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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, with the possible exception of a finite number of discontinuity points 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 simple (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 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. References. 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.: Nauka, 989.

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

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How to Solve a Differential Equation
operational calculus method?

In this lesson, we will examine in detail a typical and widespread task of complex analysis - 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 problem using the 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 out the reference material provided. 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 independent decision:

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 fellow 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 DIY solution:

Example 4

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

A short solution and 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 answer.

As a result, the operator solution:

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

Thus, a particular solution:

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 - 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.

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

, 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, considering that all the functions under consideration are equal to zero for negative values t .

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