y^2}{x^3}+\frac{5y}{2x}\quad$ and is not solution of $\quad\frac{dy}{dx}+\frac{y^2}{x^3}+\frac{y}{2x}+\frac{x}{2}=0$. So, necessarily there is a typo.


1

Hint: $|(a+b)-(c+d)|=|(a-c)+(b-d)| \leq |a-c| +|b-d|$. (Your first step makes it impossible to complete the proof). [Take $a$ and $b$ to be the first and second terms of $x(t)$ and $c$ and $d$ to be the first and second terms of $y(t)$].


0

This is really a matter of applying the multivariable chain rule, and being clear on what your notation means. Hopefully you know that $\partial_tF$ means the function you get when you differentiate $F$ with respect to its first argument, and $(\partial_tF)(t,x(t))$ means the value of this function when you evaluate it at the point $(t,x(t))$, and likewise ...


1

You could take them equal, then the right side is zero. However, this is only one special case, there is nothing in the task description that links the birth and death rate.


2

The book's answer is quite mystifying and possibly wrong. Finding the point on the graph where $dT/dt = -1$ by eye is extremely hard, especially because the two axes have different scales, so you're not looking simply for a place in the figure where the slope is $45^\circ$ visually. And even if you are looking for the right slope, the whole computation ...


1

Consider $F$ as a function of $t$ and $x$: then we have $$ \begin{split} x''(t)&=\lim_{h\to 0}\frac{x'(t+h)-x'(t)}{h}=\lim_{h\to 0} \frac{F(t+h,x(t+h))-F(t,x(t))}{h}\\ &=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))+F(t,x(t+h))-F(t,x(t))}{h}\\ &=\lim_{h\to 0} \frac{F(t+h,x(t+h))- F(t,x(t+h))}{h}+\lim_{h\to 0}\frac{F(t,x(t+h))-F(t,x(t))}{h}\\ &...


1

Use Legendre's Substitution. $$x = e^t$$ Let $\frac{d}{dx} = D_x$ and $\frac{d}{dt} = D$ Now, $\bullet \ xD_{\small{x}} =D$ $\bullet \ x^2D^2_{\small{x}} = D(D-1)$ $\bullet \ x^3D^3_{\small{x}} = D(D-1)(D-2)$ So we have $$[D(D-1)(D-2) + 6D(D-1) +4D]y = 0$$ $$\implies D^2(D+3)y = 0$$ Solve this in terms of $t$ and finally substitute $t= \ln x$


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