Tag Info

New answers tagged

0

First off, your result should have some absolute value bars around it: $$y = \ln|x+1| +C$$ Other than that you are correct. Next step is to utilize the fact that $y(0)=0$. What this is saying is that at $x = 0$, $y = \ln|x+1| +C = 0$. $$y = \ln|0+1| +C$$ $$ = \ln{1} +C$$ $$ = 0+C = 0$$ Thus $C = 0$. Furthermore since we now know $y = \ln|x+1|$, ...


4

A first good step is multiplying by $2\dot{\theta}$, to get $$ \frac{d}{dt} \dot{\theta}^2=\sin \theta \dot{\theta} $$ which can be integrated to $$\dot{\theta}^2=-\cos \theta+A,$$ where $A$ is a constant. Thus $$\frac{d \theta}{ \sqrt{A-\cos \theta}}= \pm dt $$ and a second integration gives $$F_A(\theta)=\pm(t-t_0) $$ where $F_A$ is a primitive ...


0

$$\frac{d^2v}{dx^2}-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$ This is an autonomous ODE. So, the change of function is : $\frac{dv}{dx}=f(v)$ $\frac{d^2v}{dx^2}=\frac{df}{dv}\frac{dv}{dx}=f'f$ $$f'f-\left(v^{3}-v\right)-\varepsilon\frac{1}{2}\left(1-v^{2}\right)=0$$ ...


0

The equation being $$\frac{dy}{dx}\tan(x+y)=1-\tan(x+y)$$ change, as you did $x+y=u$, that is to say $y=u-x$, $\frac{dy}{dx}=\frac{du}{dx}-1$. So, the equation rewrite $$(\frac{du}{dx}-1)\tan(u)=1-\tan(u)$$ that is to say $$\frac{du}{dx}=\cot(u)$$ in other words $$\frac{dx}{du}=\tan(u)$$ Integrating both sides $$x+C=-\log (\cos (u))$$ $$\cos(u)=Ce^{-x}$$ ...


0

The terms x,y,z (x1,x2,x3 in your case) are the coefficients of the Fourier series expansion of a mathematical model of fluid convection. The magnitude of these terms describes how the system evolves in time. Each point in phase space corresponds to a particular system with a particular convection speed and temperature gradient. You can easily find a ...


1

Define $$x(t) = C_1e^t\cos t + C_2e^t\sin t + C_3e^{-t}\cos t + C_4e^{-t}\sin t$$ and $$y(t) = C_2e^t\cos t - C_1e^t\sin t - C_4e^{-t}\cos t + C_3e^{-t}\sin t$$ for some constants $C_1,C_2,C_3,C_4 \in \mathbb{R}$. Differentiating twice yields $$x''(t) = -2C_1e^t\sin t + 2C_2e^t\cos t + 2C_3e^{-t}\sin t - 2C_4e^{-t}\cos t = 2y(t)$$ and $$y''(t) = ...


0

Hint: \begin{align*} \frac{dy}{dx}\tan(x+y) &= 1 - \tan(x+y) \ \ \ || \dot\ dx \ \ \ (1) \\\tan(x+y)dy &= 1-tan(x+y)dx \\\int\tan(x+y)dy &= \int 1-tan(x+y)dx \end{align*} $(1)$ is called Leibniz method (correct me if I'm wrong), and it is quite handy in solving ODE's.


2

Now substitute into the original equation: $y = u - x$, $y' = u' - 1$. You should end up with a separable differential equation.


0

In general case we have: $$L(f^n)=s^nL(f)-s^{n-1}f(0)-\dots-sf^{n-2}(0)-f^{n-1}(0)$$ so $$L(y')=sL(y)-y(0)=sL(y)-1$$ then $$sL(y)-1+2L(y)=\frac{s}{s^2+9}$$ $$L(y)=\frac{s^2+s+9}{(s+2)(s^2+9)}$$ $$L(y)=\frac{\frac{2}{13}s+\frac{9}{13}}{s^2+9}+\frac{\frac{11}{13}}{s+2}$$ ...


1

The general formula is $$ L(f^{(n)})(s) = s^n L(f)(s) - \sum_{k=1}^n s^{n-k} f^{(k-1)}(0) $$ In particular, $L(y')(s) = s L(y)(s) - y(0)$.


0

This is just meant to be a long comment. I just want to make you aware that this can be proved easily using the Banach fixed point theorem, which states that if $X$ is a complete metric space, then any map $\;T:X \to X$ which is Lipchitz with constant less than $1$ has a unique fixed point. (If you're taking quals then you must be familiar with this?) ...


0

Dr. MV has covered the existence statement quite well. To get uniqueness, suppose $f$ and $g$ are both functions satisfying the conditions. Then, we would have $$f(x) - g(x) = - \int_0^1 [f(y) - g(y)]K(x,y)\;dy$$ Let $x$ be such that $|f(x) - g(x)|$ is maximal on $[0,1]$. If $f(x) \not= g(x)$, then we can divide by $f(x) - g(x)$ to obtain $$1 = - ...


2

HINT: Let $f^{(n)}(x)$ be given by the recursive relationship $$f^{(n)}(x)=\sin x^2-\int_0^1K(x,y)f^{(n-1)}(y)dy$$ with $f^{(0)}=0$. Then, show that $$\begin{align} \left|f^{(n)}(x)-f(x)\right|&=\left|\int_0^1K(x,y)\left(f^{(n-1)}(y)-f(y)\right)dy\right|\\\\ &\le\int_0^1|K(x,y)|\left|f^{(n-1)}(y)-f(y)\right|dy\\\\ &<\lambda ...


1

Just note $$\frac{1}{s^2+a^2}-\frac{1}{s^2+b^2}=\frac{b^2-a^2}{(s^2+a^2)(s^2+b^2)}$$ Then, if $a^2\neq b^2$ and $ab\neq 0$, we have \begin{align*} \frac{1}{(s^2+a^2)(s^2+b^2)}&=\frac{\frac{1}{b^2-a^2}}{s^2+a^2}-\frac{\frac{1}{b^2-a^2}}{s^2+b^2}\\ ...


0

As you said, $$\mathcal{L}^{-1}\left(\frac{a}{s^2+a^2}\right)=\sin(at)\tag{1} $$ hence, assuming $a\neq b$ and $a,b\neq 0$: ...


1

Your partial fractions decomposition is off. You should get $$ \frac{1}{(s^2+a^2)(s^2+b^2)} = \frac{1}{b^2-a^2} \left( \frac{1}{s^2+a^2} - \frac{1}{s^2+b^2} \right), $$ so the inverse Laplace transform is $$ \frac{1}{b^2-a^2} \left( \frac1a \sin at - \frac1b \sin bt \right). $$


0

If it changes slowly then it is still changin hence $f'(x)$ will be small but non-zero. If it is a constant then yes $f'(x)$ will be $0$ over the constant interval. Indeed suppose $f(x)=k$, $k \in \mathbb{R}$ then $f'(x)=0$. Although as others have stated the derivative is telling us about how quickly the function changes w.r.t the variable so you need to ...


2

The question is not whether $x$ changes slowly. The derivative of $f$ tells you how $f$ changes when $x$ changes. This is independent on observations about what $x$ actually does.


3

By the limit definition of the derivative it doesn't make sense to take the derivative with respect to a constant because the denominator in the limit would always be $0$. However, if you were to take the derivative of both sides of your equation with respect to some other variable, say $t$, then you get $$\frac{dy}{dt}=f'(x)\frac{dx}{dt}=0$$ so in that case ...


0

Another example, maybe simpler : $$f_\epsilon(x)=\epsilon \sin(\epsilon^2 x)$$ It will oscillate at a speed of $\epsilon^2$ with an amplitude of $\epsilon$, and therefore will have a derivative of order $\frac{1}{\epsilon}$.


0

You also needn't use the Laplace transform here. Your Ansatz could be $Ae^{3x}$ and you would have $18Ae^{3x}+3Ae^{3x}-Ae^{3x}=e^{3x}$ $18A+3A-A=1$ which gives you the particular solution. The solution to the homogenous equation, as you know, will be a linear combination of $e^{r_1x}$ and $e^{r_2x}$. Then you can plug in the initial values. To me, this ...


1

No, consider something like $$ f(x) = e^{-x}\sin(e^{x}) $$ $$ f'(x) = -e^{-x}\sin(e^{x}) + e^{-x}\cos(e^{x})e^{x} = \cos(e^{x}) - e^{-x}\sin(e^{x}) $$ Then $f(x)$ gets small very quickly, but $f'(x)$ asymptotically oscillates faster and faster between $-1$ and $1$.


1

No. Consider $f(x)=\frac1x\sin x^3$ as $x\gg 0$.


3

No. You could have a function like $e^{-x^2}\sin(e^{x^2})$, which is very small for large values of $x$, but whose derivative can be very large.


1

No, that is not correct. There is ambiguity here because $z = F(x,y) = F(x,f(x))$ can be considered to depend on $x$ in two different ways. In order to clarify the situation, I'm going to introduce a third variable $t$ (which will in fact have the same value as $x$ but be considered a distinct variable). Then we can say that $z$ depends on the pair ...


1

$$y_1y_2'-y_1'y_2=0$$ $$\frac{y_1y_2'-y_1'y_2}{y_1^2}=0$$ $$\frac{y_2}{y_1}=C$$ Unless I'm mistaken, this appears to imply if the Wronskian is $0$, the solutions are linearly dependent.


1

no because $x$ is not dependent of $y$. Moreover $\frac{dF}{dy}$ is not correct because the differential is not total. It should be $\frac{\partial F}{\partial y}=...$


3

In general, when you do Laplace transforms, you need to do partial fraction decomposition to separate all the terms: $$L(y)=\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}$$ $$\frac{1}{(s-3)^2}\cdot\frac{1}{2s-1}\cdot\frac{1}{s+1}=\frac{A}{s-3}+\frac{B}{(s-3)^2}+\frac{C}{2s-1}+\frac{D}{s+1}$$ Clear the denominator: ...


2

Since $$ \frac{dw_1}{dt}w_1+\frac{dw_2}{dt}w_2=\frac12\frac{d}{dt}(w_1^2+w_2^2), $$ your differential equation is equivalent to: $$\tag{1} \frac{du}{dt}=\alpha u, $$ with $$ u=w_1^2+w_2^2,\quad \alpha=-2\frac{c}{I}. $$ Solving (1) we get: $$ w_1^2(t)+w_2^2(t)=\beta\exp\left(-2\frac{ct}{I}\right), $$ where $$ \beta=w_1^2(0)+w_2^2(0). $$


1

He wrote that $$\frac{1}{(s^2+1)(s^2+4)} = \frac{1}{3}\frac{1}{s^2+1} - \frac{1}{3}\frac{1}{s^2+4}$$ It's a classical method of partial fraction decomposition: https://en.wikipedia.org/wiki/Partial_fraction_decomposition


1

$e^{2\pi im} = 1$ does not imply $e^{i\pi}e^{2m} = 1$, as if it did then $e^{2\pi im} = e^{i\pi + 2m}$. In general, $e^{ix} = 1$ iff $x = 2k\pi$ for integer $k$. Hence $e^{2\pi im} = 1$ iff $m$ is an integer.


0

This is not what is expected, but a different method which allows to compare the respective results. $$\frac{d^2x}{dt^2}+\left(\frac{dx}{dt} \right)^2-2x=0$$ This is an autonomous ODE. So, the change of function is $\frac{dx}{dt}=f(x)$ and $\frac{d^2x}{dt^2}=\frac{df}{dx}\frac{dx}{dt}=f'f$ $$f'f+f^2=2x$$ With $F(x)=f(x)^2$ : $$F'+2F=4x$$ ...


1

If you are really interested in proofs and exploring calculus from first principles, I will highly recommend Michael Spivak's Calculus textbook (available here). It explains things very well in a very thorough and methodical approach. It also has easy as well as very challenging questions that would stretch the minds of even the brightest mathematicians. ...


4

I think Stewart's or Anton's Calculus are not the right choice for you...unless you have not yet completed a course in university calculus. So, the answer here necessarily depends on your background. If I assume you have take differential and integral calculus of one-variable then I recommend: Susan Colley's Vector Calculus (even the first or second edition ...


1

I think it depends on how you organize your problem. Suppose your problem takes place in a time interval $a \le t \le b$. Some things you know are $f'(t)$ and $f(b)$. Then to find $f(t)$, you may think of integrating backward from $b$. And maybe some ways to write that would involve thinking of negative $dt$.


4

What about all in one ? :-) this book covers almost all what you want. (Sorry I don't have enough reputation for comment)


0

I think this is related to classical physics, yes? It seems that you have 3 degrees of freedom or coordinates and basis. The displacement vectors or velocity or what ever contravariant vector you have exists independent of the coordinates and can change it's inherent position. This change is most often seen as continuous and t is nothing more than a ...


2

First answer: Your operator can be written in terms of the usual derivative operator. We have $L=(\frac{d}{dx})^2+3\frac{d}{dx}-4$, where $4$ stands for multiplication by the constant. From the point of view of linear algebra, $L$ is a linear operator/mapping/function from a vector space to itself. Second answer: A square matrix as we usually understand it ...


1

(1) No, there is nothing uniquely special about this $L$. (2) The notions of eigenvalues and eigenvectors are defined for any linear map from a vector space $\Bbb V$ to itself: A (nonzero) vector $v$ is an eigenvector, of eigenvalue $\lambda$, for the linear transformation $T: \Bbb V \to \Bbb V$ iff $$T(v) = \lambda v.$$ If $\Bbb V$ is finite-dimensional, ...


1

$$x(1-x)y''(x)-\lambda y(x)=0$$ is a particular case of the hypergeometric differential equation : $$x(1-x)y''+\left(c-(a+b+1)x\right)y'-aby=0$$ http://mathworld.wolfram.com/HypergeometricFunction.html In the present case : $a=\frac{1}{2}(-1+\sqrt{1-4\lambda})$ , $b=\frac{1}{2}(-1-\sqrt{1-4\lambda})$ , $c=0$ The general solution involves the Gauss ...


2

$$yy'+\frac{y}{x}+k=0 \quad\quad (1)$$ Change of function : $y(x)=\frac{1}{f(x)}$ $\frac{1}{f}\left(-\frac{f'}{f^2}\right)+\frac{1}{xf}+k=0$ $$f'=kf^3+\frac{1}{x}f^2$$ This is an Abel's differential equation of first kind which is knonw as ''non-sovable'' form, meaning that the solutions are not known on the form of a finite number of standard functions. ...


0

Start with Newton's Law of Motion: $$\sum F=ma$$ The mass of the boat and it's load is $m=\frac{2150}{g}\approx 216.16$ The sum of the forces is $110 - 6.7V$. Thus, the equation for motion becomes: $$216.16\frac{d^2x}{dt^2}+6.7\frac{dx}{dt}-110=0.$$ Solve using your favorite technique for constant coefficient nonhomogeneous ODE. (The method of ...


1

If at least one eigenvalue has positive real part, the equilibrium is unstable. If all eigenvalues have real part $\le 0$, and some have real part $0$, then the linearization does not determine stability.


1

First for general n-D case:Full classification (via linearization) in higher dimensions gets complicated. But the general idea is that you may have products of the basic equilibria. For example, in restricted three-body problem, linearization around some fixed points tells us that they are a product of a saddle and two centers (in 6D). This is said to be a ...


0

A function is in $C[a,b]$ (also written $C^{0}[a,b]$) if it is continuous on the interval $[a,b]$. A function is in $C^{n}[a,b]$ if it has $n$ derivatives, all continuous. The interval can be omitted if one means all of whatever set one is writing about. For example, $|x|$ is in $C^{0}$ because it is a continuous function, but $|x|\not\in C^{1}$ because its ...


0

Below, there is a calculus which is more a guideline to solve this hard problem, than a complete solving. Theoretically the most difficult part could be overcome, but with big equations, which is not carried out here. Nevertheless, the incomplete result suggests the general form of the function $f(R)$, in which the argument of some Bessel functions is not ...


2

Hint: $$ \left|\dfrac{d}{dt} (x-y)\right| = |(\sin(x) - \sin(y)) \sin(t) + (x - y) \cos(t)| \le |x - y| (|\sin(t)| + |\cos(t)|)$$


1

$$f(t)^2 - 2g(t)f(t)\sin(t) - 2f'(t) + g(t)^2 - 2g(t)\cos(t) + 1 = 0$$ This is a Riccati ODE. So, the usal change of function is : $f(t)=-2\frac{y'(t)}{y(t)}$ $f'=-2\frac{y''(t)}{y(t)}+2\frac{(y')^2}{y^2}$ $4\left(\frac{y'}{y}\right)^2 + 4g\frac{y'}{y}\sin(t) - 2\left(-2\frac{y''}{y}+2\frac{(y')^2}{y^2}\right)+ g^2 - 2g\cos(t) + 1 = 0$ ...


0

I think here you probability need $\Omega$ to be bounded as well. (however since $u\in W_0^{1,2}(\Omega)$, you do not need smooth boundary condition) And I will assume by $d\lambda^n$ you just mean $x\in \mathbb R^n$, the standard integration notation. For your first question, since $u\in W_0^{1,2}(\Omega)$, by Sobolev embedding you have $u\in L^{p^*}$ ...


0

$$(1-x^2)y''-2xy'+2y=0$$ Since $y=x$ is an obvious solution, we look for other solutions on the form $y=xf(x)$ $y'=f+xf'$ and $y''=2f'+xf''$ $(1-x^2)(2f'+xf'')-2x(f+xf')+2xf=0$ $$2(1-2x^2)f'+x(1-x^2)f''=0$$ $\frac{f''}{f'}=-2\frac{1-2x^2}{x(1-x^2)}$ The integration leads to : $$f'=c_1\frac{1}{x^2(1-x^2)}$$ Integrating again leads to : ...



Top 50 recent answers are included