New answers tagged

0

We have: $$ \frac{dv}{dt}=-kv \quad \rightarrow \quad v(t)=v_0e^{-kt} $$ and, from the initial condition $v_0=1000$ $$ v(t)= 1000e^{-kt} $$ also we have: $$ \frac{dv}{dr}=\frac{dv}{dt}\frac{dt}{dr}=-kv\frac{1}{v}=-k \quad \rightarrow \quad v(r)=-kr+v_0=1000-kr $$ so: $$ v(1200)=900=1000-k1200 \quad \rightarrow \quad k=\frac{1}{12} $$ and, from $v(t)$: $$ ...


1

The equation you want to solve is $$\ddot{x}(t) = -k\dot{x}(t),$$ or $$\dot{v}(t) = -kv(t)$$ a damping force proportional to the velocity. With the anzatz $v(t) = Ae^{at}$ we get that $a = -k$. Then at $t=0$, $v=1000$ so $A = 1000$, or $$v(t) = 1000e^{-kt}.$$ Integrating we get $$x(t_1) - x(t_0) = -\frac{1000}{k}\left(e^{-kt_1} - e^{-kt_0}\right).$$ ...


1

The solution is easier if you start at the right endpoint with $$y=C_1\cos k(1-x)+C_2\sin k(1-x)$$ From this form of the solution we have right away $$y(1)=C_1=0$$ So all we need do is search for solutions of the form $y=\sin k(1-x)$. Applying the boundary condition at the left endpoint $$y(0)+y^{\prime}(0)=\sin k-k\cos k=0$$ Since $\cos k=0$ does not lead ...


0

To solve $0 = x(a - bx - cy)$ and $0 = y(-d + ex - fy)$ for the non-zero solutions $x \ne 0, y \ne 0$ you have to solve $(a - bx - cy) = 0$ and $(-d + ex - fy)$. This is two equations in two unknowns so you can solve for $x, y$ (in fact they are the equations of two lines and you are looking for their intersection). But yes, you have to go through all the ...


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Compute (using the product derivation rule) $$ \dot x(t)=\frac{d}{dt}\left(\gamma^{t}\cdot\int\limits_{0}^t \gamma^{-s} u(s)ds\right)=\log(\gamma)\, \gamma^t\cdot \int\limits_{0}^t \gamma^{-s} u(s) ds+\gamma^t\cdot\left(\gamma^{-t}u(t)\right)=\log(\gamma)\cdot x(t)+u(t). $$


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The problem is that, also for a polynomial, in two unknowns, we have general methods that gives a formal solution only if the unknow that we want to find has degree $n\le 4$ ( and the cases $n>2$ are not so simple), but if the unknown has degree $n>4$ it is very difficult to find a formal solution. As an example, also a ''simple'' equation as $$ ...


1

The answer is no in general. a good specific example is the busy beaver function which is unsolvable in general for large enough x. This question seems to have a lot to do with undecidability where things can get complicated. another example is the length of the collatz function for natural number x. there are no proven answers to this, if there were it ...


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You want all the series to be written with the "same power" $x^m$. Then $$ y'=\sum_{m=0}^\infty m\,a_m\,x^{m-1}=\sum_{m=0}^\infty (m+1)\,a_{m+1}\,x^{m}, $$ $$ x\,y'=\sum_{m=0}^\infty m\,a_m\,x^{m}. $$ Putting it all together you get $$ =\sum_{m=0}^\infty\bigl(7\,(m+1)\,a_{m+1}+m\,a_m-a_m\bigr)x^m=0. $$


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I take it that you have the difference equation $$\frac{U_{i+1}-2U_i+U_{i-1}}{(\Delta x)^2}=-\lambda_nU_i$$ The Neumann boundary condition is the same as having the mirror image of the solution on the other side of $x=0$. Thus one could assume $U_{-1}=U_1$ and the left boundary difference equation reads $$\frac{2U_1-2U_0}{(\Delta x)^2}=-\lambda_nU_0$$ EDIT: ...


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We have a linear system $$\dot y = A y + u$$ with initial condition $y (0) = y_0$. What you call the complementary solution is what is usually known as the homogeneous solution, or, natural response. We now assume zero initial conditions and look for the forced response. Laplace-transforming both sides, we get $$s Y(s) = A Y (s) + U (s)$$ Hence, $$Y ...


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Ok, let us start with the second order homogeneous differential equation: $$y''-2y'+2y=0$$ I will derive a solution, but not in the traditional way. I am going to arrive at the differential equation solution in another way, and it involves substitution and solving a couple of 1st order ODEs. It kind of gives some insight as to why the traditional way works. ...


1

The matrix $A=0$ makes every initial condition to be stationary. The solution of the start edo is $y(t)=e^{At}y_0$, then for every point be stationary we need that $e^{At}=I$, this means that $At=0$, for every $t$, then the matrix $A=0$.


2

Look for solutions of the form $\mathrm{e}^{s t}$. The ODE then becomes $$(s^{100} + 100) \, \mathrm{e}^{s t} = 0$$ Hence, $s^{100} + 100 = 0$, or, $$s^{100} = 100 \, \mathrm{e}^{i (2k+1) \pi}$$ where $k \in \mathbb Z$. Taking the 100th root, we obtain $100$ solutions $$\{ \sqrt[100]{100} \, \mathrm{e}^{i (2k+1) \frac{\pi}{100}} \mid k \in ...


1

You equation is: $$8y''-6y'+y=6e^x+3x-16$$ The corresponding homogeneous linear differential equation: $$8y''-6y'+y=0$$ Its solution (do you know how to solve this equation?): $$y^*=C_1e^{\frac{1}{2}x}+C_2e^{\frac{1}{4}x}$$ Now we seek a special solution of the following inhomogeneous equation $$8y''-6y'+y=6e^x$$ The form of this special solution is ...


0

Your statement is false in general, unless $\Omega$ is simply connected. On the other hand, although you can give a more or less straightforward proof in this situation, it is a very particular case of the fact that the de Rham cohomology (which is a homotopy invariant) vanishes for contractible spaces together with Stokes' theorem.


2

First: How to construct $\mu$. I think the function $\mu$ would be constructed by "deconvolution". This is needed in a lot of experimental situations, because in many measurement processes the measuring instrument acts like a convolution filter. The most standard way of doing the deconvolution is taking the Fourier transforms of both $r(t)$ and ...


0

Assuming no friction, conservation of energy means that $$E=\frac12mV^2+mgh=\frac12mR^2\dot\theta^2-mgR\cos\theta=\frac12mv^2-mgR$$ Where $m$ is the particle's mass, $g$ is the acceleration of gravity, $\theta$ is the angle along the arc from the lowest point, $R$ is the arc's radius $V=R\dot\theta$ is the particle's current speed, and $v$ is its initial ...


1

With the help from @Robert Israel and @G Cab, \begin{align} &c_1a^b+c_2a^{-b}=0\\ &\Rightarrow c_1a^b=-c_2a^{-b}\\ &\mbox{Let}\,\,\,\,c_1a^b=-c_2a^{-b}=c_0\\ &\Rightarrow c_1=\frac{c_0}{a^b}\\ &\Rightarrow c_2=-\frac{c_0}{a^{-b}}\\ \end{align} Thanks guyz!


2

If $c_1 a^b + c_2 a^{-b} = 0$, then $c_2 = \ldots$. Now see what $c_0$ has to be to put the solution in the desired form.


0

With advice from Mattos I found the way myself: $$\frac{d}{dx}log(y)=\frac{d}{dx}c_1x^{2}+c_2$$ $$\frac{y'}{yln(10)}=2c_1x$$ $$y'=2c_1ln(10)xy$$ $$y''=2c_1ln(10)\frac{d}{dx}xy=2c_1ln(10)(y+xy')=\frac{y'}{x}+\frac{(y')^2}{y}$$ From that the answer follows: $$xyy''-yy'-x(y')^2=0$$


0

Your goal is to get rid of the constants. So $$ \eqalign{ & \ln y = c_{\,1} x^{\,2} + c_{\,2} \cr & y = e^{c_{\,1} x^{\,2} + c_{\,2} } = a_{\,2} \,e^{c_{\,1} x^{\,2} } \cr & y' = 2c_{\,1} x\,a_{\,2} \,e^{c_{\,1} x^{\,2} } = 2c_{\,1} x\,y \cr & y'' = 2c_{\,1} \,a_{\,2} \,e^{c_{\,1} x^{\,2} } + 4c_{\,1} ^2 x\,a_{\,2} ...


1

You are doing pretty much the right thing. The horizontal tangents condition was $$\frac{dy_2}{dt}=a_{21}y_1+a_{22}y_2=-4y_1+a_{22}\cdot2y_1=0$$ So you got $a_{22}=2$. Then the vertical tangents condition is $$\frac{dy_1}{dt}=a_{11}y_1+a_{12}y_2=a_{12}y_2=0$$ So that $a_{22}=0$. Now your matrix looks like ...


0

Your aswer is also right. Instead of getting (x-t)^2/sqrt(2), you will get (x-t)^2/2. simililarly K4 can be find and when we generalize it , it will will be series of exp(x-t)


0

Even if it Mathematica shows constants as C[1], C[2]... in its outputs, it does not understand them when you use them as inputs. Try the following instead: DSolve[{Derivative[1][x][t] == -(Exp[2 t a - 2 t b]/c) + x[t], x[0] == R}, x[t], t] where cis the constante C[1] in your initial code.


1


1

$$\fbox{$y^{\prime\prime}+\left(\frac{1-2a}{x}\right)y^{\prime}+\left[\left(bcx^{c-1}\right)^2+\frac{a^2-p^2c^2}{x^2}\right]y=0$}\tag{5}$$ Let $y(x)=x^\alpha Y(X)\quad$ with $X=\beta x^\gamma$ $\qquad \alpha, \beta , \gamma$ are constants. $\frac{dX}{dx}=\gamma\beta x^{\gamma-1} \quad;\quad \frac{d^2 X}{dx^2}=\gamma(\gamma-1)\beta x^{\gamma-2}$ ...


2

Here are hints for parts of the questions: 1) Note that if $\nabla f(x_0) \neq 0$ then $\dot{\gamma}(t) \neq 0 $ for all $t \ge 0$. To see this, if $\dot{\gamma}(t_0) = 0$ for some $t_0 >0$, then $t \mapsto \gamma(t_0)$ and $t \mapsto \gamma(t)$ would be two solutions passing through $\gamma(t_0)$ and by uniqueness they must be the same hence a ...


2

I think the issue is that you are conflating the Biharmonic operator $\Delta^2$ of the sphere (the first one in your question) with the one $\bar \Delta^2$ of $\mathbb R^3;$ i.e. you are assuming that when $f : \mathbb R^3 \to \mathbb R$ depends only on $\phi, \theta$ that $\Delta^2 f = \bar \Delta^2 f$. The equivalent statement for the Laplacian is true ...


0

It suffices to note that $$ \left(-\theta+\arctan\frac r\theta\right)'=0, $$ which gives $$ -1+\left(\frac r\theta\right)'\frac1{1+r^2/\theta^2}=\frac{-1-r^2/\theta^2+r'/\theta-r/\theta^2}{1+r^2/\theta^2}=0. $$ Therefore, $$ \theta r'=\theta^2+r^2+r. $$


1

It is (clearly) $x=0$ (this is due to the fact that you can show that there is a solution of the form $(0,y(t))$, where $y'=-y$, and since this "curve" is tangent to the stable space it will be the stable amnifold). Since we know that it is unique, nothing else is needed. Anyways, you can proceed as for the unstable manifold, nothing changes other than ...


0

I might be missing something, but it looks like you can just solve this. $x=Ae^{t}$ so you just solve the other equation, $\dot{y}+y = A^2e^{2t}$. Get your integrating factor as $e^t$ and solve $$d(y\cdot e^t)= A^2e^{3t}$$ integrate to get $$y\cdot e^{t}={A^2\over 3}e^{3t}+C$$ so $y= {A^2\over 3}e^{2t}+Ce^{-t}$ i.e. $$y= {1\over 3}x^2+{C'\over x}, ...


0

From $$ r=\theta \tan (\theta+c) $$ we have $$ \frac{dr}{d\theta}=\tan(\theta +c)+\theta\left[1+\tan^2(\theta+c)\right] $$ so, multiplying by $\theta$: $$ \theta r'=\theta\tan(\theta +c)+\theta^2\left[1+\tan^2(\theta+c)\right] $$ and substituting $\theta \tan (\theta+c)=r$ you find your differential equation. product rule: $$ \frac{d}{d \theta}(\theta ...


1

Yes, that's what it means. (I think it's a rather obtuse and needlessly confusing way to write it.)


3

Use Laplace transform: $$x'(t)=3x(t)+y(t)+e^{2t}\Longleftrightarrow sx(s)-x(0)=3x(s)+y(s)+\frac{1}{s-2}$$ $$y'(t)=y(t)-x(t)+e^t\Longleftrightarrow sy(s)-y(0)=y(s)-x(s)+\frac{1}{s-1}$$ Now, use the intial conditions and simplify: $$x(s)=\frac{1+y(s)+\frac{1}{s-2}}{s-3}$$ $$y(s)=\frac{\frac{1}{s-1}-x(s)}{s-1}$$ Now, substitute them into eachother: ...


2

The given ODE is a linear homogeneous equation of second order. This implies that the set of solutions is a two-dimensional complex vector space. If the equation $a\lambda^2+b\lambda +c=0$ has two different solutions $\lambda_1$, $\lambda_2\in{\mathbb C}$ then the general solution is given by $$y(t)=C_1e^{\lambda_1 t}+C_2e^{\lambda_2 t}\ .$$ If the ...


2

I can see a problem right off the bat: $$L=\dot x^2+x^2\dot y^2$$ $$p_x=\frac{\partial L}{\partial\dot x}=2\dot x$$ $$\dot p_x=2\ddot x=\frac{\partial L}{\partial x}=2x\dot y^2$$ Since your equation for $\dot p_x$ was not just a typo in my reading of your solution, you were hosed at this point. $$p_y=\frac{\partial L}{\partial\dot y}=2x^2\dot y$$ $$\dot ...


1

$$y''(t)=4\left(\dfrac1{y(t)}\right)',$$ $$y'(t)=\dfrac4{y(t)}+C.$$ If $t=0,$ then $2=\dfrac42+C,\quad C=0$, so $$\dfrac{dy}{dt}=\dfrac4y,$$ $$2ydy=8dt,$$ $$y^2=8t+C.$$ If $t=0,$ then $4=C,$ so $$y^2=8t+4,$$ $$\boxed{y=\pm\sqrt{8t+4}}$$


1

If $a = 0$ and $b = 0$, then we have $c \, y(t) = 0$, which is not even an ODE. If $c \neq 0$, then the solution is $y (t) = 0$. If $c = 0$, every function from $\mathbb R$ to $\mathbb C$ is a solution. If $a = 0$ and $b \neq 0$, then we have a 1st order ODE $$\dot y + \left(\frac{c}{b}\right) y = 0$$ whose solution is $y (t) = \beta \, \exp\left(- ...


1

From (1) we have $x_2 = \frac14 x_1' - \frac12 x_1$, and differentiating yields \begin{align} x_1'' &= 2x_1' + 4x_2'\\ &= 2x_1'+4(3x_1-2x_2)\\ &= 2x_1' + 12x_1 - 2x_1'+4x_1,\\ \end{align} so we may instead consider the second-order ODE $$x_1''-16x_1=0. \tag 3$$ Suppose $$x_1(t) = \sum_{n=0}^\infty a_n t^n.$$ Then differentiating and substituting ...


2

These two versions of the Poincaré-Bendixson Theorem are not contradicting each other. Instead, I think you overlooked some of the hypotheses. The first theorem tells you that a positively invariant, compact subset of the phase plane always contains at least one closed orbit, provided there are no fixed points in it (or provided it has just one unstable ...


0

Suppose you have a differential equation that looks like this: $$y'=F\left ( \frac{y}{x}\right )$$ then you can make a substitution $v(x)=\frac{y}{x} \iff y=vx \implies y'=v+xv'$ to transform your ODE into an ODE in $v$ $$\implies v+xv'=F(v) \iff \frac{dv}{F(v)-v}=\frac{dx}{x}$$ This equation is separated and you can solve it by the usual methods. In your ...


0

If you make $y=x z$, $y'=x z'+z$ and the equation becomes $$x z'+z=z-z^2$$ that is to say $$x z'=-z^2$$which is separable.


4

Since you did not show your work, I can only conjecture that you forgot to differentiate $a$ in $$\Delta f = \frac{1}{a^2 \sin\phi} \frac{\partial}{\partial \phi} \left( \sin\phi \frac{\partial f}{\partial \phi} \right) + \frac{1}{a^2\sin^2\phi} \frac{\partial^2 f}{\partial \theta^2} \, , $$ In the next step, if you want to take $\Delta$ again, note then ...


0

Approach $1$: Let $S=w^aT$ , Then $\dfrac{dS}{dw}=w^a\dfrac{dT}{dw}+aw^{a-1}T$ $\therefore w^aT\left(w^a\dfrac{dT}{dw}+aw^{a-1}T\right)=\dfrac{aw^{2a}T^2}{w}+\dfrac{bw^aT}{w}-c$ $w^{2a}T\dfrac{dT}{dw}+aw^{2a-1}T^2=aw^{2a-1}T^2+bw^{a-1}T-c$ $w^{2a}T\dfrac{dT}{dw}=bw^{a-1}T-c$ $T\dfrac{dT}{dw}=bw^{-a-1}T-cw^{-2a}$ Let $x=-\dfrac{bw^{-a}}{a}$ , Then ...


2

Equation $(3)$ seems to say it all, provided you fix the typo to $(x^2-1)^2$ because if $v_0>\frac1{\sqrt2}$ then $y=x^{\prime}>0$, so $x$ just keeps on going. If $v_0=\frac1{\sqrt2}$, then when $y=x^{\prime}=0$, $x=1$, so $y^{\prime}=x^{\prime\prime}=0$ also, so (unstable) equilibrium is reached, so the system never turns around. Or, put another way, ...


0

You can also do this. Rewrite your equation as: $$\psi''+k^2 \psi=0$$ $$\psi'' +ik\psi'-ik\psi'+k^2 \psi=0$$ $$\psi''+ik\psi'-ik(\psi'+ik \psi)=0$$ Suppose $z=\psi'+ik\psi.$ We now have turned our 2nd order differential equation into a 1st order one. $z'-ikz=0$ Separate variables to see: $z=De^{ikx}, D\in\mathbb{C}$ Replace this solution $z$ in ...


1

For $y'=\dfrac{(x-y)y-x-y}{(x-y)x+x+y}$ , Let $u=\dfrac{y}{x}$ , Then $y=xu$ $\dfrac{dy}{dx}=x\dfrac{du}{dx}+u$ $\therefore x\dfrac{du}{dx}+u=\dfrac{(x-xu)xu-x-xu}{(x-xu)x+x+xu}$ $x\dfrac{du}{dx}=\dfrac{(1-u)xu-1-u}{(1-u)x+1+u}-u$ $x\dfrac{du}{dx}=\dfrac{(x-1)u-xu^2-1}{(1-x)u+x+1}-u$ $x\dfrac{du}{dx}=\dfrac{xu^2-(x-1)u+1}{(x-1)u-x-1}-u$ ...


1

Changed: When you take your limit $t\to\infty$ you assume that $$ \lim_{t\to\infty} e^{5t}(c_1+c_2 e^{-{4t}})=c_1e^{5t}, $$ which is obviously not correct.


0

Here is the integrating factor method. To briefly explain, if you are solving the (inseparable) differential equation in the form $$y'+p(x)y=q(x),$$ then you multiply both sides by $\mu(x)$ and the role of this $\mu(x)$ is to make the left hand side into the form $(y\mu(x))'$ (i.e. the reverse of the derivative product rule). It turns out $\mu(x)=e^{\int ...


0

I was doing this problem again, and I have found that there is a mistake in the solution. The ODE for $r_{a,\xi}$ yields $r_{a,\xi} = -\epsilon \frac{8}{3} r_a^2$, instead of the cubic term in the rhs. Also I think there is something wrong with the $\frac{8}{3}$ coefficient; I have applied a simple multiple scale method and the solvability condition gives me ...



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