Find the original function from the image. How to find a similar picture, photograph, image on the Internet

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, condensed theoretical information about the section under consideration mathematical analysis. 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, the real function (original) according to certain rules must be converted into 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 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 deal with 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 in this case add the numbers –2 and –3. I strongly recommend that teapots do not miss this stage:

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 in left side 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:

I remind you that the further 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 do-it-yourself 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 point problem is fraction expansion 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 add an additional two from the numerator, however, after taking stock, I came to the conclusion that this step will practically not simplify the further solution.

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:

The problem is posed as follows: given a function F(p), we need to find the function /(<)>whose image is F(p). Let us formulate conditions sufficient for the function F(p) of a complex variable p to serve as an image. Theorem 12. If a function F(p) analytic in a half-plane 1) tends to zero for any half-plane Rep = a > s0 uniformly with respect to arg Finding the original from an image 2) the integral a-xu converges absolutely, then F(p) is an image some original function f(t). Tasks*. Can the function F(p) = ^ serve as an image of some original function? We will indicate some ways to find the original from an image. 3.1. Finding the original using image tables First of all, it is worth bringing the function F(p) to a simpler, “tabular” form. For example, in the case when F(p) is a fractional rational function of the argument p, it is decomposed into elementary fractions and the appropriate properties of the Laplace transform are used. Example 1. Find the original for We write the function F(p) in the form Using the displacement theorem and the linearity property of the Laplace transform, we obtain Example 2. Find the original for the function M We write F(p) in the form Hence / 3.2. Using the inversion theorem and its corollaries Theorem 13 (inversion). /Gauche function fit) is the original function with growth exponent s0 and F(p) is its image, then at any point of continuity of the function f(t) the relation is satisfied where the integral is taken along any straight line and is understood in the sense of the principal value, i.e. . as Formula (1) is called the Laplace transform inversion formula, or Mellin's formula. Indeed, let, for example, f(t) be piecewise smooth on each finite segment)