Operational calculus has now become one of the most important chapters of practical mathematical analysis. The operational method is directly used in solving ordinary problems 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 the given 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

, at t0 , where M 0, s 00 - some real constants, s 0 called growth indicator of the function f(t) .

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

satisfies condition 2 and does not satisfy condition 1, then the product will also satisfy condition 1, i.e. will be original. To simplify the notation, we will, as a rule, use the multiplier H (t) omit, assuming that all the functions under consideration are equal to zero at negative values t .

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

, is a complex parameter.

Theorem.

The Laplace integral converges absolutely in the half-plane

(that is, the image F (p) is obviously defined at ), where s 0 – growth rate f (t). we get: , but according to the property of modules .

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

(1)

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

or (2)

§2. Basic theorems of operational calculus

2.1 Rolling originals.

Rolled originals

and the function is called .

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

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

And unit function We have . while . (2.1.1)

Theorem 1. If

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

Let it be necessary to find a particular solution to a linear third-order 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 - Kharkovsky Publishing House state university, 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 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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Heaviside expansion formula

Let the image of the function be a fractional rational function.

Theorem. Let, where and are differentiable functions. Let us introduce both the poles of the function, i.e. roots (zeros) of its denominator. Then, if we get the Heaviside formula:

We carry out the proof for the case when and are polynomials of degrees T And n accordingly, while T n. Then it is a proper rational fraction. Let's present it as a sum of simple fractions:

From here we find the coefficients from identity (17.2), rewriting it in the form

Let's multiply both sides of the last equality by and go to the limit at. Considering that and, we get

whence it follows (17.1). The theorem has been proven.

Note 1. If the coefficients of polynomials are real, then the complex roots of the polynomial are pairwise conjugate. Consequently, in formula (17.1) the complex conjugate quantities will be the terms corresponding to the complex conjugate roots of the polynomial, and the Heaviside formula will take the form

where the first sum is extended to all real roots of the polynomial, the second - to all its complex roots with positive imaginary parts.

Note 2. Each term of formula (17.1) is written in complex form hesitation where. Thus, the real roots () correspond to aperiodic oscillations, complex roots with negative real parts - damped oscillations, with purely imaginary roots - undamped harmonic oscillations.

If the denominator has no roots with positive real parts, then for sufficiently large values we get the steady state:

Purely imaginary roots of a polynomial with positive imaginary parts.

Oscillations corresponding to roots with negative real parts decay exponentially at and therefore do not enter the steady state.

Example 1. Find the original image

Solution. We have. Let's write down the roots of the polynomial: .

According to formula (17.1)

Here, since the numbers are the roots of the equation. Hence,

Example 2. Find the original image

Where A 0; .

Solution. Here the function, in addition to the obvious root, has infinitely many roots, which are zeros of the function. Solving the equation, we get where

Thus, the roots of the denominator have the form and, where

Using formula (17.3) we find the original

Operator method for solving differential equations

Differential equations. Consider the Cauchy problem for a linear differential equation

(here) with initial conditions

Passing to images in (18.1), due to the linearity of the Laplace transform, we will have

Using Theorem 3 of § 16 and initial conditions (18.2), we write the images of derivatives in the form

Substituting (18.4) into (18.3), after simple transformations we obtain the operator equation

where (characteristic polynomial); .

From equation (18.5) we find the operator solution

The solution to the Cauchy problem (18.1), (18.2) is the original operator solution (18.6):

For the Cauchy problem, in the accepted notation we can write

The operator equation has the form

Let us decompose the operator solution into simple fractions:

Using the formulas obtained in § 15, we obtain the originals:

Thus, the solution to the Cauchy problem will have the form

Example 1. Solve the Cauchy problem for a differential equation with initial conditions, where.

Solution.

Its solution has the form

Using Theorem 2 of § 16, we consistently find:

Example 2. Solve the Cauchy problem for a differential equation with zero initial conditions, where is the step impulse function.

Solution. Let us write the operator equation

and his decision

From Theorem 2 of § 16 it follows

in accordance with the delay theorem (§ 15)

Finally,

Example 3. Per point mass T, attached to the spring by a stiffness With and located on a smooth horizontal plane, a periodically changing force acts. At the moment of time, the point was subjected to an impact carrying an impulse. Neglecting resistance, find the law of motion of a point if at the initial moment of time it was at rest at the origin.

Solution. We write the equation of motion in the form

where is elastic force; - Dirac function. Let's solve the operator equation

If (case of resonance), then

By the delay theorem

Finally,

Duhamel's integral (formula). Let us consider the Cauchy problem for equation (18.1) under initial conditions. The operator solution in this case has the form

Let the weight function be the original for. then by Theorem 1 of § 16 we obtain

Relation (18.7) is called Duhamel’s integral (formula).

Comment. For nonzero initial conditions, Duhamel's formula is not directly applicable. In this case, it is necessary to first transform the original problem to a problem with homogeneous (zero) initial conditions. To do this, we introduce a new function, assuming

where are the initial values ​​of the desired solution.

How easy it is to see, and therefore .

Thus, the function is a solution to equation (18.1) with the right-hand side obtained by substituting (18.8) into (18.1), with zero initial data.

Using (18.7), we find and.

Example 4. Using the Duhamel integral, find a solution to the Cauchy problem

with initial conditions.

Solution. The initial data is non-zero. We assume, in accordance with (18.8), . Then, for the definition, we obtain an equation with homogeneous initial conditions.

For the problem under consideration, a characteristic polynomial, a weight function. According to Duhamel's formula

Finally,

Systems of linear differential equations with constant coefficients. The Cauchy problem for a system of linear differential equations in matrix notation has the form

where is the vector of the required functions; - vector of right sides; - coefficient matrix; - vector of initial data.

It's a sultry time outside, poplar fluff is flying, and this weather is conducive to relaxation. For academic year Everyone is tired, but the anticipation of summer vacations should inspire successful completion 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’ll look at another one typical task 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 what does it mean to find general solution systems and a particular solution of 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.

General move and the 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 technical technique! This polynomial is better immediately try to factor it. For these purposes, one should try to solve 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 independent decision. 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 parts we will carry out bracketing, in addition, we will reduce to common denominator the right side of the second equation:

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.

Method uncertain 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 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:

How to Solve a Differential Equation
operational calculus method?

On this lesson a typical and widespread task of complex analysis will be analyzed in detail - finding a particular solution to a 2nd order DE with constant coefficients using the operational calculus method. Time and time again I rid you of the preconception that the material is unimaginably complex and inaccessible. It’s funny, but to master the examples, you may not be able to differentiate, integrate, and even not know what it is complex numbers. Application skill required method of uncertain coefficients, which is discussed in detail in the article Integration of Fractional-Rational Functions. In fact, the cornerstone of the assignment is simple algebraic operations, and I am confident that the material is accessible even to a high school student.

First, concise theoretical information about the section of mathematical analysis under consideration. The main point operational calculus is as follows: function valid variable using the so-called Laplace transform displayed in function comprehensive variable :

Terminology and designations:
the function is called original;
the function is called image;
capital letter denotes Laplace transform.

Speaking in simple language, a real function (original) according to certain rules must be converted into a complex function (image). The arrow indicates precisely this transformation. And the “certain rules” themselves are Laplace transform, which we will consider only formally, which will be quite sufficient for solving problems.

The inverse Laplace transform is also feasible, when the image is transformed into the original:

Why is all this needed? In a number of higher mathematics problems, it can be very beneficial to switch from originals to images, since in this case the solution to the problem is significantly simplified (just kidding). And we will consider just one of these problems. If you have lived to see operational calculus, then the formulation should be very familiar to you:

Find a particular solution to a second-order inhomogeneous equation with constant coefficients for given initial conditions.

Note: sometimes the differential equation can be homogeneous: , for it in the above formulation the method of operational calculus is also applicable. However, in practical examples homogeneous DE of 2nd order is extremely rare, and further we will talk about inhomogeneous equations.

And now the third method will be discussed - solving differential equations using operational calculus. Once again I emphasize the fact that we are talking about finding a particular solution, Besides, the initial conditions strictly have the form(“X’s” equal zeros).

By the way, about the “X’s”. The equation can be rewritten as follows:
, where “x” is an independent variable, and “y” is a function. It is no coincidence that I am talking about this, since in the problem under consideration other letters are most often used:

That is, the role of the independent variable is played by the variable “te” (instead of “x”), and the role of the function is played by the variable “x” (instead of “y”)

I understand that it’s inconvenient, of course, but it’s better to stick to the notations that are found in most problem books and training manuals.

So, our problem with other letters is written as follows:

Find a particular solution to a second-order inhomogeneous equation with constant coefficients for given initial conditions .

The meaning of the task has not changed at all, only the letters have changed.

How to solve this task operational calculus method?

First of all, you will need table of originals and images. This is a key solution tool, and you can’t do without it. Therefore, if possible, try to print the indicated reference material. Let me immediately explain what the letter “pe” means: a complex variable (instead of the usual “z”). Although this fact is not particularly important for solving problems, “pe” is “pe”.

Using the table, the originals need to be turned into some images. What follows is a series of typical actions, and the inverse Laplace transform is used (also in the table). Thus, the desired particular solution will be found.

All problems, which is nice, are solved according to a fairly strict algorithm.

Example 1

, ,

Solution: In the first step, we will move from the originals to the corresponding images. We use the left side.

First, let's look at the left side of the original equation. For the Laplace transform we have linearity rules, so we ignore all constants and work separately with the function and its derivatives.

Using tabular formula No. 1, we transform the function:

According to formula No. 2 , taking into account the initial condition, we transform the derivative:

Using formula No. 3, taking into account the initial conditions, we transform the second derivative:

Don't get confused by the signs!

I admit that it is more correct to say “transformations” rather than “formulas,” but for simplicity, from time to time I will call the contents of the table formulas.

Now let's look at the right side, which contains the polynomial. Due to the same linearity rules Laplace transform, we work with each term separately.

Let's look at the first term: - this is the independent variable “te” multiplied by a constant. We ignore the constant and, using point No. 4 of the table, perform the transformation:

Let's look at the second term: –5. When a constant is found alone, it can no longer be skipped. With a single constant, they do this: for clarity, it can be represented as a product: , and the transformation can be applied to unity:

Thus, for all elements (originals) of the differential equation, the corresponding images were found using the table:

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

The next task is to express operator solution through everything else, namely through one fraction. In this case, it is advisable to adhere to the following procedure:

First, open the brackets on the left side:

We present similar terms on the left side (if they exist). In this case, we add the numbers –2 and –3. I strongly recommend that teapots do not skip this step:

On the left we leave the terms that contain , and move the remaining terms to the right with a change of sign:

On the left side we put the operator solution out of brackets, on the right side we reduce the expression to a common denominator:

The polynomial on the left should be factorized (if possible). Solving the quadratic equation:

Thus:

We reset to the denominator of the right side:

The goal has been achieved - the operator solution is expressed in terms of one fraction.

Act two. Using method of uncertain coefficients, the operator solution of the equation should be expanded into a sum of elementary fractions:

Let us equate the coefficients at the corresponding powers and solve the system:

If you have any problems with please catch up on the articles Integrating a Fractional-Rational Function And How to solve a system of equations? This is very important because fractions are essentially the most important part of the problem.

So, the coefficients are found: , and the operator solution appears before us in disassembled form:

Please note that constants are not written in fraction numerators. This form of recording is more profitable than . And it’s more profitable, because the final action will take place without confusion and errors:

The final stage of the problem is to use the inverse Laplace transform to move from the images to the corresponding originals. Using the right column tables of originals and images.

Perhaps not everyone understands the conversion. The formula of point No. 5 of the table is used here: . In more detail: . Actually, for similar cases the formula can be modified: . And all the tabular formulas of point No. 5 are very easy to rewrite in a similar way.

After the reverse transition, the desired partial solution of the DE is obtained on a silver platter:

Was:

Became:

Answer: private solution:

If you have time, it is always advisable to perform a check. The check is carried out according to the standard scheme, which has already been discussed in class. Inhomogeneous differential equations of the 2nd order. Let's repeat:

Let's check the fulfillment of the initial condition:
– done.

Let's find the first derivative:

Let's check the fulfillment of the second initial condition:
– done.

Let's find the second derivative:

Let's substitute , and to the left side of the original equation:

The right side of the original equation is obtained.

Conclusion: the task was completed correctly.

A small example for your own solution:

Example 2

Using operational calculus, find a particular solution to a differential equation under given initial conditions.

An approximate sample of the final assignment at the end of the lesson.

The most common guest in differential equations, as many have long noticed, is exponentials, so let’s consider a few examples with them, their relatives:

Example 3

, ,

Solution: Using the Laplace transformation table (left side of the table), we move from the originals to the corresponding images.

Let's look at the left side of the equation first. There is no first derivative there. So what? Great. Less work. Taking into account the initial conditions, using tabular formulas No. 1, 3 we find the images:

Now look at the right side: – the product of two functions. In order to take advantage linearity properties Laplace transform, you need to open the brackets: . Since the constants are in the products, we forget about them, and using group No. 5 of tabular formulas, we find the images:

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

Let me remind you that the next task is to express the operator solution in terms of a single fraction.

On the left side we leave the terms that contain , and move the remaining terms to the right side. At the same time, on the right side we begin to slowly reduce the fractions to a common denominator:

On the left we take it out of brackets, on the right we bring the expression to a common denominator:

On the left side we obtain a polynomial that cannot be factorized. If the polynomial cannot be factorized, then the poor guy must immediately be thrown to the bottom of the right side, his legs concreted in the basin. And in the numerator we open the brackets and present similar terms:

The most painstaking stage has arrived: method of undetermined coefficients Let us expand the operator solution of the equation into a sum of elementary fractions:


Thus:

Notice how the fraction is decomposed: , I’ll soon explain why this is so.

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

In the two lower transformations, formulas Nos. 6 and 7 of the table were used, and the fraction was pre-expanded just to “fit” it to the table transformations.

As a result, a particular solution:

Answer: the required particular solution:

A similar example for a do-it-yourself solution:

Example 4

Find a particular solution to a differential equation using the operational calculus method.

Quick Solution and the answer at the end of the lesson.

In Example 4, one of the initial conditions is zero. This certainly simplifies the solution, and the most ideal option, when both initial conditions are zero: . In this case, the derivatives are converted to images without tails:

As already noted, the most difficult technical aspect of the problem is the expansion of the fraction method of undetermined coefficients, and I have quite labor-intensive examples at my disposal. However, I won’t intimidate anyone with monsters; let’s consider a couple more typical variations of the equation:

Example 5

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

Solution: Using the Laplace transform table, we move from the originals to the corresponding images. Considering the initial conditions :

There are no problems with the right side either:

(Remember that multiplier constants are ignored)

Let's substitute the resulting images into the original equation and perform standard actions, which, I hope, you have already worked well:

We take the constant in the denominator outside the fraction, the main thing is not to forget about it later:

I was thinking about whether to remove an additional two from the numerator, however, after taking stock, I came to the conclusion that this step would practically not simplify the further decision.

The peculiarity of the task is the resulting fraction. It seems that its decomposition will be long and difficult, but appearances are deceptive. Naturally, there are difficult things, but in any case - forward, without fear and doubt:

The fact that some odds turned out to be fractional should not be confusing; this situation is not uncommon. If only the computing technology did not fail. In addition, there is always the opportunity to check the response.

As a result, the operator solution is:

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

Thus, a particular solution: