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0

You are starting with 2 second order equations in two unknowns. In effect, you have a single vector ODE of the form v'' = some function of v, where v = (p,q). What you want is something of the form w' = a function of w. Try letting w = (p, q, u, s), where the new variables u and s are u = p' and s = q'.


0

As has been stated the first condition is $100 = T(0)$. From the solution of the first order equation it is known that $T(t) = c_{0} \, e^{-k \, t}$. Now, $T(0) = 100 = c_{0}$ for which $T(t) = 100 \, e^{- k \, t}$. Now using the the given time and value it is determined that $T(4) = 60 = 100 \, e^{- 4 k}$ which provides $e^{4k} = \frac{5}{3}$ and taking the ...


0

As pointed out in the comments above, there is one piece of information you haven't used, yet: that the water was boiling initially. From this, we have $100=T(0)=C,$ and knowing $C$ allows us to then solve for $k$ by the work you've done so far, whence we can find $T(8).$ Note that we can (and should) find an exact value of $k$, instead of a numerical ...


0

To some extent, the operator norm is just a way to define a useful structure on the set of linear operators. And, as you've already mentioned, this structure resembles usual Euclidean space: you can add and subtract two operators, multiply them by scalar and measure "how big" is this operator. This is just called a normed vector space. Why one might need ...


0

The solution can be expressed on the form of the antiderivative of a particular hypergeometric function. The most likely, there is no simpler closed form in terms of a finite number of elementary and standard special functions.


0

(1) gives $\begin{align} &p< \frac{X(p+i)-X(p)}{X(p+i)-X(p)-(Y(p+i)-Y(p))} < p+i\Leftrightarrow \\ &p< \dfrac{\frac{X(p+i)-X(p)}i}{\frac{X(p+i)-X(p)}i-\left(\frac{Y(p+i)-Y(p)}i\right)} < p+i \end{align}$ : $$X^\prime(p)/(X^\prime(p)-Y^\prime(p))=p \text{ as } i\rightarrow 0$$ $$(2)\Leftrightarrow \frac{X(p+i+j)-X(p+i)}{X(p+i)-X(p)} ...


1

Given your definition of y(x), we have that: $$y'(x):$$$$=-4x^3, x < 0$$ $$=4x^3, x \geq 0$$ substitute these values into $xy'=4y$ for $x<0$ you have $x(-4x^3)=4(-x^4)$ which reduces to $x^4=x^4$ you take the same steps to verify for $x>0$.


2

Here's a hint: Let $(\lambda,x)$ be an eigenpair of $A$ and consider $$ \exp(A)x = \sum_{n=0}^\infty \frac{A^n x}{n!} = \sum_{n=0}^\infty \frac{\lambda^n x}{n!} $$


2

$\phi$ is actually defined on $\mathbb R^2$; the author could have written $$ \phi: \mathbb R^2 \to \mathbb R: (a, b) \mapsto a^2 + a b^2 $$ The problem is that our notation for derivatives depends on the choice of "dummy variables" that we use in denoting a function (sigh...on the good side, it often makes the chain rule work out in really nifty ways). ...


0

It's saying that if x and y (or their derivatives) are multiplied together in any way, it's not considered a linear differential equation because it's not solvable in the usual ways that linear ODE's are. This relates to normal linear equations in that if you have an equation where x and y are multiplied or otherwise modify each other in a way that prevents ...


0

(1)For any $x$ with $||x||\leq 1$ there exists $y$ with $||y||=1$ and $|T(x)|\leq |T(y)|$. PROOF.If $x=0$ let $y$ be any vector of norm $1$ because $T(0)=0$. If $ 0<||x||\leq 1 $ let y=x/||x|| (Because $||x||\leq 1 \implies |T(y)|=|T(x)|/||x|| \geq |T(x)|$.) .....................(2) Therefore $$||T||= \sup \{|T(x)| : ||x||\leq 1\} \leq \sup \{|T(y)| : ...


0

Firstly, you define the Hamiltonian of the system, which corresponds to the total energy, that is, kinetic energy plus potential energy. In your case it reads $H(p,q) = \frac{p^2}{2} + U(q)$, where $p$ is the momentum and $q=\binom{x_1}{x_2}$ is the position (note that $p$ and $q$ are vectors). Then, you derive the equation of motion by computing $\dot{p} ...


3

There are multiple ways, but I'll give you one. From writing out the Taylor expansions of $f(x+\Delta x)$ and $f(x-\Delta x)$ you can show that we can approximate the second derivative in $\mathbb{R}^2$ by $$\frac{h(x+\Delta x,t)-2h(x,t)+h(x-\Delta x,t)}{(\Delta x)^2}=h_{xx}(x,t)+O(\Delta x^2)$$ So, substituting in we get $$\frac{h(x+\Delta ...


1

Fourier transforms is a perfectly fine tool to solve the system. The advantage of using it is that you avoid having to solve two equations, one for $x<0$ and one for $x>0$, and then matching at $x=0$. The drawback is that you will have to compute the inverse transform which is usually a bit more work than solving and performing the matching if you have ...


0

Let's assume that $f$ has, at least locally, an inverse function $g$ so that $g(f(x))=x$ and $f(g(y))=y$, so that $g'(f(x))·f'(x)=1$ and $f'(g(y))·g'(y)=1$. Then $$ \frac{d}{dy}\,z(g(y))=z'(g(y))·g'(y)=f''(g(y))·\frac{1}{f'(g(y))}=\frac{d}{dx}\,\ln(f'(x))\Big|_{x=g(y)} $$ if indeed this is what you were aiming for.


1

$x$-absent second order differential equation is solved by the substitution $y′=u$ and $y′′=u\frac{du}{dy}$. Actually, this is false. First of all, if $y'=u$, then $y'' = (y')' = u'$, not what you wrote. Also, the substitution $y'=u$ is useful if the differential equation is $y$ absent, not $x$ absent. For example, the equation $y'' = y'x$ can be ...


0

$$ \left\{\begin{matrix} x=\rho \cos(\theta) \rightarrow x' =\rho'\cos(\theta) -\rho \sin(\theta) \theta' & \\ y=\rho \cos(\theta) \rightarrow y' =\rho'\sin(\theta) +\rho \cos(\theta) \theta' & \end{matrix}\right. $$ $$ \left\{\begin{matrix} x'= (1-x^2-y^2)x - y = \rho'\cos(\theta) -\rho \sin(\theta) \theta' = (1-\rho^2)\rho \cos(\theta) - \rho ...


-1

From the Schrodinger equation $$\alpha \psi''(x) + (E + \delta(x)) \psi(x) = 0$$ and \begin{align} &\mathcal{F}\{ \alpha \psi''(x) \} = -4\alpha \pi^2 p^2 \hat{\psi}(p)\\ &\mathcal{F}\{ E \psi(x) \} = E \hat{\psi}(p) \\ &\mathcal{F}\{ \delta(x)\psi(x) \} = \psi(0) \\ &\mathcal{F} \{0 \} = 0 \end{align} then it is seen that: $$(1 + E -4 \alpha ...


0

$x^{'} (t) = 3(x (t)-2) ; \quad x(0)=-1 $ The differential equation can be written equiavalently as $$ x^{'} (t) -3x(t)=-6$$ Now multiplying both sides of the above with the "integrating factor" $ e^{-3t}$ we obtain: $$ x^{'} (t) e^{-3t} -3e^{-3t}x(t)= -6e^{-3t} $$ $$ (x(t) e^{-3t})^{'} = (2 e^{-3t})^{'} $$ Hence, $x(t)e^{-3t} = 2e^{-3t} +c$, where $c ...


2

Hint. What you have done is not correct. Observe that $$ \int 3 \:dt= 3t+C, \quad C \, \text{is a constant}. $$ Here you have $$ \int\frac{dx}{x-2}=\int 3 \:dt \tag1 $$ or $$ \ln|x-2|=3t+C $$$$ x(t)-2=Ke^{3t}. \tag2 $$ You obtain $K$ by putting $t=0$ in $(2)$, $$ x(0)-2=K $$$$ -1-2=K. $$ Then $$ x(t)=2-3e^{3t}. $$


1

To verify that $x=(t+1)e^{2t}$ is a solution to $$x'=2x+e^{2t},\hspace{10mm}x(0)=1$$ you don't need to do any integration. All you have to do is verify the initial condition and that the two sides of the differential equation are the same: Verifying $x(0)=0$: $$x(0)=(0+1)e^{2(0)}=e^0=1$$ Calculation of $x'$: ...


1

$$x(t)=(t+1)(e^{2t})$$ $$x'(t)=e^{2t}+(2t+2)e^{2t}=2(t+1)e^{2t}+e^{2t}=2x(t)+e^{2t}$$ $$x(0)=1\times e^{0}=1$$


0

Because this is a 2nd order system, you probably cannot find an $u$ that transfers the states to the desired value in 1 step, which you have failed to do. Since this system is reachable and 2nd order, you can reach any state at most 2 steps, i.e. using 2 constant values of $u$ (and I think this is implied by "piecewise constant"). In order to find such $u$, ...


0

Clearly this bound may not exist if the $e_{ij}$ are too large since the equations may explode, so the suitable condition to prevent this is suppose to be $\Pi e_{ij} < \Pi c_{ij}$. What that should do is somehow create enough push from the negative terms to prevent explosion. The condition $\Pi e_{ij} < \Pi c_{ij}$ is not sufficient. For ...


0

I would suggest doing it one time interval at a time to keep your work neat. It gets quite messy with all the partial fractions sometimes. Start with the interval $0<t<3$, with $y(0)=0$ and $y'(0)=1$ as initial conditions. $\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y=0$ $(s^2Y-sy(0)-y'(0))-3(sY-y(0))+2Y=0$ $Y={1 \over s^2-3s+2}=\frac{1}{s-2}-\frac{1}{s-1}$ ...


0

I don't agree with your $Y(s)$. Here comes a suggestion on how to get $Y(s)$ and on how to get your function back in the end: Taking Laplace transform of the left-hand side, invoking the initial value conditions, gives $$ (s^2-3s+2)Y(s)-1. $$ The right-hand side can be written $$ 12\theta(t-3)e^{-2t}. $$ The Laplace transform of $12\theta(t)$ is $12/s$, and ...


2

Differential-algebraic equations are important for mathematical modeling and scientific computation. If you write down the mathematical laws for some chemical, electrical, or physical system, you often will just end up with a system of equations involving parameters, various partial derivatives and purely algebraic quantities. Maybe you also get some ...


0

This is already solved in a satisfactory way, by viewing $x$ as a function of $y$ instead. Here is a solution that does not. If we let $$ z=(y-x)^2, $$ (the square fit well with the $2$) then, by differentiation, and by using the differential equation $$ z'=2(y-x)(y'-1)=\frac{y^2}{x}-2(y-x)=\frac{1}{x}(y^2-2yx+x^2)+x=\frac{z}{x}+x. $$ This differential ...


3

$$y'=\frac { y^{ 2 } }{ 2x(y-x) } \\$$ Solve respect to the $x $ $$\\ \\ { y }^{ 2 }{ x }^{ \prime }-2x(y-x)=0\\ \\ { x }^{ \prime }-2\frac { x }{ y } +2\frac { { x }^{ 2 } }{ { y }^{ 2 } } =0\\ x=zy\\ { x }^{ \prime }=z+y{ z }^{ \prime }\\ z+y{ z }^{ \prime }-2z+2{ z }^{ 2 }=0\\ y{ z }^{ \prime }-z+2{ z }^{ 2 }=0\\ y{ z }^{ \prime }=z\left( 1-2z ...


1

Maple solves this one using Heun C functions: $$ y \left( x \right) =c_{{1}} \left( {x}^{2}-1 \right) {\it HeunC} \left( 0,-1/2,1,-1/4,1/2,{x}^{2} \right) +c_{{2}} \left( {x}^{3}-x \right) {\it HeunC} \left( 0,1/2,1,-1/4,1/2,{x}^{2} \right) $$ I doubt that there are closed-form solutions in elementary functions.


2

First, we search one particular solution easy to find, on the form $\Phi_p(x)$ : $$x\frac{d\Phi_p}{dx}+(\alpha+1-x)\Phi_p(x)=0$$ $$\Phi_p(x)\: e^x x^{-\alpha-1}$$ Then we change of function : $\Phi(x,y)= F(x,y)\Phi_p(x)= F(x,y)e^x x^{-\alpha-1}$ $\frac{\partial \Phi}{\partial x} =\frac{\partial F}{\partial x}e^x x^{-\alpha-1}+F(x,y)e^x ...


1

One of the simple methods is separation of variables which looks for a solution of the form $$ \phi(x,y) = f(x)\, g(y) $$ Putting this into the PDE gives $$ x\, f'(x)\, g(y) + y\, f(x)\, g'(y) + (\alpha + 1 - x)\, f(x)\, g(y) = 0 $$ Next is assuming $f(x) \ne 0$ and $g(y) \ne 0$ and for these $(x,y)$ sorting the dependencies on $x$ to the left and those on ...


1

Problem 1. Observe from the initial equation that a constant $f$ is a solution. Now if $f$ is not constant then $f'(r) \ne 0$ for some $r>0$ and hence $f'(s) \ne 0$ for all $s$ in some open interval $J$ containing $ r$. So proceed to work within the largest possible such $J$ to find the the consequences of a non-constant $f$. Problem 2 is not a ...


1

A slightly different approach (in attachment) leads to : $$f(r)=\frac{1}{c_1}\ln \left| c_1r \pm \sqrt{c_1^2r^2-1}\right|+c_2$$ which includes all real solutions, not forgeting $f(r)=c_2$ at limit $c_1 \rightarrow \infty.$


0

Assuming we're talking about complex vector spaces: The eigenvalues of $A$ have real parts all positive. The eigenvalues of $A$ have real parts all greater $1$. The proof is by rewriting it using the Jordan normal form. It would reduce to a system of independent equation in one dimension (unless you have eigenvalues with multiple independent ...


0

So I'll admit, I didn't know nearly enough to answer this question. However curiosity got the best of me, so here it is. Indeed, the given ODE is a Bessel's equation of the first kind, we can write this in sturmian form as $$ \begin{cases} (xy)'+(x-\frac{1}{x})y = 1\\ y(0)=y(b)=0 \end{cases} $$ In order to construct a Green's Function for the problem, ...


0

In polar coordinates $(r,\phi)$ the equation becomes $\partial u/\partial r=0$, so any function of $\phi$ alone will be a solution. Now it's only a matter of finding some function of $\phi$ which is well-behaved on the whole punctured plane and translating it back to the $(x,y)$ variables. Can you take it from there?


0

As commented by Michael, the equation is separable and can write $$\frac{dt}{dy}=-\frac {81}{5b}\times\frac{1}{y^3-12 y^2-6 y+\frac{4}{5}}$$ Using the trigonometric resolution of the cubic, the roots of $y^3-12 y^2-6 y+\frac{4}{5}=0$ are given by $$y1=4+6 \sqrt{2} \cos \left(\frac{1}{3} \tan ^{-1}\left(\frac{1}{7}\right)\right)$$ $$y_2=4+3 \sqrt{6} \sin ...


0

You can always check if your answer is a solution to an ODE (or PDE for that case) by simply taking the appropriate derivatives and plugging them into the initial ODE. In your case $$ y(x) = 2 \implies y'(x) = 0 $$ Therefore, your ODE becomes $$ y' + y = 2 \implies 0 + 2 = 2 $$ which is true for all $x$. Thus the solution you found is correct.


1

The discretization comes from the solution to the full first-order ODE. You have $$ v' + \epsilon v = f$$ Using an integrating factor, the general solution to this has the form $$ v(t) = e^{-\epsilon t} \left( v(0) + \int_0^{t} e^{\epsilon \tau} f(\tau) \, d\tau \right) $$ For small time, we might approximate $f$ as a constant, in which case the above ...


1

Subtracting $ v(t) $ from either side gives: $$ v(t+1) - v(t) = v(t) \exp(-\epsilon) - v(t) + \frac{f(t)}{\epsilon}( 1 - \exp(- \epsilon ) ) $$ As $ \epsilon \rightarrow 0 $, taylor expanding the exponentials, $ \exp(-\epsilon) = 1 - \epsilon + \epsilon^2 / 2 \cdots $, gives: $$ v(t+1) - v(t) = -\epsilon v(t) + f(t) + O(\epsilon) $$ The term $ v(t+1) - v(t) ...


1

The general first order linear equation can be solved with quadratures. (But the result is possibly given implicitly.) The general second order linear equation cannot be solve with quadratures. Look for the subject "differential algebra" to find where such things may be proved.


1

Remember that an expression with $n$ arbitrary constants will yield a differential equation of order $n$. So to get the $n^{th}$ order derivative you'll have to differentiate the expression $n$ times, and in that process you'll obtain $n$ more relations so that now you have a total of $n+1$ relations from which you can eliminate the $n$ arbitrary constants ...


2

The zero function does not satisfy that differential equation, since $$\frac{dy}{dx} = 0 \neq x$$ on all open intervals that include $0$. It satisfies the differential equation $$\frac{dy}{dx} = y$$ but that's a whole other story...


0

HINT $dz = (6x+3y)dx+(3x-4y)dy=0,\quad \frac{\partial (6x+3y)}{\partial y} = \frac{\partial (3x-4y)}{\partial x}$ $\Rightarrow \,dz\,$ is the total differential of function $z = f(x,y),$ I would say, find the function $z = f(x,y)$: $\frac{\partial z}{\partial x}= 6x+3y \Rightarrow \, z = \int (6x+3y)\,dx=3x^2+3xy+\phi(y)$ $\frac{\partial z}{\partial y}= ...


0

Without the forced vibration, the period would be $T_1=\frac{2\pi}n$. With the forced vibration the period becomes $T_2=\frac{2\pi}p$. This is a simplification, but close enough to be useful. Set $\frac{T_2}{T_2}=\frac52$ and square. Squaring is necessary because we know about the magnitude; $p$ or $n$ could be negative.


3

I found out why I am wrong according to http://lpsa.swarthmore.edu/BackGround/PartialFraction/RootsRepeat.html For any multiply repeated roots (n>2), the coefficient $A_k$ should be: $$ A_k=\frac{1}{(k-1)!}\Big[\frac{d^{(k-1)}}{ds^{(k-1)}}(s+a)^nF(s)\Big]_{s=-a} $$ Therefore, $C$ should be: $$ C=\frac{1}{2}\Big[\frac{d^2}{ds^2}s^3X(s)\Big]_{s=0} $$


0

Coming back to this, I wanted to say that this problem can actually be solved by doing an odd reflection through the origin, as you appear to be trying to solve the heat equation in the half open domain $[0,\infty)$. Consider first letting $v(y,t) = u(y,t) - u_0$. Here I'll be assuming $u_0 \in \mathbb{R}$ since you didn't specify. Then our new $v$ can solve ...


2

Let's look at the second order case first: $x''+ax'+bx=0$ and $x_{1}$ is a solution. Then, we seek a $v$ such that $x_{2}=x_{1}v$ is a solution. The requirement is then that $(x_{1}'v+x_{1}v')'+a(x_{1}'v+x_{1}v')+bx_{1}v=0\Rightarrow $ $x_{1}''v+x_{1}v'+x_{1}'v'+x_{1}v''+a(x_{1}'v+x_{1}v')+bx_{1}v=0\Rightarrow $ ...


1

In the second method, $$-4u(t-6)=\begin{cases} 0 , & t<6 \\ -4 , & t\ge 6\end{cases}$$ but you want a function behaves like $$\begin{cases} -4 , & t<6 \\ 0 , & t\ge 6\end{cases}$$ It sholud be ...



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