Let's consider the operational method for solving differential equations using the example of a third-order equation.

Suppose we need to find a particular solution to a third-order linear differential equation with constant coefficients

satisfying the initial conditions:

c 0, c 1, c 2 - given numbers.

Using the property of differentiation of the original, we write:

In equation (6.4.1), let's move from originals to images

The resulting equation is called operator or an equation in images. Resolve it relative to Y.

Algebraic polynomials in a variable r.

The equality is called the operator solution of the differential equation (6.4.1).

Finding the original y(t), corresponding to the found image, we obtain a particular solution to the differential equation.

Example: Using the operational calculus method, find a particular solution to a differential equation that satisfies the given initial conditions

Let's move from originals to images

Let's write the original equation in images and solve it for Y

To find the original of the resulting image, we factorize the denominator of the fraction and write the resulting fraction as a sum of simple fractions.

Let's find the coefficients A, B, And WITH.

Using the table, we record the original of the resulting image

Particular solution of the original equation.

The operational method is similarly applied to solve systems of linear differential equations with constant coefficients

Unknown functions.

Let's move on to the images

We obtain a system of representing equations

We solve the system using Cramer's method. We find the determinants:

Finding a solution to the imaging system X(p), Y(p) , Z(p).

We obtained the required solution of the system

Using operational calculus, you can find solutions to linear differential equations with variable coefficients and partial differential equations; calculate integrals. At the same time, solving problems is greatly simplified. It is used in solving problems of mathematical physics equations.

Questions for self-control.

1. Which function is called the original?

2. What function is called the image of the original?

3. Heaviside function and its image.

4. Obtain an image for the functions of the originals using the image definition: f(t) =t , .

5. Obtain images for functions using the properties of Laplace transforms.

6. Find the functions of the originals using the table of images: ;

7. Find a particular solution to a linear differential equation using operational calculus methods.

Literature: pp. 411-439, pp. 572-594.

Examples: pp. 305-316.

LITERATURE

1. Danko P.E. Higher mathematics in exercises and problems. In 2 parts. Part I: Textbook. manual for colleges/P.E. Danko, A.G. Popov, T.Ya. Kozhevnikova - M.: Higher. school, 1997.– 304 p.

2. Danko P.E. Higher mathematics in exercises and problems. In 2 parts. Part II: Textbook. manual for colleges./ P.E. Danko, A.G. Popov, T.Ya. Kozhevnikova - M.: Higher. school, 1997.– 416 p.

3. Kaplan I.A. Practical classes in higher mathematics. Part 4./ I.A. Kaplan - Kharkov State University Publishing House, 1966, 236 p.

4. Piskunov N.S. Differential and integral calculus. In 2 volumes, volume 1: textbook. manual for colleges./ N.S. Piskunov - M.: ed. “Science”, 1972. – 456 p.

5. Piskunov N.S. Differential and integral calculus for colleges. In 2 volumes, volume 2: textbook. A manual for colleges../ N.S. Piskunov – M.: ed. “Science”, 1972. – 456 p.

6. Written D.T. Lecture notes on higher mathematics: complete course.–4th ed./ D.T. Written – M.: Iris-press, 2006.–608 p. – (Higher education).

7. Slobodskaya V.A. Short course of higher mathematics. Ed. 2nd, reworked and additional Textbook manual for colleges./ V.A. Slobodskaya - M.: Higher. school, 1969.– 544 p.

© Irina Aleksandrovna Dracheva

Lecture notes Higher mathematics

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Operational calculus has now become one of the most important chapters of practical mathematical analysis. The operational method is directly used in solving ordinary differential equations and systems of such equations; it can also be used to solve partial differential equations.

The founders of symbolic (operational) calculus are considered to be Russian scientists M.E. Vashchenko - Zakharchenko and A.V. Letnikov.

Operational calculus attracted attention after the English electrical engineer Heaviside, using symbolic calculus, obtained a number of important results. But distrust of symbolic calculus persisted until Georgi, Bromwich, Carson, A. M. Efros, A. I. Lurie, V. A. Ditkin and others established connections between operational calculus and integral transformations.

The idea of ​​solving a differential equation using the operational method is that from the differential equation with respect to the desired original function f ( t ) move on to an equation for another function F ( p ), called image f ( t ) . The resulting (operational) equation is usually already algebraic (which means simpler than the original one). Solving it relative to the image F ( p ) and then moving on to the corresponding original, they find the desired solution to this differential equation.

The operational method for solving differential equations can be compared to the calculation of various expressions using logarithms, when, for example, when multiplying, calculations are carried out not on the numbers themselves, but on their logarithms, which leads to the replacement of multiplication with a simpler operation - addition.

Just like with logarithm, when using the operational method you need:

1) table of originals and corresponding images;

2) knowledge of the rules for performing operations on an image that correspond to the actions performed on the original.

§1. Originals and images of Laplace functions

Definition 1.Let's be a real function of a real argument f (t) call original, if it meets three requirements:

1) f (t) 0 , at t 0

2) f ( t ) increases no faster than some exponential function

3) On any finite segment  a , bpositive semi-axis Ot function f (t) satisfies the Dirichlet conditions, i.e.

a) limited,

b) is either continuous or has only a finite number of discontinuity points of the first kind,

c) has a finite number of extrema.

Functions that satisfy these three requirements are called in operational calculus represented by Laplace or originals .

The simplest original is the Heaviside unit function

If the function

Laplace integral for the original f (t) is called an improper integral of the form

Theorem.

The Laplace integral converges absolutely in the half-plane

Note that by definition of the original

Let's calculate this integral:

That is, we get that F (p) exists when

Comment . From the proof of the theorem the following estimate follows:

Definition 2 . Image according to Laplace functions f (t) is called a function of a complex variable p = s + iσ, determined by the relation

The fact that the function F (t) is an image of the original f (t), symbolically it is written like this:

§2. Basic theorems of operational calculus

2.1 Rolling originals.

Rolled originals

Functions f (t) And g (t) are called convolution components .

Let us find, for example, a convolution of an arbitrary original

Theorem 1. If

It's a sultry time outside, poplar fluff is flying, and this weather is conducive to relaxation. During the school year, everyone has accumulated fatigue, but the anticipation of summer vacations/holidays should inspire them to successfully pass exams and tests. By the way, the teachers are also dull during the season, so soon I will also take a time out to unload my brain. And now there’s coffee, the rhythmic hum of the system unit, a few dead mosquitoes on the windowsill and a completely working condition... ...oh, damn it... the fucking poet.

To the point. Who cares, but today is June 1st for me, and we will look at another typical problem of complex analysis - finding a particular solution to a system of differential equations using the operational calculus method. What do you need to know and be able to do to learn how to solve it? First of all, highly recommend refer to the lesson. Please read the introductory part, understand the general statement of the topic, terminology, notation and at least two or three examples. The fact is that with diffuser systems everything will be almost the same and even simpler!

Of course, you must understand what it is system of differential equations, which means finding a general solution to the system and a particular solution to the system.

Let me remind you that the system of differential equations can be solved in the “traditional” way: by elimination or using the characteristic equation. The method of operational calculus that will be discussed is applicable to the remote control system when the task is formulated as follows:

Find a particular solution to a homogeneous system of differential equations , corresponding to the initial conditions .

Alternatively, the system can be heterogeneous - with “add-on weights” in the form of functions and on the right sides:

But, in both cases, you need to pay attention to two fundamental points of the condition:

1) It's about only about a private solution.
2) In parentheses of initial conditions are strictly zeros, and nothing else.

The general course and algorithm will be very similar to solving a differential equation using the operational method. From the reference materials you will need the same table of originals and images.

Example 1

, ,

Solution: The beginning is trivial: using Laplace transform tables Let's move on from the originals to the corresponding images. In a problem with remote control systems, this transition is usually simple:

Using tabular formulas No. 1, 2, taking into account the initial condition, we obtain:

What to do with the “games”? Mentally change the “X’s” in the table to “I’s”. Using the same transformations No. 1, 2, taking into account the initial condition, we find:

Let's substitute the found images into the original equation :

Now in the left parts equations need to be collected All terms in which or is present. To the right parts equations need to be “formalized” everyone else terms:

Next, on the left side of each equation we carry out bracketing:

In this case, the following should be placed in the first positions, and in the second positions:

The resulting system of equations with two unknowns is usually solved according to Cramer's formulas. Let us calculate the main determinant of the system:

As a result of calculating the determinant, a polynomial was obtained.

Important technique! This polynomial is better immediately try to factor it. For these purposes, one should try to solve the quadratic equation , but many readers with a second-year trained eye will notice that .

Thus, our main determinant of the system is:

Further disassembly of the system, thank Kramer, is standard:

As a result we get operator solution of the system:

The advantage of the task in question is that the fractions usually turn out to be simple, and dealing with them is much easier than with fractions in problems finding a particular solution to a DE using the operational method. Your premonition did not deceive you - the good old method of uncertain coefficients, with the help of which we decompose each fraction into elementary fractions:

1) Let's deal with the first fraction:

Thus:

2) We break down the second fraction according to a similar scheme, but it is more correct to use other constants (undefined coefficients):

Thus:

I advise dummies to write down the decomposed operator solution in the following form:
- this will make the final stage clearer - the inverse Laplace transform.

Using the right column of the table, let's move from the images to the corresponding originals:

According to the rules of good mathematical manners, we will tidy up the result a little:

Answer:

The answer is checked according to a standard scheme, which is discussed in detail in the lesson. How to solve a system of differential equations? Always try to complete it in order to add a big plus to the task.

Example 2

Using operational calculus, find a particular solution to a system of differential equations corresponding to the given initial conditions.
, ,

This is an example for you to solve on your own. An approximate sample of the final form of the problem and the answer at the end of the lesson.

Solving a non-homogeneous system of differential equations is algorithmically no different, except that technically it will be a little more complicated:

Example 3

Using operational calculus, find a particular solution to a system of differential equations corresponding to the given initial conditions.
, ,

Solution: Using the Laplace transform table, taking into account the initial conditions , let's move from the originals to the corresponding images:

But that's not all, there are lonely constants on the right-hand sides of the equations. What to do in cases where the constant is completely alone on its own? This was already discussed in class. How to solve a DE using the operational method. Let us repeat: single constants should be mentally multiplied by one, and the following Laplace transform should be applied to the units:

Let's substitute the found images into the original system:

Let us move the terms containing , to the left, and place the remaining terms on the right sides:

In the left-hand sides we will carry out bracketing, in addition, we will bring the right-hand side of the second equation to a common denominator:

Let's calculate the main determinant of the system, not forgetting that it is advisable to immediately try to factorize the result:
, which means the system has a unique solution.

Let's move on:



Thus, the operator solution of the system is:

Sometimes one or even both fractions can be reduced, and, sometimes, so successfully that there is practically no need to expand anything! And in some cases, you get a freebie right away, by the way, the following example of the lesson will be an indicative example.

Using the method of indefinite coefficients we obtain the sums of elementary fractions.

Let's break down the first fraction:

And we achieve the second one:

As a result, the operator solution takes the form we need:

Using the right column tables of originals and images We carry out the inverse Laplace transform:

Let us substitute the resulting images into the operator solution of the system:

Answer: private solution:

As you can see, in a heterogeneous system it is necessary to carry out more labor-intensive calculations compared to a homogeneous system. Let's look at a couple more examples with sines and cosines, and that's enough, since almost all types of the problem and most of the nuances of the solution will be considered.

Example 4

Using the operational calculus method, find a particular solution to a system of differential equations with given initial conditions,

Solution: I will also analyze this example myself, but the comments will concern only special moments. I assume you are already well versed in the solution algorithm.

Let's move on from the originals to the corresponding images:

Let's substitute the found images into the original remote control system:

Let's solve the system using Cramer's formulas:
, which means the system has a unique solution.

The resulting polynomial cannot be factorized. What to do in such cases? Absolutely nothing. This one will do too.

As a result, the operator solution of the system is:

Here's the lucky ticket! There is no need to use the method of indefinite coefficients at all! The only thing is, in order to apply table transformations, we rewrite the solution in the following form:

Let's move on from the images to the corresponding originals:

Let us substitute the resulting images into the operator solution of the system: