ordinary differential equations – Solve: $ e ^ {- y} frac { partial z} { partial x} + e ^ {- x} frac { partial z} { partial y} = e ^ {x + y} $

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Lyapunov functions for ordinary differential equations

I am trying to solve exercise 3.4 (among others) of "ordinary differential equations: qualitative theory" by Barreira e Vall. My goal is to cover my handicap on this theory in order to follow my course in applied mathematics. But I have a hard time guessing the functions of Lyapunov (and noting the phases of the portrait).
For example, is there a Lyapunov function for the zero solution of the system below?
$$ left { begin {matrix} x & # 39; & = & -x + x ^ 2 + y ^ 2 \ y & # 39; & = & 2x-3y + y ^ 3 end {matrix} right. $$
I would like a hint on how to solve these kinds of problems, so that I can solve the other elements. It would also be great to have the portrait phase of the above equation, as I also have trouble solving these kinds of problems.

Thanks in advance,

ordinary differential equations – Does y (x) equal 1, a solution of y & # 39; & # 39; + 2y & # 39; + y = x

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A doubt about the ordinary points of the differential equations (linear)

We say that $ x_0 $ ( $ neq infty) $ is a $ { bf ordinary} $ point of
linear equation

$$ y ^ {(n)} + p _ {(n-1)} (x) y ^ {(n-1)} + … + p_1 (x) y & # 39; + p_0 y = 0 $$

if the coeficient functions of $ p_0, .., p_ {n-1} $ are all analytical in a
neighborhood $ x_0 $ in $ mathbb {C} $.

Now, I have a doubt about the differential equation $ y = = | x | y $

Then we see that the only point where there can be any derivative is $ x = $ 0so it's not true that this equation has all the ordinary points except $ x_0 = $ 0?

My book, however, says that the equation does not have ordinary points because | x | Nowhere is analytic.

What do I miss here?

ordinary differential equations – Existence and uniqueness of a solution of an implicit ODE linear, inhomogeneous

Problem: Given the differential equation $$ x ^ 2 y & # 39; (x) + 4x , y (x) +2 , y (x) = r (x) $$ for a function $ r en C ^ 2 ( mathbb R) $and the initial values $$ y (0) = y_0, qquad y (0) = y_1, $$ how to analyze if a solution $ y in C ^ 2 ( mathbb R) $ exists locally around $ x_0 = $ 0 and / or is unique locally, depending on the choice of $ y_0 $ and $ y_1 $?

Some ideas

  • Obviously, the Picard-Lindelöf theorem is not applicable around $ x_0 = $ 0 because the ODE is implicit. For $ x_0 neq $ 0however, we have an interval that does not contain $ 0 $, which gives an explicit ODE of the form $ y & # 39; + frac 4x y + frac {2} {x ^ 2} y = frac {r} {x ^ 2} $, where we observe that the homogeneous equation associated with the general solution $ y (x) = frac {c_1} {x} + frac {c_2} {x ^ 2} $ for $ c_1, c_2 in mathbb R $ be sure $ (0, + infty) $ or $ (- infty, 0) $.
  • Yes $ c_1 neq 0 neq c_2 $then the solution is unlimited. However, this does not mean that the general solution of the inhomogeneous problem is not $ x_0 = $ 0 (which would have resulted in a contradiction with the existence of a solution).
  • $ y (0) = y_0 $ implies that $ y_0 = frac {r (0)} {2} $.
  • The first term of the ODE $ x ^ 2 , y & # 39; $ is differentiable with $ frac { mathrm} { mathrm dx} x ^ 2 y & # 39; (x) = lim_ {h to 0} frac {h ^ 2 y & # 39; (h) -0} {h} = lim_ {h to 0} h , y & # 39; (h) = $ 0 (using the continuity of $ y & # 39; $ and therefore the delimitation on a certain interval around $ 0 $). Although we do not know if $ y & # 39; & # 39; $ exists or not we still have an expression for the derivative of $ x ^ 2 y & # 39; $ at $ x_0 = $ 0. This means that after differentiating the ODE to $ x_0 = $ 0, we have $ 6 , y (0) = r & # 39; (x) $ or $ y_1 = frac {r ($)} {6} $.
  • The last two points imply that ODE has no solution for $ y_0 neq r (0) / $ 2 and $ y_1 neq r $ (0) / $ 6. This reduces the problem to the case analysis $ y_0 = r (0) / $ 2, $ y_1 = r $ (0) / $ 6.

Ordinary Differential Equations – Someone can help me solve this ODE

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set theory – Is there a term that refers to a "set of sets" in math to distinguish an ordinary mathematical object?

In mathematics, a set is a well-defined collection of distinct objects, considered an object in its own right.

for example

A = {2, 1, 3}
B = {blue, white, red}
C = {3, 5, 6}
D = {A, C}

all the above is a set, A, B, C are normal sets and D is a set. Is there a term referring to this type of set (D), in order to distinguish a set of other sets (such as A, B, C)?

Problem in ordinary second order differential equations with a second or second coupler

I came across a system of ordinary second-order differential equations and I was perplexed for a few days.
enter the description of the image here
It was originally easy to solve, but since the coupling term is second-order, it can not be solved directly by Mma.
Here is the code that I tried

eqs = {q(1)''(t) + q(1)(t) -Sin(Pi t) (9.81 - q(1)''(t) Sin( Pi t) - q(2)''(t) Sin( 2*Pi t)) == 0 ,
  q(2)''(t) + q(2)(t) - Sin(2*Pi t) (9.81 - q(1)''(t) Sin( Pi t) - q(2)''(t) Sin( 2*Pi t)) == 0}
int = {q(1)(0) == 0, q(2)(0) == 0, q(1)'(0) == 0, q(2)'(0) == 0};
var = {q(1), q(2)};
exp = eqs~Join~int;
NDSolve(exp, var, {t, 0, 1})

enter the description of the image here
I do not know if this system can be solved with mma and I do not know how to solve it. I really hope you can help me.

How to create an APFS volume in an ordinary file?

If you have a file containing a raw image of a disk, you can use the following command in the terminal:

hdiutil attach -imagekey diskimage-class=CRawDiskImage myimage.raw

If you do not have an existing image, you can create a new empty file named myimage.raw then fix it without mounting:

hdiutil attach -imagekey diskimage-class=CRawDiskImage -nomount myimage.raw

You can then create APFS volumes on these disks, as if you had a physical disk, and then mount them. This can be done via the command line with a command such as:

diskutil apfs create disk5 MyNewVolumeName

or disk5 must be replaced by the device name that you received previously from the attach command.

ordinary differential equations – Add the output of a function to a matrix in matlab

For context, I've already asked a question here about this same problem. I believe that my logic is now correct, even if I struggle to get the output in the format that I want. I would like to enter the initial conditions for this problem ($ x (0), dot {x} (0), y (0), dot {y} (0) $ and time information (initial time, end time, no time) and output a matrix containing the following values:

$$ P = begin {pmatrix}
z_1 (t_0) & z_1 (t_1) & z_1 (t_2) & … \
z_2 (t_0) & z_2 (t_1) & z_2 (t_2) & … \
z_3 (t_0) & z_3 (t_1) & z_3 (t_2) & … \
z_4 (t_0) & z_4 (t_1) & z_4 (t_2) & …
end {pmatrix} tag {1} $$

I use the Runge-Kutta 4th order method and, until now, I have created the following functions:

function (P) = ivpSolver (t0, dt, tends, x_0, xdot_0, y_0, ydot_0)

z (1,1) = x_0;
z (2.1) = xdot_0;
z (3.1) = y_0;
z (4.1) = ydot_0;

t (1) = t0;
n = 1;
while t (n) <= tends

t (n + 1) = t (n) + dt;
(z (1, n + 1); z (2, n + 1); z (3, n + 1);> z (4, n + 1)) = step RK4 (dt, z (1, n) z (2, n), z (3, n), z (4, n));

n = n + 1
end

function (znext) = stepRK4 (dt, z1, z2, z3, z4)

k1 = dtstateDeriv (z1, z2, z3, z4);
k2 = dt
stateDeriv (z1 + k1 (1) / 2, z2 + k1 (2) / 2, z3 + k1 (3) / 2, z4 + k1 (4) / 2);
k3 = dtstateDeriv (z1 + k2 (1) / 2, z2 + k2 (2) / 2, z3 + k2 (3) / 2, z4 + k2 (4) / 2);
k4 = dt
stateDeriv (z1 + k3 (1), z2 + k3 (2), z3 + k3 (3), z4 + k3 (4));

znext = (z1; z2; z3; z4) + (1/6) * (k1 + 2 * k2 + 2 * k3 + k4);

end

function (dz) = stateDevice (z1, z2, z3, z4)

H = 1500 * 10 ^ 3; % desired scientific orbit above the surface in m m = 2000; %
mass of the spacecraft in kg A = 12; % of the cross section of the spacecraft in
m Cd = 1.5; % drag coefficient of the spacecraft R = 2574.73 * 10 ^ 3; %
radius of the planet in m M = 1.3452 * 10 ^ 23; % mass of the planet in kg
G = 6.673 * 10 (- 11); % of the gravitational constant in Nm ^ 2kg ^ -2

r = sqrt (z1 ^ 2 + z3 ^ 2); % this is equivalent to sqrt (x ^ 2 + y ^ 2)
modv = sqrt (z2 ^ 2 + z4 ^ 2); % this is equivalent to sqrt (xdot ^ 2 + ydot ^ 2)
rho = profileTitan (r-R); % bring our rho value, taking into account
This is the area of ​​grav = GM / Rd (3/2); % c is the
"magnitude" part of the friction part of the ODE
fric = (rho
Cd * A) / (2 * m) * modv; % c is the "magnitude" part of the
severity part of the ODE

dz (1) = z2; % z1 = z2 dz (2) = – gravZ1-moneyz 2; % z2 = ODE part for x
dz (3) = z4; % z3 = z4 dz (4) = – gravZ3-moneyz4; % z4 = ODE part for there

end

My problem is that the function does not seem to like that I'm trying to get it out in a new column of the matrix. Any ideas how can I change that?