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1 Solving differential equations using the Laplace transform (operational method) Operational calculus is one of the most economical methods for integrating linear differential equations with constant coefficients and is very popular among engineers. The method was proposed by the famous American electrical engineer and physicist O. Heaviside (892). He proposed formal rules for handling the operator d dx and some functions from this operator, using which he solved a number of important problems in electrodynamics. However, operational calculus did not receive a mathematical justification in the works of O. Heaviside (“his mathematics arose in a physical context from which it was not easy to isolate” [, p. 8]), many of his results remained unproven. Only in the 2nd years of the 20th century the method received justification in the works of Bromwich (T. J. I A. Bromwich) and Carson (J. R. Carson) 2.. The concept of the original and the image according to Laplace Definition. An original function is any complex-valued function f(x) of a real argument x that satisfies the following conditions:) f(x) is continuous for x, except, perhaps, for a finite number of points of discontinuity of the -th kind; 2) for all x< f(x) = ; 3) существуют такие постоянные M >and a >, for which f(x) M e ax for x. () Differential and integral equations: a textbook for students of the Faculty of Physics and Technology: in 3 hours. Part 2 / comp. : N. Yu. Svetova, E. E. Semyonova. Petrozavodsk: PetrSU Publishing House, Attempts at a rigorous justification and “mathematically acceptable” presentation of calculus resembled a “general assault”: the English mathematician Bromwich (96), the American engineer Carson (925), the Dutch electrical engineer Van der Pol () attracted the results of various theories, connected Heaviside's calculus with the Laplace transform, with the theory of functions of a complex variable.

2 2 The infimum a of all numbers a for which inequality () holds is called the growth exponent of the function f(x). Note that for any bounded function the growth index a =. The simplest original is the Heaviside function (, x ; χ(x) =, x<. Очевидно, для любой функции ϕ(x) { ϕ(x), x, ϕ(x) χ(x) =, x <. Если при x функция ϕ(x) удовлетворяет условиям и 3 определения, то функция ϕ(x)χ(x) является оригиналом. В дальнейшем для сокращения записи будем, как правило, записывать ϕ(x) вместо ϕ(x)χ(x), считая, что рассматриваемые нами функции продолжены нулем для отрицательных значений аргумента x. Определение 2. Функция F (p) комплексного переменного p (p C), определяемая интегралом F (p) = e px f(x) dx, () называется преобразованием Лапласа, или изображением по Лапласу 3, функции f(x). Для указания соответствия между оригиналом и изображением будем использовать следующую запись 4: f(x) F (p). 3 В мемуарах П. Лапласа (782 82) современные оригинал и изображение именуются fonction determinant и fonction generatrice «определяющая функция» и «производящая». Эти названия, хотя и признанные неудачными, сохранились до XX в. Хевисайд употреблял названия «подоператорная функция» (892). Оператор он обозначал буквой p, которая употребляется в современном исчислении . 4 Названия original и image и знак предложил Ван дер Поль в статьях гг. В русской литературе термин изображение и символ, по-видимому, впервые появились в книге харьковских математиков А. М. Эфроса и А. М. Данилевского «Операционное исчисление и контурные интегралы» (937), а термин оригинал только в 953 г. . Используются и другие варианты записи соответствия между оригиналами и изображениями. Например, f(x) F (p) или L{f(x)} = F (p).

3 For any original f(x), its image F (p) is defined in the half-plane Re p > a (a is the growth index of the function f(x)), where the improper integral () converges. Example. Using the definition, find the image of the function f(x) = sin 3x. Solution. For the function f(x) = sin 3x we have a =. Therefore, the image F (p) will be defined in the half-plane Re p >. Let us apply formula () to the given function, using the rule of integration by parts and a restriction on the set of values of the variable p, ensuring the convergence of the integral: F (p) = + e px sin 3x dx = = p e px sin 3x x= = 3 p p e px cos 3x = 3 p 2 9 p 2 We get the equality: Where do we find + x=+ + 3 p x=+ x= + 3 p e px cos 3x dx = + e px sin 3x dx = 3 p 2 9 p 2 F (p ). F (p) = 3 p 2 9 p 2 F (p). F (p) = 3 p Thus, the following correspondence is true: sin 3x 3 p 2, Re p >. + 9 e px sin 3x dx = 3

4 4 2. Properties of the Laplace transform In practice, when constructing images, various techniques are used based on the properties of the Laplace transform. Let us list the main properties, the validity of which can be easily established using the definitions of the image and the original. The property of linearity. If f(x) F (p), g(x) G(p), then for any α, β C αf(x) + βg(x) αf (p) + βg(p), Re p > max( a, b). Here and below, a and b are growth indicators for the functions f(x) and g(x), respectively. 2. Similarity theorem. If f(x) F (p), then for any α > f(αx) α F (p α), Re p > αa. 3. Displacement theorem. If f(x) F (p), then for any λ C e λx f(x) F (p λ), Re p > a + Re λ. 4. Differentiation of the original. Let the function f(x) be differentiable n times. Then f (x) pf (p) f(+), f (x) p 2 F (p) pf(+) f (+), f (n) (x) p n F (p) p n f(+). .. pf (n 2) (+) f (n) (+), where f (k) (+) = lim x + f (k) (x), k =, n. Comment. When constructing images of derivatives of functions continuous at zero, the plus sign is omitted in writing the argument of a function and its derivatives. 5. Image differentiation. If f(x) F (p), then In particular, for n = we have F (n) (p) (x) n f(x), Re p >. F (p) xf(x).

5 5 6. Integrating the original. If f(x) F (p), then x f(ξ) dξ F (p) p, Re p > α. 7. Image integration. If the integral and F (p) f(x), then p F (p) dp f(x) x, Re p > α. p F (p) dp converges 8. Image multiplication theorem (convolution theorem) If f(x) F (p), g(x) G(p), then F (p)g(p) x f(t) g(x t) dt = x f(x t)g(t) dt, when Re p > max(a, b). The integrals on the right side of the correspondence are called the convolution of the functions f(x) and g(x). 9. Delay theorem. If f(x) F (p), then for any ξ > f(x ξ)χ(x ξ) e ξp F (p), Re p > α. The original is restored from the image in a unique way, accurate to the values at the break points. In practice, ready-made tables of originals and images 5 are usually used. The table lists the main originals and images often found in applications. Example 2. Using the properties of the Laplace transform and the table of basic originals and images, find images of the following functions:) f(x) = e 4x sin 3x cos 2x; 3) f(x) = x 2 e 3x ; 2) f(x) = e (x 2) sin (x 2); 4) f(x) = sin2 x x. 5 Ditkin V. A., Prudnikov A. P. Handbook of operational calculus. M., 965.

6 6 Table. Basic originals and images Original Image Original Image p cos ωx p p 2 + ω 2 x n n! p n+ e λx p + λ sin ωx x cos ωx x n e λx n! (p + λ) n+ x sin ωx ω p 2 + ω 2 p 2 ω 2 (p 2 + ω 2) 2 2pω (p 2 + ω 2) 2 Solution.) Transform the expression for the function f(x) as follows: f(x) = e 4x sin 3x cos 2x = 2 e 4x (sin 5x + sin x) = = 2 e 4x sin 5x + 2 e 4x sin x. Since sin x 5 p 2 and sin 5x + p, then, using the linearity property and the displacement theorem, to depict the function f(x) we will have: F (p) = () 5 2 (p + 4) (p + 4 )) Since sin x p 2 +, ex sin x (p) 2 +, then, using the delay theorem, we will have f(x) = e x 2 sin (x 2) F (p) = e 2p (p)) So as x 2 2 p 3, then by the displacement theorem we have: f(x) = x 2 e 3x F (p) = 2 (p 3) 3.

7 For comparison, we present a method for constructing an image of the function f(x) = x 2 e 3x using the property of image differentiation: We obtained the same result. 4) Since e 3x p 3 ; xe 3x d () = dp p 3 (p 3) 2 ; x 2 e 3x d () 2 dp (p 3) 2 = (p 3) 3. sin 2 x = 2 2 cos 2x 2p 2 p p 2 + 4, then, using the property of image integration, we will have: sin 2 x ( x 2p) 2 p p 2 dp = + 4 p (= 4 ln p2) 4 ln(p2 + 4) = p 4 ln p 2 p p = 4 ln p2 + 4 p Restoring the original from the image Let the image Y (p) be proper rational fraction (is a rational function). If a fraction is decomposed into a sum of simplest (elementary) fractions, then for each of them the corresponding original can be found using the properties of the Laplace transform and a table of originals and their images. Indeed, A p a A eax ; A (p a) n A (n)! xn e ax.

8 8 After converting the fraction Ap + B A(p a) + aa + B A(p a) (p a) 2 = + b2 (p a) 2 + b 2 = (p a) 2 + b 2 + aa + B (p a) 2 + b 2, we get Ap + B (p a) 2 + b 2 A eax cos bx + aa + B e ax sin bx. b To construct the original corresponding to the fraction Ap + B ((p a) 2 + b 2) n, you can use the multiplication theorem. For example, for n = 2 we have Ap + B ((p a) 2 + b 2) 2 = Ap + B (p a) 2 + b 2 (p a) 2 + b 2. Since and then For n = 3: Ap + B (p a) 2 + b 2 A eax cos bx + aa + B e ax sin bx = h (x) b (p a) 2 + b 2 b eax sin bx = g(x), Ap + B ((p a) 2 + b 2) 2 = x Ap + B ((p a) 2 + b 2) 2 (p a) 2 + b 2 g(x t) h (t) dt = h 2 (t). x g(x t) h 2 (t) dt, Similarly, we can consider the restoration of the originals for n > 3. The denominator of the rational function Y (p) is a polynomial of order k. If it has k distinct zeros p i, i =, k, then, expanding

9 denominator by factors (p p i), the corresponding original for Y (p) can be found by the formula: y(x) = k (Y (p)(p p i)e px) p=pi. (2) i= The product Y (p)(p p i) gives a rational function, the denominator of which does not contain a factor (p p i), and calculated at p = p i determines the coefficient with which the fraction is included in the p p i expansion of the function Y (p) into the sum of elementary fractions. Example 3. Find the original corresponding to the image: Y (p) = p 3 p. Solution. Having expanded the given image into a sum of elementary fractions: p 3 p = p(p)(p +) = p + 2(p) + 2(p +), we find the original Answer: y(x) = + ch x. y(x) = + 2 ex + 2 e x = + ch x. Example 4. Find the original for the image: Y (p) = p(p 2 +). Solution. Since p 2 sin x, then, applying the integration property of the original, + we obtain: p(p 2 +) x Answer: y(x) = cos x. sin t dt = cos t x = cos x. Example 5. Find the original corresponding to the image: Y (p) = (p 2 + 4) 2. 9

10 Solution. Applying the convolution image property, we have: Y (p) = (p 2 + 4) 2 = p p x sin 2(x t) sin 2t dt. Having calculated the integral, we obtain the desired expression for the original. Answer: y(x) = 6 sin 2x x cos 2x. 8 Example 6. Find the original corresponding to the image: Y (p) = p p 3 p 2 6p. Solution. Since p 3 p 2 6p = p(p 3)(p + 2), then the denominator of the fraction Y (p) has three simple roots: p =, p 2 = 3 and p 3 = 2. Let’s construct the corresponding original using the formula (2): y(x) = (p2 + 2)e px (p 3)(p + 2) + (p2 + 2)e px p= p(p + 2) + (p2 + 2)e px p =3 p(p 3) = p= 2 = e3x e 2x. Example 7. Find the original corresponding to the image: Y (p) = e p 2 p(p +)(p 2 + 4). Solution. Let's imagine the fraction included in the expression in the form of simple fractions: p(p +)(p 2 + 4) = A p + B p + + Cp + D p Applying the method of indefinite coefficients to the expansion, we obtain: The image will look like: A = 4 ; B = D = 5 ; C = 2. Y (p) = e p 2 4 p 5 e p 2 p + pe p 2 2 p e p 2 5 p (a)

11 Using the relations: p χ(x), p + e x χ(x), p p cos 2x χ(x), p sin 2x χ(x) 2 and taking into account the retardation theorem, we obtain the desired original for image (a). Answer: y(x) = (4 5 e (x 2) cos (2x) sin (2x) 2) χ (x) Solution of the Cauchy problem for a differential equation with constant coefficients The method of solving various classes of equations using the Laplace transform is called operational method. The property of the Laplace transform, differentiation of the original, allows us to reduce the solution of linear differential equations with constant coefficients to the solution of algebraic equations. Consider the Cauchy problem for an inhomogeneous equation with initial conditions y (n) + a y (n) a n y + a n y = f(x) (3) y() = y, y () = y,..., y (n) ( ) = y n. (4) Let the function f(x) and the required solution satisfy the conditions for the existence of the Laplace transform. Let us denote by Y (p) the image of the unknown function (original) y(x), and by F (p) the image of the right side of f(x): y(x) Y (p), f(x) F (p). By the rule of differentiation of the original we have y (x) py (p) y, y (x) p 2 Y (p) py y, y (n) (x) p n Y (p) p n y p n 2 y... y n.

12 2 Then, due to the linearity property of the Laplace transform, after applying it to the left and right sides of equation (3), we obtain the operator equation M(p)Y (p) N(p) = F (p), (5) where M(p) characteristic polynomial of equation (3): M(p) = p n + a p n a n p + a n y, N(p) polynomial containing the initial data of the Cauchy problem (vanishes when the initial data is zero): N(p) = y (p n + a p n a n) + + y (p n 2 + a p n a n 2) y n 2 (p + a) + y n, F (p) image of the function f(x). Solving the operator equation (5), we obtain the Laplace image Y (p) of the desired solution y(x) in the form Y (p) = F (p) + N(p). M(p) Restoring the original for Y (p), we find a solution to equation (3) that satisfies the initial conditions (4). Example 8. Find a solution to the differential equation: y (x) + y(x) = e x, satisfying the condition: y() =. Solution. Let y(x) Y (p). Since y (x) py (p) y() = py (p), e x p +, then by applying the Laplace transform to the given equation, using the linearity property, we obtain an algebraic equation for Y (p): py (p) + Y (p) = p +. Where do we find the expression for Y (p):

13 Since then we have Y (p) = p + e x, (p +) 2 + p +. (p +) 2 xe x, Y (p) y(x) = e x x + e x. Verification: Let us show that the found function is indeed a solution to the Cauchy problem. We substitute the expression for the function y(x) and its derivative into the given equation: y (x) = e x x + e x e x = e x x e x x + e x x + e x = e x. After bringing similar terms on the left side of the equation, we obtain the correct identity: e x e x. Thus, the constructed function is a solution to the equation. Let's check whether it satisfies the initial condition y() = : y() = e + e =. Consequently, the found function is a solution to the Cauchy problem. Answer: y(x) = e x x + e x. Example 9. Solve the Cauchy problem y + y =, y() =, y() =. Solution. Let y(x) Y (p). Since 3 y (x) p 2 Y (p) py() y (), /p, then, applying the Laplace transform to the equation, taking into account the initial conditions we obtain (p 2 +)Y (p) = p = Y ( p) = p(p 2 +). Let's decompose the fraction into simpler fractions: Y (p) = p From the table we find y(x) = cos x. p p 2 +.

14 4 You can also restore the original from an image by applying the property of integrating the original (see example 4). Answer: y(x) = cos x. Example. Solve the Cauchy problem y +3y = e 3x, y() =, y() =. Solution. Let y(x) Y (p). Since y py (p) y(), y (x) p 2 Y (p) py() y (), and e 3x p + 3, then, taking into account the initial conditions, we obtain the operator equation (p 2 + 3p) Y (p) + = p + 2 = Y (p) = p + 3 (p + 3) 2 p. Let's decompose the rational function into simple fractions: p + 2 (p + 3) 2 p = A p + B p C (p + 3) 2 = A(p2 + 6p + 9) + B(p 2 + 3p) + Cp p (p + 3) 2. Let’s create a system of equations to find the coefficients A, B and C: A + B =, 6A + 3B + C =, 9A = 2, solving which we find A = 2/9, B = 2/9, C = /3. Therefore, Y (p) = 2 9 p p (p + 3) 2. Using the table we get the answer. Answer: y(x) = e 3x 3 xe 3x. Example. Find a solution to the differential equation: y (x) + 2y (x) + 5y (x) =, satisfying the conditions: y() =, y () = 2, y () =. Solution. Let y(x) Y (p). Since, taking into account the given conditions, we have y (x) p Y (p) y() = py (p) () = py (p) +, y (x) p 2 Y (p) p y() y () = = p 2 Y (p) p () 2 = p 2 Y (p) + p 2, y (x) p 3 Y (p) p 2 y() p y () y () = = p 3 Y ( p) p 2 () p 2 = p 3 Y (p) + p 2 2p,

15 then after applying the Laplace transform to the given equation we obtain the following operator equation: p 3 Y (p) + p 2 2p + 2p 2 Y (p) + 2p 4 + 5pY (p) + 5 = or after transformations: Y (p) (p 3 + 2p 2 + 5p) = p 2. Solving this equation for Y (p), we obtain Y (p) = p 2 p(p 2 + 2p + 5). Let us decompose the resulting expression into simple fractions: p 2 p(p 2 + 2p + 5) = A p + Bp + C p 2 + 2p + 5. Using the method of indefinite coefficients, we find A, B, C. To do this, we reduce the fractions to the general denominator and equate the coefficients for equal powers of p: p 2 p(p 2 + 2p + 5) = Ap2 + 2Ap + 5A + Bp 2 + Cp p(p p + 5) We obtain a system of algebraic equations for A, B, C: the solution of which will be: A + B =, 2A + C =, 5A =, A = 5, B = 4 5, C = 2 5. Then Y (p) = 5p + 5 4p + 2 p 2 + 2p + 5. To find the original of the second fraction, we select the complete square in its denominator: p 2 + 2p + 5 = (p +) 2 + 4, then in the numerator we select the term p+: 4p+2 = 4(p+)+6 and decompose the fraction into the sum of two fractions : 5 4p + 2 p 2 + 2p + 5 = 4 5 p + (p +) (p +) Next, using the displacement theorem and the correspondence table between images and originals, we obtain a solution to the original equation. Answer: y(x) = e x cos 2x e x sin 2x.

16 6 Using the operational method, a general solution to equation (3) can be constructed. To do this, it is necessary to replace the specific values y, y,..., y (n) of the initial conditions with arbitrary constants C, C 2,..., C n. Bibliography. Aleksandrova N.V. History of mathematical terms, concepts, notations: Dictionary-reference book. M.: Publishing house LKI, p. 2. Vasilyeva A. B. Differential and integral equations, calculus of variations in examples and problems / A. B. Vasilyeva, G. N. Medvedev, N. A. Tikhonov, T. A. Urazgildina. M.: FIZ-MATLIT, p. 3. Sidorov Yu. V. Lectures on the theory of functions of a complex variable / Yu. V. Sidorov, M. V. Fedoryuk, M. I. Shabunin. M.: Science, 989.

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T A Matveeva V B Vetlichnaya D K Agisheva A Zotova SPECIAL CHAPTERS OF MATHEMATICS: OPERATIONAL STUDY FEDERAL AGENCY FOR EDUCATION VOLZHKY POLYTECHNIC INSTITUTE BRANCH OF STATE EDUCATION

INTEGRAL CALCULUS INDEMNITE INTEGRAL Antiderivative function and indefinite integral of the antiderivative Function F() is called antiderivative for the function f() on the interval X if F / () = f() X.

5. 4 Basic methods of integration Direct integration. Calculation of integrals based on reducing the integrand to tabular form and using the properties of the indefinite

Lecture 3 Mathematical description of control systems In control theory, when analyzing and synthesizing control systems, we deal with their mathematical model. The mathematical model of the automatic control system is the equation

Integrating a system of differential equations by eliminating variables One of the main methods for integrating a system of differential equations is as follows: from the equations of normal

Partial differential equations of the first order Some problems of classical mechanics, continuum mechanics, acoustics, optics, hydrodynamics, radiation transfer are reduced to partial differential equations

The simplest indefinite integrals Examples of problem solving The following integrals are reduced to tabular ones by identically transforming the integrand. 1. dx = dx = 2x 2/3 /3 + 2x 1/2 + C. >2.

PRACTICAL LESSON Integrating rational fractions A rational fraction is a fraction of the form P Q, where P and Q are polynomials. A rational fraction is called proper if the degree of the polynomial P is lower than the degree

[F] Filippov AV Collection of problems on differential equations Moscow-Izhevsk: Scientific Research Center "Regular and Chaotic Dynamics" 00 URL: http://librarbsaz/kitablar/846pf [M] Matveev NM Collection of problems and exercises on

E occupation. Taylor series. Summation of power series Mat. analysis, appl. mathematics, 3rd semester Find the expansion of a function into a power series in powers, calculate the radius of convergence of the power series: A f()

Task 1.1. Find in the indicated region non-identically zero solutions y = y(x) of the differential equation that satisfy the given boundary conditions (Sturm-Liouville problem) Solution: Consider

9. Antiderivative and indefinite integral 9.. Let the function f() be given on the interval I R. The function F () is called the antiderivative of the function f () on the interval I if F () = f () for any I, and the antiderivative

~ ~ Indefinite and definite integrals The concept of antiderivative and indefinite integral. Definition: A function F is called antiderivative of a function f if these functions are related as follows

Lecture 5 7 Hilbert-Schmidt theorem We will consider an integral operator A whose kernel K(satisfies the following conditions: K(s) is symmetric, continuous in the set of variables on [, ]

Ministry of Education of the Republic of Belarus Belarusian State University Faculty of Physics Department of Higher Mathematics and Mathematical Physics O A Kononova, N I Ilyinkova, N K Filippova Linear

Topic 9 Power series A power series is a functional series of the form where the numbers... are the coefficients of the series, and the expansion point of the series.,...,... R... is called the center Power series The general term of the power series

SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS Reduction to one equation of the th order From a practical point of view, linear systems with constant coefficients are very important

Integrals and differential equations Module 1. Indefinite integral Lecture 1.2 Abstract Rational fractions. Decomposition of a proper rational fraction into the simplest sum. Integrating the simplest

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 you 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” other 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 At once try to factor it. For these purposes, one should try to solve the quadratic equation , but many readers with a trained second-year 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 that corresponds 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 that corresponds 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 you don’t even 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:

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.

Lecture notes Higher mathematics

for students of direction 6.070104 “Sea and river transport”

specialty "Operation of ship power plants"

full-time and part-time courses 2nd year

Circulation______ copies Signed for publication ______________

Order No.__________. Volume__2.78__p.l.

Publishing house "Kerch State Marine Technological University"

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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.

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.

Example 1

, ,

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” other 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.

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:

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:

Example 2

Using operational calculus, find a particular solution to a system of differential equations that corresponds 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 that corresponds 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:

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:

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:

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:

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

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

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 P accordingly, while T P. 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) represents an oscillation written in complex form, where. Thus, real roots () correspond to aperiodic oscillations, complex roots with negative real parts correspond to damped oscillations, and purely imaginary roots correspond to undamped harmonic oscillations.

If the denominator does not have roots with positive real parts, then for sufficiently large values we obtain a 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 retardation 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 a moment in 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 of coordinates.

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.

Operational method for solving linear differential equations and their systems. Operator method for solving differential equations Solving differential equations using the operational method

Transcript

Heaviside expansion formula

Operator method for solving differential equations

Payments for environmental pollution

Operator method for solving differential equations Solving differential equations using the operational method