Автор: Nabiyeva Djamale O.

электризации при механизированном заряжании взрывных полостей россыпными взрывчатыми веществами //Горный информационно -аналитический бюллетень, 2012, - №4. - С. 342-345.

5.Белин В.А., Кутузов Б.Н., Ачеева Э.А. Снижение интенсивности и вероятности воспламенения взрывчатых аэровзвесей при пневмозаряжании //Горный информационно-аналитический бюллетень, 2014, - №5. - С. 356-361.


DOI: 10.31618/ESU.2413-9335.2020.6.71.613

1Djamale O.Nabiyeva, 2Afsuna X.Xankishiyeva
1Baku Biznes Unversity,
2Azerbaijan State Pedagogical University

In the present case, the solution of the ordinary differential equation in the Scilab software package is shown. Key words: Differential equation, Scilab package.

It is known that,

H(t, x, x&, x", ..., x(n)) = 0 (1)

the equation in the form is called the n-order differential equation.

The solution of the differential equation x (t), converts the equation into the same. System of n-order differential equations

X &l=fl(t,Xl,X2,...,Xn) x&2=f2(t,xi,x2,...,xn) (2)

X n= fn(t,Xl,X2,...,Xn)

system. (2) system solution

X(t)=(Xl (t),X2(t),...,Xn (t))

It is a vector and converts the system&s equations into one.

Differential equations and systems have an infinite number of solutions, which are constants. Additional prerequisites and boundary conditions must be provided for a single solution. The number of such conditions must be consistent with the design of the differential equation or system. Differential equations, depending on the type of additional conditions, on the Koshi problem - all additional conditions are given at one (usually the first) point of the interval. Additional terms of the boundary value problems are provided within the range boundaries.

In many cases there is an exact solution of the equations. However, sometimes it is not possible to find the exact solution of the equation, especially the system of equations. In these cases, the solution is found using numerical methods. Numerical methods are also used to find the numerical solution of the equation known as the analytical solution.

To solve the differential equation and system, Scilab has the following function.

[y,w,iw]=ode([type],y0,t0,t [,rtol [,atol]],f[,jac] [,w,iw])

Here y0 - initial condition vector; t0 - the starting point of the integration interval; t - coordinates of the network nodes; f - (2) an external function that determines the right side of the equation ; y - (3) solutions vector.

= f(t, y)

y(t0) = y0

to solve the ordinary differential equation in the


need to call the function. ode Let&s take a look at the non-essential parameters of the function.

type - the parameter used to select the solution method or the type of problem to be solved with one of

the lines; adams - used in solving Adams method of differential equations or systems; stiff- is used for solving tough issues; rk - used when solving differential equations or systems; rkf - displayed when selecting a five-step four-way Rungee-Kutt method; fix - Apply the Rungee-Kutt method to the intended steps;

rtol, atol - relative and absolute errors of calculations, vector with the size y, in case of silence using rtol = 0.00001, atol = 0.0000001, rkf and fix parameters - rtol=0.001, atol=0.0001 happens; jac is a matrix showing the Jacobian side of the rigid system; The matrix is represented by an external function in the form of J = jack (t, y); w, iw are vectors that store information about integration settings.

Let&s look at using functions to solve the following issues.

Problem 1.

—— + x = sin(xt) ,x(0) = 1.5 dt

let&s solve the Koshi problem. il = -x + sin(xt), x(0) = 1.5.

Solution dt

Let&s write equation above the way it is The graphic modeling the process described in this

equation is as follows:

Then accept it as an external function Let&s use the function y = ode (x0, t0, t, f). Parameters we use:

f - reference to created function f (t, x); t - network coordinates; x0, t0 - prerequisites x (0) = 1.5 y is the result of the function; Listinq 1

-->function yd=f(t,x),yd=-x+sin(t*x),end


-->x0=1.5;t0=0;t=0:1:35; -->y=ode(x0,t0,t,f); -->plot(t,y) Problem 2.

x&=cos (xy)

Below are the Siclab commands required for both numerical and graphical solution of functions and problems that describe the system of ordinary differential equations. Listinq 2

// Function describing the system of differential equations

function dy=syst(t,y) dy=zeros(2,1); dy(1)=cos(y(1)*y(2)); dy(2)=sin(y(1)+y(2)*t); end function

// Solution of the system of differential equations x0=[0;0] ;t0=0 ;t=0 :*0.1*:10 ;y=ode(x0,t0,t,syst); // Graphical solution of the problem plot(t,y)

y&=sin(x+ty) x(0)=0y(0)=0

solve the Koshi problem in [0.10].

Figure 2. Problem 2&s solution schedule

Problem 3. Açagida verilmiç sistem ûçûn Koçi masalasini hall edin

System solution Listinq 3. Solution of Problem 3 -->B=[119.46 185.38 126.88 121.03;-10.395 -10.136 -3.636 8.577;

-->-53.302 -85.932 -63.182 -54.211;-115.58 -181.75 -112.8 -199];

-->function dx=syst1(t,x), dx=B*x, end function

-->function J=Jac(t,y),J=B, end function -->x0=[1;1;1;1]; t0=0; t=0:0.01:5; -->y=ode("stiff",x0,t0,t,syst1,Jac); -->plot(t,y); xgrid();

Figure 3 shows the graphical solution of the problem.

Figure 3. Problem 3 graphical solution Problem 4. Solve the system of nonlinear rigid differential equations

Figure 4 shows the solution of the system at the interval [0,2].


00 0.2 04 06 OA 1.0 12 14 1.0 1.8 2.0

Figure 4. Problem 4 graphical solution

Listinq 4.

function dx=syst2(t,x) // Function defining a system of ODE dx=zeros(3,1); dx(1)=-7*x(1)+7*x(2);

dx(2)=157*x(1)+x(2)-1.15*x(1)*x(3); dx(3)=0.96*x(1)*x(2)-8.36*x(3); end function

-->>// Solution-->x0=[-1;0;1]; t0=0; t=0:0.01:2;y=ode("stiff",x0,t0,t,syst2);

-->plot(t,y); xgrid();

Problem 5. Consider the following boundary problem [0.25; 2] Solve in the interval.

Since y=— let us convert the equation to the

system by substituting it.

dy — ■ m dx

x(0.25)= -1.

± = - 4y-13x +esin (t) , ^ = y y(0.25)=1,

Let us design and solve the system computing function.

Listinq 5 Problem 5 solution

function F=FF(t,x) F=[-4*x(1)-13*x(2)+exp (t);x(1)]; end function

-->// Solution of the system of differential equations-->X0=[1;-1];t0=0.25;t=0.25:0.05:2; -->y=ode("stiff&,X0,t0,t,FF); -->// Graphic representation of the solution -->plot(t,y); xgrid();

The graphical solution of the problem is shown in Figure 5.

Figure 5. Problem 5&s solution schedule


l.Verjbitski V.M. Fundamentals of Numerical Methods. — M.: Vishaya Shkola, 2002. — 840 p.(in Russion)

2.Goloskov D.P. Equations of mathematical physics. Problem solving in Maple system. — Spb.: Piter, 2004. — 539 p.(in Russion)
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differential equation scilab package
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