# Tag Info

0

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 ...

3

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

0

$$4 dx + 2 {\cos(y)\over \sin(y)} dy = 0$$ 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 ...

0

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

1

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{...

4

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(...

0

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 ...

1

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,...

1

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 ...

31

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.

4


0

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 ...

0

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 $2(\gamma_x(t),\gamma_y(t))=P_x(t)+P_y(t)$, ...

1

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))+\...

3

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 ...

3

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 $B=0,E=0,C=1,A=-1,D=1$

1

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 ...

0

1

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 $$w'_x = \frac{1}{y}y'+\frac{x}{y}y''-\frac{x}{y^2}(y')^2,$$ $$w'_x = w_z'\frac{dz}{dx}=w'_z (3\frac{x^2}{y^3}-3\frac{x^3}{y^4}y'_x),$$ $$xw_x'=3w_z'(z-zw),$$ $$xw_x'=\frac{x}{y}y'+\frac{x^2}{y}y''-\frac{x^2}{y^2}(y')^2=w+z-w^2,$$...

1

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$is an algebraic equation. The unknown is a number$...

1


0

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 $j\ne k,$ ...

1

$$(h-1)h'=2$$ 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.

1

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{...

Top 50 recent answers are included