Methodology for compilation and solution applied problems ordinary theory differential equations
Drawing up a differential equation according to the conditions of the problem (mechanical, physical, chemical or technical) consists in determining the mathematical relationship between variable quantities and their increments.
In a number of cases, the differential equation is obtained without considering the increments - due to their preliminary consideration. For example, when representing speed by expression , we do not involve the increments ∆s and ∆t, although they are actually taken into account due to the fact that
.
Acceleration at some point in time t expressed by dependence:
.
When composing differential equations, the increments are immediately replaced by the corresponding differentials. The study of any process comes down to:
1) to determine its individual moments;
2) to establish common law his progress.
A separate moment of the process (the so-called elementary process) is expressed by an equation connecting variables process with their differentials or derivatives - differential equation; law general progress process is expressed by an equation connecting the variable quantities of the process, but without the differentials of these quantities.
There are no comprehensive rules for composing differential equations. In most cases, the technique for solving technical problems using the theory of ordinary differential equations comes down to the following:
1. Detailed analysis of the conditions of the problem and drawing up a drawing explaining its essence.
2. Drawing up a differential equation for the process under consideration.
3.Integration of the compiled differential equation and determination of the general solution of this equation.
4. Determination of a particular solution to the problem based on the given initial conditions.
5. Determination, as necessary, of auxiliary steam
meters (for example, proportionality coefficient, etc.),
using additional conditions of the problem for this purpose.
6. Derivation of the general law of the process under consideration and the number
a great definition of the sought-after greatness.
7. Analysis of the answer and verification of the initial position of the problem.
Some of these recommendations depending on the nature
tasks may be missing.
As in compiling algebraic equations, when solving applied problems using differential equations, much depends on the skills acquired by the exercise. However, here still in to a greater extent it requires ingenuity and a deep understanding of the essence of the processes being studied.
Let's consider the process of solving the following problems:
Task 3.1.
The temperature of the bread removed from the oven for 20 minutes. drops from 100 0 to 60 0 (Fig. 3.1). The air temperature is 25 0. How long after the start of cooling will the temperature of the bread drop to 30 0?
Solution:
By virtue of Newton's law, the rate of cooling of a body is proportional to the difference in body temperature and environment. This is an uneven process. As the temperature difference changes during the process, the rate of cooling of the body also changes. The differential equation for cooling bread will be:
where T is the temperature of the bread;
t – ambient air temperature (in our case 25 0);
k – proportionality coefficient;
Bread cooling speed.
Let be the cooling time.
Then, separating the variables, we get:
or for the conditions of this problem:
Seeing that
integrating, we get:
Potentiating both sides of the last equality, we have:
then finally
We determine an arbitrary constant C based on the initial condition: at min, T = 100 o.
or C=75.
The value is determined based on this additional condition: at min, T = 60 o.
We get:
And 
.
Thus, the equation for cooling bread under the conditions of our problem will take the form:
. (2)
From equation (2) we easily determine the required time at bread temperature T = 30 o:
Or.
Finally we find:
min.
So, after 1 hour 11 minutes. The bread is cooled to a temperature of 30 o C.
Problem 3.2. The heating main pipeline (diameter 20 cm) is protected with 10 cm thick insulation; thermal conductivity coefficient k=1.00017. Pipe temperature 160o; the temperature of the outer cover is 30° (Fig. 8). Find the temperature distribution inside the insulation, as well as the amount of heat given off by one linear meter pipes.
Solution. If the body is in a stationary thermal state and the temperature T at each point is a function of only one coordinate x, then according to Fourier’s law of thermal conductivity, the amount of heat emitted per second.
Often just a mention differential equations makes students feel uncomfortable. Why is this happening? Most often, because when studying the basics of the material, a gap in knowledge arises, due to which further study of difurs becomes simply torture. It’s not clear what to do, how to decide, where to start?
However, we will try to show you that difurs are not as difficult as it seems.
Basic concepts of the theory of differential equations
From school we know the simplest equations in which we need to find the unknown x. Essentially differential equations only slightly different from them - instead of a variable X you need to find a function in them y(x) , which will turn the equation into an identity.
D differential equations have huge applied value. This is not abstract mathematics that has no relation to the world around us. Differential equations are used to describe many real natural processes. For example, string vibrations, movement harmonic oscillator, using differential equations in mechanics problems, the speed and acceleration of a body are found. Also DU are widely used in biology, chemistry, economics and many other sciences.
Differential equation (DU) is an equation containing derivatives of the function y(x), the function itself, independent variables and other parameters in various combinations.
There are many types of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, first and higher order differential equations, partial differential equations, and so on.
The solution to a differential equation is a function that turns it into an identity. There are general and particular solutions of the remote control.
A general solution to a differential equation is a general set of solutions that transform the equation into an identity. A partial solution of a differential equation is a solution that satisfies additional conditions, specified initially.
The order of the differential equation is determined highest order derivatives included in it.
Ordinary differential equations
Ordinary differential equations are equations containing one independent variable.
Let's consider the simplest ordinary differential equation of the first order. It looks like:
Such an equation can be solved by simply integrating its right-hand side.
Examples of such equations:
Separable equations
IN general view this type of equation looks like this:
Here's an example:
When solving such an equation, you need to separate the variables, bringing it to the form:
After this, it remains to integrate both parts and obtain a solution.
Linear differential equations of the first order
Such equations look like:
Here p(x) and q(x) are some functions of the independent variable, and y=y(x) is the desired function. Here is an example of such an equation:
When solving such an equation, most often they use the method of varying an arbitrary constant or represent the desired function as a product of two other functions y(x)=u(x)v(x).
To solve such equations, certain preparation is required and it will be quite difficult to take them “at a glance”.
An example of solving a differential equation with separable variables
So we looked at the simplest types of remote control. Now let's look at the solution to one of them. Let this be an equation with separable variables.
First, let's rewrite the derivative in a more familiar form:
Then we divide the variables, that is, in one part of the equation we collect all the “I’s”, and in the other - the “X’s”:
Now it remains to integrate both parts:
We integrate and get general solution of this equation:
Of course, solving differential equations is a kind of art. You need to be able to understand what type an equation belongs to, and also learn to see what transformations need to be made with it in order to lead to one form or another, not to mention simply the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if you have at the moment you don’t have time to figure out how differential equations are solved, or the Cauchy problem has stuck like a bone in your throat, or you don’t know, contact our authors. In a short time we will provide you with a ready-made and detailed solution, the details of which you can understand at any time convenient for you. In the meantime, we suggest watching a video on the topic “How to solve differential equations”:
Differential called an equation that relates an argument X, the required function at and its derivatives at’ ,y” , ...,y (n) of various orders. In general, the differential equation can be written:
F (x, y, y’ ,y” , ...,y (n) ) = 0 .
The order of a differential equation is determined by the highest order of its derivative.
An example of a differential equation is Newton's second law, which determines the force F as the product of body mass m to the acceleration acquired under the influence of force a: F = ma.
Considering that acceleration is the first derivative of speed v, Let's write Newton's second law in the form of a first-order differential equation:
Or, since acceleration is the second derivative of the path S this law can be represented as a second order differential equation:
If the specific nature of the acting force is known, then by solving equation (2), we will establish the type of motion, i.e., we will find how for this case the path depends on time: S = f(t).
By decision of a differential equation is a function that turns this equation into an identity.
Example. Solve the equation: at’ - x = 0(3)
Let's rewrite original equation in the form:
(4)
In equation (4), a separation of variables is carried out, consisting in the fact that the desired function and its differential are placed in one part of the equation, and the argument and its differential are placed in the other.
To obtain a solution, it is necessary to get rid of differentials in equation (4), so we integrate its left and right sides:
(5)
When finding indefinite integrals, arbitrary constants appear WITH 1 And WITH 2 . They should be combined into one constant WITH. Finally:
Formula (6) is a general solution to the differential equation (3), containing as many derivatives of constants as the order of the differential equation.
It is easy to prove that function (6) is indeed a solution to equation (3), since its substitution in equation (3) turns the latter into an identity.
Arbitrary constant WITH can be determined if, along with the original differential equation, some additional information is given - they are called initial conditions.
For example: when x = 0y = 1. This initial condition, when substituting it into the general solution (6), allows us to find the constant WITH:
1 = 0+ CC = 1.
Then from the general solution (6) for a given initial condition we obtain a particular solution to equation (3) that does not contain an arbitrary constant:
2. Stages of solving problems when using differential equations
Differential equations are a mathematical apparatus that allows you to solve not only purely mathematical or physical problems, but also quantitatively describe a wide variety of processes (medico-biological, economic, social, etc.). Despite the variety of phenomena under consideration, the use of the apparatus of differential equations to study them must occur in a certain general logical sequence.
2.1. Drawing up a differential equation. This stage is the most difficult and responsible. Here it is necessary to take into account all the factors that influence the course of the process under study, perhaps make some assumptions, and determine the initial conditions. In this case, the researcher must be based on firmly established experimental facts or logical premises. For example, when creating mathematical models of the heart, their practical usefulness (obtaining new information to improve the diagnosis of cardiovascular diseases and increase the effectiveness of their treatment) will be determined by the completeness and correctness of the mathematical accounting of physiological data and clinical practice.
2.2. Solution of the equation. This stage can be considered simpler than the first, since it involves performing purely mathematical operations. If it is impossible to obtain a solution to a differential equation in analytical form, then it can be solved by calculation using modern computer technology.
2.3. Evaluation and analysis of the result. Having obtained a solution to a differential equation (or system of equations), it is necessary to evaluate the theoretical and practical usefulness of the results obtained - whether new patterns have been established in the course of, for example, physiological processes; Has the influence of selected factors on, for example, the degree of development and nature of the pathology, etc. been quantified?
In addition, the results obtained should be compared with the existing established facts. If unexpected and previously unknown information follows from a mathematical description of a physiological process, this may mean: 1) a new phenomenon has actually been established, which can subsequently be confirmed by experimental research; 2) the obtained result arose due to the fact that at the stage of drawing up the differential equation all the necessary factors were not taken into account or too rough assumptions were made.
Solving problems in physics or mechanics using differential equations breaks down, in accordance with what was said in paragraph 1, into the following stages:
a) drawing up a differential equation;
b) solution of this equation;
c) study of the obtained solution.
1. Establish the quantities that change in a given phenomenon and identify the physical laws connecting them.
2. Select an independent variable and the function of this variable that we want to find.
3. Based on the conditions of the problem, determine the initial or boundary conditions.
4. Express all quantities appearing in the problem statement
through the independent variable, the desired function and its derivatives.
5. Based on the conditions of the problem and physical law to whom he is subordinate this phenomenon, create a differential equation.
6. Find a general solution or general integral differential equation.
7. Using the initial or boundary conditions, find a particular solution.
8. Investigate the resulting solution.
In many cases, the construction of a first-order differential equation is based on the so-called “linearity of the process in the small,” i.e., on the differentiability of functions expressing the dependence of quantities. As a rule, we can assume that all quantities involved in a particular process change with constant speed. This makes it possible to apply the laws known from physics that describe uniformly occurring phenomena to draw up a relationship between the values, i.e., the quantities involved in the process, and their increments. The resulting equality is only approximate, since quantities change even over a short period of time, generally speaking, unevenly. But if we divide both sides of the resulting equality by and go to the limit where it tends to zero, we get an exact equality. It contains time t, changing over time physical quantities and their derivatives, i.e. is a differential equation that describes this phenomenon. The same equation in differential form can be obtained by replacing the increment with the differential and the increment of the functions with the corresponding differentials.
Thus, when drawing up a differential equation, we take a kind of “snapshot” of the process
at a given moment in time, and when solving the equation using these instantaneous snapshots, we restore the course of the process. So, the basis of the solution physical problems using differential equations lies general idea linearization - replacement of functions on small intervals of argument change linear functions. And although there are processes (for example, Brownian motion), for which linearization is impossible because there is no rate of change of some quantities at a given time, in the vast majority of cases the method of differential equations works flawlessly.
Example 1. A small hole of area 5 is made in the bottom of a cylindrical vessel filled with water and having a height H and a base radius R (Fig. 2). In what period of time will all the water flow out through the hole if a third of the water flows out?
Solution. If the flow of water occurred uniformly, then solving the problem would not present any difficulties - all the water would flow out in 3 s. But observations show that at first the water flows out quickly, and as the water level in the vessel decreases, the speed of its flow decreases. Therefore, it is necessary to take into account the relationship between the outflow velocity v and the height h of the liquid column above the hole. Experiments carried out by the Italian physicist Torricelli showed that the speed v is approximately expressed by the formula, where g is the acceleration free fall and k is a “dimensionless” coefficient depending on the viscosity of the liquid and the shape of the hole (for example, for water in the case of a round hole.
Let's take a “snapshot” of the flow process over a period of time. Let at the beginning of this interval the height of the liquid above the hole be equal to , and at the end of it it decreased and became , where is the “increment” of the height (which, obviously, is negative). Then the volume of liquid flowing out of the vessel is equal to the volume of a cylinder with height and base area, i.e.
This liquid poured out in the form of a cylindrical stream with a base area S. Its height is equal to the path traveled by the liquid flowing out of the vessel in a period of time. At the beginning of this period of time, the outflow velocity was equal to Torricelli’s law, and at the end it was equal to .
If it is very small, then it is also very small, and therefore the resulting expressions for the speed are almost the same. Therefore, the path traveled by the liquid over a period of time is expressed by the formula
Where . This means that the volume of liquid spilled over a period of time is calculated by the formula
We have obtained two expressions for the volume of liquid poured out of the vessel over a period of time. Equating these expressions, we obtain the equation
The disadvantage of equation (1) is that we do not know the expression for a. To eliminate this drawback, we divide both sides of equation (1) by and move to the limit at Since , we obtain the differential equation
Physicists usually talk more briefly. They study the process over an “infinitesimal period of time and believe that over the period of time the rate of flow of liquid from the vessel does not change. Therefore, instead of the approximate equation (1), they obtain the exact equation
which is nothing more than differential form equation (2).
To solve the resulting equation, we separate the variables and denote the fraction by A for brevity. Integrating both sides of the resulting equation, we obtain the answer in the form
We have obtained a relationship between t and , which includes two constants A and C. Constant A depends on the size and shape of the hole, the viscosity of the liquid and others
physical parameters, and the constant C arose during the solution of the problem. Their values are unknown to us, but they can be found taking into account the conditions of the problem that have not yet been used.
First, let's find the value of C. For this we use initial conditions. According to the conditions of the problem, at the beginning of the outflow the vessel was filled, i.e., the height of the liquid column was equal to . In other words, when we have: . Substituting the values into formula (3) we get: and therefore equality (3) can be rewritten in the form
To find the value of A, remember that in the first minutes a third of the total liquid flowed out. This corresponds to a decrease in the liquid level by . In other words, when we have: . From here we find that
Now it’s easy to find the time for emptying the vessel: we need to find a value of t at which:
Received value times greater value, which was obtained under the assumption that the liquid flows out evenly.
Of course, this solution is not perfectly accurate - we neglected, for example, the phenomena
capillarity (and they are significant if the hole diameter is small), liquid vortices, the so-called boundary layer(a layer of liquid near the walls of the hole at which the transition of velocity values from zero to u occurs) and many other factors. But it is still more accurate than a solution based on the assumption of uniform fluid flow.
Let us finally examine the resulting solution. To do this, we substitute the value into equality (4), find and obtain that
It is clear that the larger the values of R and H (the dimensions of the vessel), the longer the liquid will flow out of it, as follows from the answer received. Further, the larger S, i.e. the area of the hole, the faster the liquid will flow out of the vessel. An increase in acceleration g, as well as the coefficient k, acts in the same direction (the greater k, the greater the speed of fluid flow according to Bernoulli’s formula).
Thus, the resulting formula passed the “test for common sense" It still needs to be tested for size. Note that in Bernoulli’s formula the coefficient k is dimensionless and therefore we have:
The control carried out confirms that the problem was solved correctly.
In many cases, drawing up a differential equation according to the conditions of the problem is facilitated by the fact that the corresponding law of physics connects the values of a certain quantity and the rate of its change, or connects the values of a quantity, the rate of its change and acceleration with each other.
Example 2. A skydiver falls under the influence of gravity. Let us find the law for changing the parachutist’s height above the earth’s surface if air resistance is proportional to the speed of his fall, and at the beginning of the fall he was at height H, and was at rest.
Solution. According to Newton's second law we have: . If you choose a direction coordinate axis so, as shown in Figure 3, then (gravity is directed in negative direction, and the air resistance force is directed to the side, opposite speed falls). Therefore, the equality takes the form: Since acceleration is a derivative of speed, we obtain a differential equation, i.e.
The initial condition has the form: ( initial speed drop is zero).
Separating the variables in equation (5) and integrating, we obtain:
Since when we have: , then and therefore
From here we find:
We have obtained the law of speed change over time. Let us now find the law of change in altitude A of the parachutist. To do this, we note that , and therefore we obtain the differential equation
It follows from it that
By condition, we have: . Substituting these values into (8), we obtain that and therefore
For small values of t we have:
Retaining only the first two terms, we obtain from formula (7) that This shows that at the beginning of the fall the parachutist moves almost uniformly. However, in further influence air resistance becomes noticeable, and at we have: therefore tends to . In other words, the movement becomes almost uniform with the speed directed downward. This speed is proportional to the force of gravity acting on the parachutist, and inversely proportional
coefficient k, showing the force of air resistance.
From formula (9) you can approximately find the time during which the parachutist will fall earth's surface. To do this, we take into account that we write an approximate equality using formula (9): From it we find that Note that the term is equal to the time it would take for a parachutist to fall at a constant speed, and the addition occurred because at first the fall was slower.
