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## New answers tagged differential-equations

0

It depends on what you call a solution. For example, if you are looking for a continuously differentiable function $x$, then there is no such solution. You can, however, consider a merely continuous solutions to the following integral equation $$x(t)=x_0+\int_{t_0}^t h(x(\tau),g(\tau))d\tau.$$ Then it can be proved (pick up almost any textbook on optimal ...

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Comment to the question (v3): Since OP mentions the Euler-Lagrange equations, it seems relevant to mention the following elementary derivation of the 3-dimensional isoperimetric inequality $$\tag{1} 36\pi V^2 \leq A^3,$$ although it additional assumes that the region is star-shaped. Sketched proof: We can use spherical coordinates $(r,\theta,\varphi)$. ...

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Solution 1 (direct): In this case it is easier to solve the equation directly: $$\frac{\partial}{\partial y} \frac{\partial Z}{\partial x} = 0$$ implies that $\frac{\partial Z}{\partial x}$ does not depend on $y$: $$\frac{\partial Z}{\partial x} = f(x).$$ with an arbitrary function $f(x)$. From here: $$Z(x,y) = \int f(x) dx + G(y),$$ where $G(y)$ is ...

2

Wikipedia references: Streamlines, streaklines, and pathlines Stream function <quote> Streamlines are a family of curves that are instantaneously tangent to the velocity vector of the flow. These show the direction a massless fluid element will travel in at any point in time. </quote> Consider the velocity field $(u,v)$ of a two-dimensional ...

1

for the homogenous equation solve the characteristic equation $$\lambda^3-\lambda^2-\lambda+1=0$$ and for a particluar solution make the ansatz $$e^x(A+Bx+Cx^2)+\sin(x)(Ex+F)+\cos(x)(Gx+H)$$

2

When you have a point mass $m$ that can move in the (horizontal) $y$-direction, but is drawn to the origin by a spring then it moves according to the ODE $$m\>\ddot y=-f\> y\ ,$$ where $f>0$ denotes the spring constant. Assume now that you have two such masses with origins at a certain distance of each other. Then they independently move according ...

2

Hint Just rewrite $$(1+x^2)y'=\frac{1}{y}$$ as $$y \space y'=\frac{1}{1+x^2}$$ and now integrate both sides remembering that $(y^2)'=2 y \space y'$. I am sure that you can take from here.

0

Having been quite a bit of time I thought about this some more! If we cut our original radical at a finite time we have (as pointed out by Aryabhata) we have $$y = n!\frac{d^n}{dx^n}$$ Solutions to this include $$y= C_1 e^{x\sqrt[n]{n!}_1} +C_2 e^{x\sqrt[n]{n!}_2}... C_n e^{x\sqrt[n]{n!}_n}$$ For all possible nth roots of unity. Now consider ...

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As Amzoti commented, compute each derivative and simplify it as much as you can before going to the next derivative (otherwise, it could be a nightmare). Starting with $$y= c_1+c_2e^x+c_3e^{-x}+(\frac{1}{12}+\frac{9\cos2x - 7\sin2x}{520})e^{2x}$$ and simplifying, you should get $$y'={c_2} e^x-{c_3} e^{-x}+\frac{1}{390} e^{2 x} (-24 \sin (2 x)+3 \cos (2 ... 0 Hint: Let x=-\dfrac{u'}{tu} , Then x'=-\dfrac{u''}{tu}+\dfrac{u'}{t^2u}+\dfrac{(u')^2}{tu^2} \therefore-\dfrac{u''}{tu}+\dfrac{u'}{t^2u}+\dfrac{(u')^2}{tu^2}=\dfrac{(u')^2}{tu^2}-\dfrac{u'}{tu}-t^3 \dfrac{u''}{tu}-\dfrac{u'}{tu}-\dfrac{u'}{t^2u}-t^3=0 tu''-(t+1)u'-t^5u=0 1 The pathlines are curves in space, and parametric functions of t. So if you eliminate t, you get an expression for \frac{dy}{dx}.$$ \frac{dy}{dx} = \frac{y}{x} $$On integrating this, you have y=cx, so the pathlines are straight lines through the origin. 0 Note that x'=0 when x^2+y^2-2=0, i.e., on the circle x^2+y^2=2. So outside the circle, x'>0 and inside the circle, x'<0. This tells you the "east-west" behavior of trajectories in those regions. On the other hand, y'=0 on y=x^2 so above the parabola, y'>0 and below it y'<0. This tells you the "north-south" behavior of ... 0 There is always a singularity at x=0 for Bessel Equation If you have P(x)y''+Q(x)y'+R(x)y=0 then it has a singularity when \frac{Q(x)}{P(x)} and \frac{R(x)}{P(x)} are not defined here that is only when when x=0 3 Call P = 2xy^2-y and Q = y^2+x+y. Unfortunately,$$\frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x},$$so the equation is not exact. I'll show you how can we find an integrating factor. Suppose that we have a function \mu such that$$\mu P \ {\rm d}x + \mu Q \ {\rm d}y = 0$$is an exact equation. So we must have: ... 5 Hint This is not an exact differential equation -- but almost. You need to find an integrating factor \mu, such that$$\mu(2xy^2-y)dx+\mu(y^2+x+y)dy=0$$is exact 2 What is expected of you here is to compare the lefthand side to a total differential d\Phi of some potential function \Phi(x,y); it is known that we can express d\Phi=\frac{\partial\Phi}{\partial x}dx+\frac{\partial\Phi}{\partial y}dy and similarly d\Phi=0\implies\Phi(x,y)=C,C\in\mathbb{R} i.e. a total differential of zero suggests our function is ... 1 Strictly speaking, you get \ln|y(x)|=x^2+C, hence y(x) = \pm e^{x^2}e^C. Or y\equiv 0, which is a solution we lost when dividing by y. A neat way to put all these together is to write y(x) = C_1e^{x^2} where C_1 is a new constant, C_1=\pm e^C or C_1=0 (in other words, C_1 can be any real number). This is done all the time with ODE that ... 1$$\ln(y(x))=x^2+Ce^{\ln(y(x))}=e^{x^2+C}y(x)=e^{x^2}e^{C}$$e^{C} is just a number, so let C_{1}=e^{C}:$$y(x)=C_1e^{x^2}$$C_1 is just another way of expressing our constant of integration. 4 I'll give the idea, since the calculations really seems a pain. Let:$$X(t) = \begin{pmatrix} x(t) \\ y(t)\end{pmatrix}, \quad \mbox{and} \quad A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}.$$Now your system is X'(t) = A \cdot X(t). The idea is to diagonalize A. Since A is symmetric, we're good. The characteristic polynomial is p(t) = t^2 ... 1 Your reasoning is not valid, though your conclusion is correct. It is true that y=\frac{1}{(x+c)^2} is a solution for all c, and since if C=\frac{1}c you get y=\frac{C^2}{(Cx+1)^2}, that must also be a solution. However, this only covers cases where C\neq 0, since as you note, 0 cannot be written as \frac{1}{\text{something}}. There is no rule ... 1 Yes, you are right, C can take the value 0. The solution to the original DE involves dividing both sides by y^{3/2}, before integrating. They should consider the special case when y=0 at that point. The limit of 1/(x+c)^2 as c\to\infty is 0 at every point x, so the new formula is a limit of old formulas. 0 Do you know how to convert an (uncoupled) second order scalar equation into a first order system? For example, can you convert x''+x'+x=f(t) into a first order system? The key there is to let u=x, v=x' so that u'=v and v'=u'=x'' and then$$x''=-x'+x+f(t)$$becomes v'=-v+u+f(t) and so on. So in this coupled situation, just use that same ... 2 In terms of$$Z = x^2 + x'^2 - 1$$the equation becomes$$Z' = -2x'^2Z$$which has the solution Z(t) = Z(0)e^{-2\int_0^t x'^2 dt}. Now if x is periodic then Z must be periodic, but this is only possible (since \int_0^t x'^2 dt is an increasing function) if x'\equiv 0 or Z(0) = 0. The only non-constant periodic solutions therefore satisfy ... 0 Following definitions are taken from Tenenbaum's ODE book. Consider a linear differential equation:$$y^{(n)}+F_{n-1}(x)y^{(n-1)}+\dots+F_0(x)y=Q(x)$$A point x=x_0 is called an ordinary point of the linear differential equation if each function (F_0, F_1, \dots, Q) is analytic at x=x_0. By analytic, we mean that there is a Taylor series for the ... 5 Rearrange the equation to yield$$ A^2y'' + AA'y' + y = 0 = A^2y'' + \left(\frac{A^2}{2}\right)'y' + y $$mutiply by y' we find$$ A^2y''y' + \left(\frac{A^2}{2}\right)'y'^2 + yy' = 0\\ \frac{A^2}{2}\left(y'^2\right)' + \left(\frac{A^2}{2}\right)'y'^2 + yy' = 0 $$the last equation can be written as$$ \frac{1}{2}\dfrac{d}{dx}\left(A^2y'^2\right) + ...

1

$$(x^3)dt+(1-t)[(x^2+1)t-1]dx=0$$ $X=\frac{1}{x^2}$ $$X\:dt-(1-t)[(1+X)t-X]\:dX=0$$ $Y=t-1$ $$X\:dY+Y[(1+X)(Y+1)-X]\:dX=0$$ $$X\frac{dY}{dX}+(1+X)Y^2+Y=0$$ This is a Bernoulli EDO. Let : $F=\frac{1}{Y}$ $$X\frac{dF}{dX}-F=X+1$$ This linear EDO is easy to solve for $F(X)$

6

since $$2x^3yy'+(1-y^2)(x^2y^2+y^2-1)=0$$ then we have $$2x^3yy'=(y^2-1)^2+(y^2-1)x^2y^2=(y^2-1)^2+(y^2-1)x^2(y^2-1)+x^2(y^2-1)$$ so $$x^3(y^2-1)'=(y^2-1)^2+(y^2-1)^2x^2+x^2(y^2-1)$$ let $y^2-1=u$,then we have $$\dfrac{du}{dx}=\left(\dfrac{1}{x}+\dfrac{1}{x^3}\right)u^2+\dfrac{u}{x}$$ this is Bernoulli equation ...

1

Yes, it's a tough one. There is an integrating factor $$\mu = \dfrac{1}{x^2 (y^2-1)^2}$$ which leads to the implicit solution $${\frac {{x}^{2} y^{2}- y^{2}+1}{x \left( y ^{2}-1 \right) }} = c$$ but I don't know how you would find these by hand (I used Maple).

1

Method 1: $$f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f^{(3)}(0)}{3!}x^3+\cdots$$ You have: $$f(0)=2$$ $$f'(0)=0$$ Replacing that in your differential equation: $$f''(0) + 5f'(0) + 4f(0) = 0$$ $$f''(0)+5\cdot 0 +4\cdot 2=0$$ $$f''(0)=-8$$ If you take the derivative of your differential equation: \frac{d^3f}{dx^3} + ... 0 Answer: Yes it does, and there is a much simpler method using translational symmetry. We begin as before, but then substitute \zeta=\xi-x to remove the x-dependence of the boundary: \begin{align} \frac{\partial \hat T_D}{\partial t} &= \frac{1}{m(D)}\int_{x+D} \frac{\partial T}{\partial t}\!(\xi,t)\,d\xi \\ &= \frac{1}{m(D)}\int_{x+D} ... 3 The underlying assumption is that each atom's decay is treated as an independent event, and that the probability of it decaying in any given second is a constant. (In other words, a Poisson process.) So, the more atoms you have around to decay, the more decays per second you'll get. 0 Hint: You can solve this separable equation to get the general solution (in the implicit form: x a function of y). There is only one value of the constant that makes x \to 0 as y \to 0+. EDIT:\int \dfrac{dy}{1+\sqrt{y}} = 2 \sqrt{y} - 2 \ln(1+ \sqrt{y}) + C$$EDIT: More generally, for y' = f(y) where f is continuous and positive for y \ge ... 0 I think a maximal domain of solution is an interval (also called the interval of validity of the ODE) by definition, thus it is I = (-1,+\infty). 3 The equation is$$ \begin{bmatrix} x'\\y' \end{bmatrix} = \begin{bmatrix} 2+\cos(t)&1-\sin(t)\\ -1-\cos(t)&2+\sin(t) \end{bmatrix} \begin{bmatrix} x\\y \end{bmatrix}\tag{1} $$Although the trace of the square matrix above is 4+\sqrt2\sin(t+\pi/4) which is positive for all t, Artem has pointed out an article in which an example is given where some ... 0 Your differential equation can be solved as:$$mu'' + \gamma u' + ku = F(t)$$Since there is no forcing function (F(t) = 0), And it is critically damped, meaning \gamma = 2\sqrt{km} = 2\sqrt{9} = 6, Our differential equation becomes:$$u'' + 3u' + 9u = 0$$Since the differential equation is homogeneous, we can construct our characteristic ... 3 First question Multiply the first equation by x', the second by y' and add:$$ x\,x'+y\,y'=2(x^2+y^2)+x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t. $$If 0\le t\le\pi/2 then 0\le\cos t\le1, 0\le\sin t\le1 and$$\begin{align} x^2\cos t+y^2\,\sin t-2\,x\,y\cos t\sin t&\ge x^2\cos^2 t+y^2\,\sin^2 t-2\,x\,y\cos t\sin t\\ &=(x\cos t-y\sin t)^2\\ ...

2

We are given: $$\frac{dT_i}{dt}=\frac{1}{RC}(T_a-T_i)+\frac{1}{C} \Phi_h$$ This is a separable equation and separating yields: $$\displaystyle \int \dfrac{1}{\frac{1}{RC}(T_a-T_i)+\frac{1}{C} \Phi_h}~dT_i = \int dt$$ Integrating each side yields: $$-RC \ln(T_a + \Phi_h R - T_i) = t + k$$ We can re-write this as: $$\ln(T_a + \Phi_h R - T_i) = ... 0$$ \begin{align} b = &\frac{1}{RC}\left(T_a +R\Phi_h\right)&\\ a = &\frac{1}{RC}& \end{align} $$you end up with an equation similar to what you know :) i,e,$$ y' = b - a\cdot y $$that you stated you know above 1 I do not know which stability criterion should be applied, so I try a simple argument: An object at the origin (0,0) which is disturbed by a small positive displacement \epsilon in x-direction will experience a velocity x' = (2 + \cos t) \epsilon > 0, it will move away from the origin. For a small disturbance \epsilon in y-direction it will ... 0 The equations you quote are already in the desired form. Those methods generally want y'=f(x,y,stuff). Note that the independent variable can appear in any form on the right. In your problem that is t. Your suggestion is for dy/dt=y'=y'+f(x,y) but you don't want y' on the right. Both equations separate: you have A=\int_0^t 635.14 \tanh (0.6033t)\ ... 1 If x = y, then$$(x^2 + y^2)(y - 1) + yx - y^2 = 2x^2(x - 1) < 0since x < 1. Now suppose x \neq y. If x and y are positive, then 0 < x^2 < x and y^3 < y^2, thus \begin{align}(x^2 + y^2)(y - 1) + yx - y^2 &= x^2y + y^3 - x^2 - y^2 + yx - y^2\\ & < x^2y - x^2 - y^2 + yx \\ & < 2xy - (x^2 + y^2)\\ & < ... 2y\dfrac{dy}{dx}=(x+7)(y^2+6)\implies \dfrac{y}{y^2+6}\dfrac{dy}{dx}=(x+7)\implies \dfrac{y}{y^2+6} dy=(x+7) dx\implies \int\frac{y}{y^2+6}dy=\int x+7 \,dx\implies \frac12\log|y^2+6|=\dfrac{x^2}2+7x\implies \log|y^2+6|=x^2+14x+c\implies y^2+6=e^{x^2+14x+c}\implies y=\sqrt{e^{x^2+14x+c}-6}$$2$$y \dfrac{dy}{dx} = (x+7)(y^2 + 6)$$Divide both sides by y^2 + 6 to get$$\dfrac{y}{y^2+6}\,dy = (x+7)\,dx$$Now, it's just a matter of integrating each side of the equation:$$\int \dfrac{y}{y^2+6}\,dy = \int(x+7)\,dx$$2 You have$$\frac{y}{y^2 + 6} \frac{dy}{dx} = (x + 7)$$Now integrate both sides with respect to x and you have$$\int \frac{y}{y^2 + 6} dy = \int (x+7) \ dx$$2 First, a quick but very important warning: In basic calculus, the symbol combination$$ydy - x dx = 0$$does not really make sense. It is a dirty trick because it is the quickest way of transforming$$y\frac{dy}{dx} - x = 0,$$which is a differential equation, into$$\int ydy - \int xdx = 0,$$which is the next step in solving the differential ... 1 -\dfrac{d}{dx} \left(p(x) \dfrac{d}{dx} y\right) +q(x) y = y'' + k y where p(x) = -1 and q(x) = k. 0$$\dfrac{1}{a}\int \dfrac{dy}{y+\dfrac{b}{a}}=\dfrac{1}{a}\int\dfrac{d\left(y+\dfrac{b}{a}\right)}{y+\dfrac{b}{a}}=\int dx$$1 Hint: When expanding the power series solution about an ordinary point x_0, the radius of convergence is at least as large as the distance from x_0 to the nearest singular point in the complex plane. 3 If you have$$y'(x)=ay(x)+b(x),$$for constant a, then the simpliest way is to multiply the equation by e^{-ax} (nonzero everywhere) and deduce that$$(y(x)e^{-ax})' = e^{-ax}b(x),$$which implies that$$y(x)e^{-ax} - y(0) = \int_0^x e^{-as}b(s)ds.$$Finally,$$y(x) = y(0)e^{ax} + \int_0^x e^{-a(x-s)}b(s)ds.

2

As a separable equation: \begin{align} {dy\over dx}&=ay+b\\ {dy\over ay+b}&=dx\\ \int {dy\over ay+b}&=\int\,dx\\ {1\over a}\ln|ay+b|&=x+C\\ \ln|ay+b|&=ax+aC\\ e^{\ln|ay+b|}&=e^{ax+aC}\\ |ay+b|&=e^{ax}e^{aC}\\ ay+b&=\pm e^{aC}e^{ax}\\ ay+b&=De^{ax}, \quad D\not= 0\\ y&=Ke^{ax}-{b\over a} \end{align} You can also solve ...

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