New answers tagged ordinary-differential-equations
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Approximating solution to vector recurrence relation with element-wise exponential
This turned out to be quite long and complicated, maybe someone can come up with a shorter solution. You have the system of ODEs:
$$\frac{d\vec{x}(t)}{dt} = A \exp(-\gamma \odot \vec{x(t)})
$$
...
- 301
1
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Can we convert the following integral equation to a differential equation:$h(r)= \int_0^\infty\frac{f(x)}{e^{r x} + 1} dx$?
Alternatively, you can consider a little bit more general problem by introducing an auxiliary variable and defining the following function :
$$
g(r,s) := \int_0^\infty \frac{f(sx)}{e^{rx}+1} \,\mathrm{...
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Global existence of an ODE
To guarantee the existence of a global solution (you can't in general), you need to look at the forward dynamics $u'=f(u)$ and the reverse dynamics $u'=-f(u)$.
Let $f(u) = u-u^3$, note that $f(u) = 0 $...
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Finding singular solution to a Lagrange equation
A differential equation of type
$$y = x\varphi \left( {y'} \right) + \psi \left( {y'} \right),$$
where φ (y' ) and ψ (y' ) are known functions differentiable on a certain interval, is called the ...
2
votes
Stochastic differential equations with only time integrals
OK - time for an update. Thanks to everyone who commented.
As pointed out by some commentators, these types of differential equations are called Random (Ordinary) Differential Equations. They have ...
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Change of variables by scaling.
You need to take the transformation rule into account. Because of the scaling $\mathrm{d}{z}\not = \mathrm{d}r$ but rather $\mathrm{d}z=\sqrt{\lambda}~\mathrm{d}r$ and if I did not mix up my ...
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How would I prove or disprove this statement?
There is nothing that prevents this equation or its explicit form
$$
y^{(n)}=\frac{\sum_{k=0}^{n-1}y^{(k)}}{\prod_{k=0}^{n-1}y^{(k)}-1}
$$
from having a local, unique and analytic solution. Obviously ...
- 128k
3
votes
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The space of solutions for an ODE on an interval containing a regular singular point
Try $x^2 y''(x) - 6 x y'(x) + 12 y(x) = 0$. $y(x) = a x^3 + b x^4$ are solutions, and since all of these have $y''(0) = 0$ you can have a solution with different $a$'s and $b$'s on $(0,+\infty)$ and ...
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Asymptotic stability and Lyapunov functions
That is not possible because on any closed annulus $\overline B_{\epsilon_1}(x_0)\setminus B_{\epsilon_2}(x_0)$ with $\epsilon_2<\epsilon_1$, $\Lambda'(t)\le -\alpha$ for some $\alpha>0$ (here $...
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Global existence of an ODE
Since $f(u)=u-u^3$ is locally Lipschitz, the IVP $u'=f(u),u(t_0)=u_0$ has a unique solution $u=u(t)$ in $(t_0,t_0+\delta)$ for any $u_0\in \mathbb R$ for small $\delta>0$. Extend $u=u(t)$ to the ...
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4
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Does Solution to Banach Space ODE $\dot{x} = Ax$ with Non-Positive Spectrum Converges to Kernel?
No. Consider the case $X = \mathbb{R}^2$ and the system
$$
\begin{aligned}
x_1' &= x_2 \\
x_2' &= 0.
\end{aligned}
$$
The kernel of the operator $A = \begin{pmatrix}0 & 1 \\ 0 & 0 \end{...
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2
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Can we convert the following integral equation to a differential equation:$h(r)= \int_0^\infty\frac{f(x)}{e^{r x} + 1} dx$?
You can go part of the way if you substitute $x$ with $u=rx$.
You then have $x=u/r$ and $dx=\frac{1}{r}du$.
If $r>0$
$h(r)=\frac{1}{r}\int_{0}^{\infty}\frac{f(\frac{u}{r})}{e^u+1}du \\
\Rightarrow ...
- 36
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Maximal interval of existence of a first order IVP
Since $f$ is bounded, say $|f(s)|\le M$ and therefore $|y'(x)|\le M$, you know that any local solution issued from $(0,0)$ will be contained in the sector $|y|\le Mx$. Therefore you can extend the ...
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votes
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Differential inequation $f'(x) > f(x)/x$
"Is this condition sufficient?" It is almost sufficient.
"Can I already conclude? Or am I missing something?" You are missing negative numbers.
Consider this Image :
We do not ...
- 12.3k
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A brick sliding in an horizontal plane after an initial push (under Coulomb's dry friction) - closed form solutions validation?
There is nothing particularly obstructive about the use of $\mathrm{sgn}$ in the DE. The DE remains locally Lipschitz almost everywhere. To start, the DE in question, $$x'' = -k\,g\,\mathrm{sgn}(x')$$ ...
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Using the Laplace transform, solve the system of linear differential equations with constant coefficients
You can rewrite the Laplace transforms of the two ODEs as
\begin{align}
Y(s)+Z(s)&=\frac{1}{s^3}+\frac{3}{s}, \tag{1} \\
s^2Y(s)-Z(s)&=3s-2+\frac{1}{s+1}. \tag{2}
\end{align}
Adding $(1)$ and $...
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0
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What is the definition of a differential equation's general solution?
I like the way you combined the Solutions !
There are ways to combine 2 (or more) Constants into one , though that is not Simple Constant , In other words , I think Constants should be Simple ...
- 12.3k
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Solving ODE system with less equations
Yes, in this case of a linear equation you do not need to solve the original equation as prerequisite to solve the sensitivity equations.
This is of course different for non-linear equations, as each ...
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What is the definition of a differential equation's general solution?
An ODE, to be "ordinary", has a domain, an open set in the time-state space, that contains no singular points for the DE. Inside the domain the derivative can be expressed as a continuous ...
- 128k
2
votes
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Given is the system of differential equations
a) and b) $H=3x^2+4y^2$ is a solution. So the phase portraits are $ 3x^2+4y^2=C$ with different constant 'C'. c) You solve the eigenvalue of this system，and for a linear system, if all the eigenvalues ...
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Find solution of the IVP $ y' = y+ \frac12 |\sin(y^2)|,\,\, x>0,\,\, y(0) = -1$
You have $0 \leqslant y' - y \leqslant \frac{1}{2}$.
Multiplying by $e^{-x}$ yields $0 \leqslant y'(x)e^{-x} - y(x)e^{-x} \leqslant \frac{1}{2}e^{-x}$.
Integrating on $[0,x]$, after noticing that $y'(...
- 20k
1
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About solution to homogeneous ODE $u'' + u = 0$
Another approach, let $f(x) = u(x)^2+(u'(x))^2$. Then $f(0) = 0$ and $f'(x) = 0$ hence $f(x) = 0$.
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Find solution of the IVP $ y' = y+ \frac12 |\sin(y^2)|,\,\, x>0,\,\, y(0) = -1$
You should check first that the right hand side is locally Lipschitz in $y$. Picard-Lindelöf implies existence and uniqueness.
Under these circumstances, solutions to first order autonomous ODEs (like ...
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About solution to homogeneous ODE $u'' + u = 0$
We can check that $u''+u=0$ has the Solution $u=c_1e^{-x}+c_2e^{+x}$ :
$u'=-c_1e^{-x}+c_2e^{+x}$
$u''=c_1e^{-x}+c_2e^{+x}$
[[ Solving Characteristic Polynomial $D^2+1=0$ , we get $D=+1,D=-1$ , hence ...
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Why are some differential equations unsolvable?
The simple pendulum is also unsolvable analytically, unless you allow an Elliptic Integral, a special function.
This is in direct relation to Liouville's theorem on analytical integration. ODEs are as ...
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Global existence of solution of ODE - Gray-Scott model
Even easier/more intuitive and with some qualitative remarks on the behaviour of solutions.
It can be seen that the positive cone $K_+:=[0,+\infty)\times [0,+\infty)$ is forward invariant (if we start ...
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Poincaré-Bendixson applied to region enclosed by homoclinic orbit
I think your assumption is generally wrong. Consider the following system of differential equations: $x'=x^2-y^2,\; y'=2xy.$ The origin is the only equilibrium position. Note that the equation of the ...
- 21
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Solving Laplace equation
In order to use the method of separation of variables, we need homogeneous boundary conditions. Defining $v:=u-1$, we can rewrite the problem as
\begin{align}
v_{xx}+v_{yy}&=0, \tag{1} \\
v(x,0)&...
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votes
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Global existence of solution of ODE - Gray-Scott model
Assume by contradiction, there exists a local in time solution $(x,y)$, which explodes in finite time $T$ (this is the opposite of having a global solution by the explosion in finite time theorem).
...
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What does ordinary differential equation mean, based on the word itself?
Elaborating the "WHY" :
Etymology/History of the term can be traced back by the HSM Post which user Moo commented.
The semantic meaning can be considered like this :
We can have Linear ...
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Green function of a differential equation
It is to be noted that the operator $\Lambda = \partial_{yy} \circ L$ generates a PDE by acting on a bivariate function $u(x,y)$. In consequence, the associated fundamental solution $G(x,y;s,t)$ will ...
- 10.3k
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Using the Laplace transform, solve the system of linear differential equations with constant coefficients
Now you face the following linear system :
$$
\begin{bmatrix} s & s \\ s^2 & -1 \end{bmatrix}
\begin{bmatrix} Y(s) \\ Z(s) \end{bmatrix}
=
\begin{bmatrix} 3+\frac{1}{s^2} \\ 3s-2+\frac{1}{s+1} ...
- 10.3k
4
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Find $\sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\left(e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\cdots-\frac{x^n}{n!}\right)\right)$
Another way to solve the problem is to use:
How to integrate $ \int x^n e^x dx$?
or the regularized gamma function to find:
$$\sum_{j=n+1}^\infty\frac{x^j}{j!}=\frac{e^x}{n!}\int_0^x e^{-t}t^ndt$$
...
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5
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Find $\sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\left(e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\cdots-\frac{x^n}{n!}\right)\right)$
Some thoughts. (We need to prove that the function is differentiable etc.)
We have
$$A'(x) = A(x) + \sum_{n=1}^{\infty}\left(n\sin\left(\frac{\pi n}{2}\right)\cdot \frac{x^n}{n!}\right)
= A(x) + x\cos ...
- 40.3k
2
votes
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Showing $M = i\sum\limits_{n=-\infty}^{\infty} u_{n}u_{n+1}^{\ast} - u_{n}^{\ast}u_{n+1}$ is conserved
Your last equation has a few mistakes. Since $i \dot{u}_{n}^* = (-i\dot{u}_n)^*=(u_{n+1}^*+u_{n-1}^*)(1+|u_{n}|^2)$, it should be
\begin{align}
\frac{dM}{dt}&= \sum\limits_{n=-\infty}^{\infty}[-\,(...
- 11.3k
3
votes
Using the Laplace transform, solve the system of linear differential equations with constant coefficients
Hint: From your first equation $sY(s) + sZ(s) -3 = 1/s^2$, solve for $Z(s)$ in terms of $s$ and $Y(s)$. Then, plug this into the second equation $s^2Y(s) - 3s + 2 - Z(s) = \frac 1 {(s+1)}$ and solve ...
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How to solve linear differential equation with polynomial coefficients?
As this equation has an essential singularity at $x=0$ it might help to shift that singularity to infinity. Thus insert $y(x)=u(t)=u(1/x)$,
$$
y'(x)=-\frac1{x^2}u'(1/x)\\
y''(x)=\frac1{x^4}u''(1/x)+\...
- 128k
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How to linearize first order ODE
Assuming you need to linearize
$$
f' + g(f) = 0
$$
around $f_0(x)$ we have
$$
f' + \left(\frac{\partial g}{\partial f}\right)_{f=f_0}(f-f_0) + O(|f-f_0|^2)=0
$$
so we follow with
$$
f' + \left(\frac{...
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vote
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Solutions to $y'=y$
This approach gives you all the solutions:
First, note that $e^x$ is a solution. Suppose that $f$ is any other solution and consider $g(x)=f(x)e^{-x}$. Then $g'(x)=f'(x)e^{-x}-f(x)e^{-x} = 0$, so $g(x)...
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How to linearize first order ODE
\begin{equation}
\frac{df}{dx} + \frac{f^2}{a} + \frac{f}{2a} - \frac{3}{2a} =0\quad \text{Condition : }f(a)=1
\end{equation}
This is a Riccati ODE which can solved easily for the general solution.
...
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2
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Numerical methods derivation
In one-step or Runge-Kutta methods you would explore their properties by inserting abstract Taylor series. This includes computing compositions of power series with symbolic variables as coefficients, ...
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How to solve linear differential equation with polynomial coefficients?
Writing the ode as (for non zero $x$)
\begin{equation}
x^{2}y^{\prime\prime}-\frac{y}{x^{2}}=0\tag{1}%
\end{equation}
Comparing the above to the generalized form of the Bessel ode given by Bowman (...
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5
votes
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Periodic solutions of differential equation
The constant $y=0$ is technically a periodic solution, but I'm assuming you're excluding that from consideration.
Suppose $y$ is a periodic solution of your equation that is not identically $0$, and ...
- 455k
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How is this calculation of solution to damped oscillator done in MIT OCW "Vibrations and Waves" course?
There is a very silly oversight in the OP that was pointed out by user Semiclassical. Let me show what that oversight was.
We posited a trial solution $z(t)=e^{i\alpha t}$.
We found two roots
$$\alpha=...
- 5,021
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$\frac{\text {d}y}{\text {d}x} = e^y$ general solution $y = -\ln(-x+C)$ or $y = -\ln|-x+C|$?
Checking the given answer:
\begin{align*}
\frac{\mathrm{d}}{\mathrm{d}x} \left( -\ln |-x-C| \right)
&= \frac{-1}{|-x-C|}\frac{\mathrm{d}}{\mathrm{d}x}\left( |-x-C| \right) \\
&= \frac{...
- 67.8k
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votes
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$\frac{\text {d}y}{\text {d}x} = e^y$ general solution $y = -\ln(-x+C)$ or $y = -\ln|-x+C|$?
Your answer is right and the given answer is wrong. One can check for example that $y=-\ln|{-}x|$ is not a solution to the differential equation when $x>0$ (where $y=-\ln x$).
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votes
How to solve linear differential equation with polynomial coefficients?
In TeX
With the differential equations for the chain rule
$$\frac{d}{dx}= \frac{dt}{dx}\ \frac{d}{dt} = \frac{1}{\frac{dx}{dt}} \ \frac{d}{dt} $$
$$x^4 y''(x) \ - \ y(x) \quad
\rightarrow
\...
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votes
Non-trivial solution to $x'(t)=2\cdot t\cdot x+t^2\cdot y, y'(t)=t^3\cdot x+4\cdot t\cdot y$ via by hand possible?
We face a system characterized by the following matrix :
$$
\begin{cases}
\dot{x} = -2tx + t^2y \\
\dot{y} = t^3x + 4ty
\end{cases}
\quad\Longrightarrow\quad
A =
\begin{pmatrix}
-2t & t^...
- 10.3k
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vote
Existence and uniqueness of homoclinic orbit
H(x,y) in the solution is not a first integral. It is easy to check that the trajectories along H are not constant.
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vote
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Determining $t(x)$ from $\frac{dx}{dt}$?
Numerically, the calculation of $x(t)$ is a table with two rows, $t$, and $x$, and it simultaneously calculates $t(x)$: calculating $t(x)$ from is simply a matter of reading the table inversely. ...
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