Solution of inhomogeneous differential equations of the first order. Linear and homogeneous differential equations of the first order

Often just a mention differential equations makes students feel uncomfortable. Why is this happening? Most often, because when studying the basics of the material, a gap in knowledge arises, due to which further study of difurs becomes simply torture. It’s not clear what to do, how to decide, where to start?

However, we will try to show you that difurs are not as difficult as it seems.

Basic concepts of the theory of differential equations

From school we know the simplest equations in which we need to find the unknown x. In fact differential equations only slightly different from them - instead of a variable X you need to find a function in them y(x) , which will turn the equation into an identity.

D differential equations are of great practical importance. This is not abstract mathematics that has no relation to the world around us. Many real natural processes are described using differential equations. For example, the vibrations of a string, the movement of a harmonic oscillator, using differential equations in problems of mechanics, find the speed and acceleration of a body. Also DU are widely used in biology, chemistry, economics and many other sciences.

Differential equation (DU) is an equation containing derivatives of the function y(x), the function itself, independent variables and other parameters in various combinations.

There are many types of differential equations: ordinary differential equations, linear and nonlinear, homogeneous and inhomogeneous, first and higher order differential equations, partial differential equations, and so on.

The solution to a differential equation is a function that turns it into an identity. There are general and particular solutions of the remote control.

A general solution to a differential equation is a general set of solutions that transform the equation into an identity. A partial solution of a differential equation is a solution that satisfies additional conditions specified initially.

The order of a differential equation is determined by the highest order of its derivatives.

Ordinary differential equations

Ordinary differential equations are equations containing one independent variable.

Let's consider the simplest ordinary differential equation of the first order. It looks like:

This equation can be solved by simply integrating its right-hand side.

Examples of such equations:

Separable equations

In general, this type of equation looks like this:

Here's an example:

When solving such an equation, you need to separate the variables, bringing it to the form:

After this, it remains to integrate both parts and obtain a solution.

Linear differential equations of the first order

Such equations look like:

Here p(x) and q(x) are some functions of the independent variable, and y=y(x) is the desired function. Here is an example of such an equation:

When solving such an equation, most often they use the method of varying an arbitrary constant or represent the desired function as a product of two other functions y(x)=u(x)v(x).

To solve such equations, certain preparation is required and it will be quite difficult to take them “at a glance”.

An example of solving a differential equation with separable variables

So we looked at the simplest types of remote control. Now let's look at the solution to one of them. Let this be an equation with separable variables.

First, let's rewrite the derivative in a more familiar form:

Then we divide the variables, that is, in one part of the equation we collect all the “I’s”, and in the other - the “X’s”:

Now it remains to integrate both parts:

We integrate and obtain a general solution to this equation:

Of course, solving differential equations is a kind of art. You need to be able to understand what type of equation it is, and also learn to see what transformations need to be made with it in order to lead to one form or another, not to mention just the ability to differentiate and integrate. And to succeed in solving DE, you need practice (as in everything). And if you currently don’t have time to understand how differential equations are solved or the Cauchy problem has stuck like a bone in your throat, or you don’t know, contact our authors. In a short time, we will provide you with a ready-made and detailed solution, the details of which you can understand at any time convenient for you. In the meantime, we suggest watching a video on the topic “How to solve differential equations”:

A first-order equation of the form a 1 (x)y" + a 0 (x)y = b(x) is called a linear differential equation. If b(x) ≡ 0 then the equation is called homogeneous, otherwise - heterogeneous. For a linear differential equation, the existence and uniqueness theorem has a more specific form.

Purpose of the service. An online calculator can be used to check the solution homogeneous and inhomogeneous linear differential equations of the form y"+y=b(x) .

To obtain a solution, the original expression must be reduced to the form: a 1 (x)y" + a 0 (x)y = b(x). For example, for y"-exp(x)=2*y it will be y"-2 *y=exp(x) .

Theorem. Let a 1 (x) , a 0 (x) , b(x) be continuous on the interval [α,β], a 1 ≠0 for ∀x∈[α,β]. Then for any point (x 0 , y 0), x 0 ∈[α,β], there is a unique solution to the equation that satisfies the condition y(x 0) = y 0 and is defined on the entire interval [α,β].
Consider the homogeneous linear differential equation a 1 (x)y"+a 0 (x)y=0.
Separating the variables, we get , or, integrating both sides, The last relation, taking into account the notation exp(x) = e x , is written in the form

Let us now try to find a solution to the equation in the indicated form, in which instead of the constant C the function C(x) is substituted, that is, in the form

Substituting this solution into the original one, after the necessary transformations we obtain Integrating the latter, we have

where C 1 is some new constant. Substituting the resulting expression for C(x), we finally obtain the solution to the original linear equation
.

Example. Solve the equation y" + 2y = 4x. Consider the corresponding homogeneous equation y" + 2y = 0. Solving it, we get y = Ce -2 x. We are now looking for a solution to the original equation in the form y = C(x)e -2 x. Substituting y and y" = C"(x)e -2 x - 2C(x)e -2 x into the original equation, we have C"(x) = 4xe 2 x, whence C(x) = 2xe 2 x - e 2 x + C 1 and y(x) = (2xe 2 x - e 2 x + C 1)e -2 x = 2x - 1 + C 1 e -2 x is the general solution of the original equation. In this solution y 1 ( x) = 2x-1 - motion of the object under the influence of force b(x) = 4x, y 2 (x) = C 1 e -2 x - proper motion of the object.

Example No. 2. Find the general solution to the first order differential equation y"+3 y tan(3x)=2 cos(3x)/sin 2 2x.
This is not a homogeneous equation. Let's make a change of variables: y=u v, y" = u"v + uv".
3u v tg(3x)+u v"+u" v = 2cos(3x)/sin 2 2x or u(3v tg(3x)+v") + u" v= 2cos(3x)/sin 2 2x
The solution consists of two stages:
1. u(3v tan(3x)+v") = 0
2. u"v = 2cos(3x)/sin 2 2x
1. Equate u=0, find a solution for 3v tan(3x)+v" = 0
Let's present it in the form: v" = -3v tg(3x)

Integrating, we get:

ln(v) = ln(cos(3x))
v = cos(3x)
2. Knowing v, Find u from the condition: u"v = 2cos(3x)/sin 2 2x
u" cos(3x) = 2cos(3x)/sin 2 2x
u" = 2/sin 2 2x
Integrating, we get:
From the condition y=u v, we get:
y = u v = (C-cos(2x)/sin(2x)) cos(3x) or y = C cos(3x)-cos(2x) cot(3x)

In this topic, we will talk about ways to solve linear inhomogeneous differential equations of the form y " = P (x) · y = Q (x). Let's start with the method of variation of an arbitrary constant and show how to use this method to solve the Cauchy problem. Let's continue by considering the method that assumes representation of an arbitrary constant y as the product of two functions u (x) and v (x).In this section we present a large number of problems on the topic with a detailed analysis of the solution.

In case the terms and concepts used in the analysis of the topic turn out to be unfamiliar to you, we recommend looking at the section “Basic terms and definitions of the theory of differential equations.”

Variation method for an arbitrary constant for solving first-order LPDEs

For brevity, we will denote a linear inhomogeneous differential equation by the abbreviation LNDE, and a linear homogeneous differential equation (LODE).

LNDE of the form y " = P (x) y = Q (x) corresponds to LDE of the form y " = P (x) y = 0 , with Q(x)=0. If you look at the differential equation y " = P (x) y = 0, it becomes clear that we are dealing with an equation with separable variables. We can integrate it: y " = P (x) y = 0 ⇔ d y y = - P (x) d x , y ≠ 0 ∫ d y y = - ∫ P (x) d x ⇔ ln y + C 1 = - ∫ P (x) d x ⇔ ln y = ln C - ∫ P (x) d x , ln C = - C 1 , C ≠ 0 ⇔ e ln y = e ln C - ∫ P (x) d x ⇔ y = C e - ∫ P (x) d x

We can claim that the value of the variable y = 0 is also a solution, since with this value of the variable the equation y " = P (x) y = 0 becomes an identity. This case corresponds to the solution y = C e - ∫ P (x ) d x at value C=0.

It turns out that y = C e - ∫ P (x) d x is the general solution of the LODE, where WITH– arbitrary constant.

y = C · e - ∫ P (x) d x is the solution to the LOD y " = P (x) · y = 0 .

In order to find a general solution to the inhomogeneous equation y " = P (x) y = Q (x), we will consider C not a constant, but a function of the argument x. In fact, we will take y = C (x) e - ∫ P (x) d x by the general solution of the LNDE.

Let us substitute y = C (x) e - ∫ P (x) d x into the differential equation y " = P (x) y = Q (x). It becomes the identity:

y " = P (x) y = Q (x) C x e - ∫ P (x) d x + P (x) C (x) e - ∫ P (x) d x = Q (x)

Now let's turn to the product differentiation rule. We get:

C " (x) e - ∫ P (x) d x + C (x) e - ∫ P (x) d x + P (x) C (x) e - ∫ P (x) d x = Q ( x)

The derivative of a complex function e - ∫ P (x) d x " is equal to e - ∫ P (x) d x · - ∫ P (x) d x " .

Now let us recall the properties of the indefinite integral. We get:

e - ∫ P (x) d x · - ∫ P (x) d x " = - e - ∫ P (x) d x · P (x)

Now let's make the transition:

C " (x) e - ∫ P (x) d x + C (x) e - ∫ P (x) d x " + P (x) C (x) e - ∫ P (x) d x = Q (x) C " (x) e - ∫ P (x) d x - P (x) C (x) e - ∫ P (x) d x + P (x) C (x) e - ∫ P (x) d x = Q (x) C " (x) e - ∫ P (x) d x = Q (x)

So we came to the simplest differential equation of the first order. In solving this equation we will define the function C(x). This will allow us to write the solution to the original first-order LPDE as follows:

y = C (x) e - ∫ P (x) d x

Summarize

The method of varying an arbitrary constant when solving an LPDE involves three stages:

finding a general solution to the corresponding LOD y " + P (x) · y = 0 in the form y = C · e - ∫ P (x) d x ;
variation of an arbitrary constant C, which consists of replacing it with a function C(x);
substituting the function y = C (x) e - ∫ P (x) d x into the original differential equation, from which we can calculate C(x) and write down the answer.

Now let's apply this algorithm to solve the problem.

Example 1

Find the solution to the Cauchy problem y " - 2 x y 1 + x 2 = 1 + x 2 , y (1) = 3 .

Solution

We need to find a particular solution to the LNDDE y " - 2 x y 1 + x 2 = 1 + x 2 with the initial condition y (1) = 3.

In our example P(x) = - 2 x 1 + x 2 and Q(x)=x2+1. Let's start by finding a general solution to the LOD. After this, we will apply the method of variation of an arbitrary constant and determine the general solution of the LPDE. This will allow us to find the desired particular solution.

The general solution of the corresponding LOD y " - 2 x y 1 + x 2 = 0 will be the family of functions y = C · (x 2 + 1), where C is an arbitrary constant.

We vary an arbitrary constant y = C (x) · (x 2 + 1) and substitute this function into the original equation:
y " - 2 x y 1 + x 2 = 1 + x 2 C x · (x 2 + 1 " - 2 x · C (x) · (x 2 + 1) 1 + x 2 = 1 + x 2 C " ( x) · (x 2 + 1) + C (x) · 2 x - 2 x · C (x) = 1 + x 2 C " (x) = 1,

whence C (x) = ∫ d x = x + C 1 , where C 1– arbitrary constant.

This means that y = C (x) · (x 2 + 1) = (x + C 1) · (x 2 + 1) is the general solution to the inhomogeneous equation.

Now let's start finding a particular solution that will satisfy the initial condition y (1) = 3.

Since y = (x + C 1) · (x 2 + 1) , then y (1) = (1 + C 1) · (1 2 + 1) = 2 · (1 + C 1) . Turning to the initial condition, we obtain the equation 2 · (1 + C 1) = 3, whence C 1 = 1 2. Therefore, the desired solution to the Cauchy problem has the form y = x + 1 2 · (x 2 + 1)

Now consider another method for solving linear inhomogeneous differential equations y " + P (x) · y = Q (x) .

Another method for solving first-order LPDEs

We can represent the unknown function as the product y = u ⋅ v, where u and v– argument functions x.

We can substitute this function into a first-order LNDE. We have:

y " + P (x) y = Q (x) (u v) " + P (x) u v = Q (x) u " v + u v " + P (x) u v = Q (x) u " v + u (v " + P (x) v) = Q (x)

If we find such v that it is a non-zero partial solution of the differential equation v " + P (x) v = 0, then u can be determined from the separable equation u " · v = Q (x) .

Let's consider this solution algorithm using the previous example. This will allow us to focus on the main thing without being distracted by minor details.

Example 2

Find the general solution to the linear inhomogeneous differential equation y " - 2 x y 1 + x 2 = 1 + x 2 .

Solution

Let y = u ⋅ v, then
y " - 2 x y x 2 + 1 = x 2 + 1 ⇔ (u v) - 2 x u v x 2 + 1 = x 2 + 1 u " v + u v " - 2 x u v x 2 + 1 = x 2 + 1 u " v + u v " - 2 x v x 2 + 1 = x 2 + 1

We find this v, other than zero, so that the expression in parentheses becomes zero. In other words, we find a particular solution to the differential equation v " - 2 x · v x 2 + 1 = 0.
v " - 2 x · v x 2 + 1 = 0 ⇔ d v d x = 2 x · v x 2 + 1 ⇒ d v v = 2 x d x x 2 + 1 ⇔ d v v = d (x 2 + 1) x 2 + 1 ∫ d v v = ∫ d ( x 2 + 1) x 2 + 1 ln v + C 1 = ln (x 2 + 1) + C 2

Let's take a particular solution v = x 2 + 1, corresponding to C 2 – C 1 = 0.

For this particular solution we have
u " v + u v " - 2 x v x 2 + 1 = x 2 + 1 ⇔ u " (x 2 + 1) + u 0 = x 2 + 1 ⇔ u " = 1 ⇔ u = x +C

Therefore, the general solution to the original linear inhomogeneous differential equation is y = u v = (x + C) (x 2 + 1)

The answers in both cases are the same. This means that both solution methods that we presented in the article are equivalent. It is up to you to choose which one to use to solve the problem.

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I think we should start with the history of such a glorious mathematical tool as differential equations. Like all differential and integral calculus, these equations were invented by Newton in the late 17th century. He considered this particular discovery of his to be so important that he even encrypted a message, which today can be translated something like this: “All laws of nature are described by differential equations.” This may seem like an exaggeration, but it is true. Any law of physics, chemistry, biology can be described by these equations.

Mathematicians Euler and Lagrange made a huge contribution to the development and creation of the theory of differential equations. Already in the 18th century they discovered and developed what they now study in senior university courses.

A new milestone in the study of differential equations began thanks to Henri Poincaré. He created the “qualitative theory of differential equations”, which, combined with the theory of functions of a complex variable, made a significant contribution to the foundation of topology - the science of space and its properties.

What are differential equations?

Many people are afraid of one phrase. However, in this article we will outline in detail the whole essence of this very useful mathematical apparatus, which is actually not as complicated as it seems from the name. In order to start talking about first-order differential equations, you should first become familiar with the basic concepts that are inherently associated with this definition. And we'll start with the differential.

Differential

Many people have known this concept since school. However, let’s take a closer look at it. Imagine the graph of a function. We can increase it to such an extent that any segment of it will take the form of a straight line. Let’s take two points on it that are infinitely close to each other. The difference between their coordinates (x or y) will be infinitesimal. It is called the differential and is denoted by the signs dy (differential of y) and dx (differential of x). It is very important to understand that the differential is not a finite quantity, and this is its meaning and main function.

Now we need to consider the next element, which will be useful to us in explaining the concept of a differential equation. This is a derivative.

Derivative

We all probably heard this concept at school. The derivative is said to be the rate at which a function increases or decreases. However, from this definition much becomes unclear. Let's try to explain the derivative through differentials. Let's return to an infinitesimal segment of a function with two points that are at a minimum distance from each other. But even over this distance the function manages to change by some amount. And to describe this change they came up with a derivative, which can otherwise be written as a ratio of differentials: f(x)"=df/dx.

Now it’s worth considering the basic properties of the derivative. There are only three of them:

The derivative of a sum or difference can be represented as a sum or difference of derivatives: (a+b)"=a"+b" and (a-b)"=a"-b".
The second property is related to multiplication. The derivative of a product is the sum of the products of one function and the derivative of another: (a*b)"=a"*b+a*b".
The derivative of the difference can be written as the following equality: (a/b)"=(a"*b-a*b")/b 2 .

All these properties will be useful to us for finding solutions to first-order differential equations.

There are also partial derivatives. Let's say we have a function z that depends on the variables x and y. To calculate the partial derivative of this function, say, with respect to x, we need to take the variable y as a constant and simply differentiate.

Integral

Another important concept is integral. In fact, this is the exact opposite of a derivative. There are several types of integrals, but to solve the simplest differential equations we need the most trivial ones

So, let's say we have some dependence of f on x. We take the integral from it and get the function F(x) (often called the antiderivative), the derivative of which is equal to the original function. Thus F(x)"=f(x). It also follows that the integral of the derivative is equal to the original function.

When solving differential equations, it is very important to understand the meaning and function of the integral, since you will have to take them very often to find the solution.

Equations vary depending on their nature. In the next section, we will look at the types of first-order differential equations, and then learn how to solve them.

Classes of differential equations

"Diffurs" are divided according to the order of the derivatives involved in them. Thus there is first, second, third and more order. They can also be divided into several classes: ordinary and partial derivatives.

In this article we will look at first order ordinary differential equations. We will also discuss examples and ways to solve them in the following sections. We will consider only ODEs, because these are the most common types of equations. Ordinary ones are divided into subspecies: with separable variables, homogeneous and heterogeneous. Next, you will learn how they differ from each other and learn how to solve them.

In addition, these equations can be combined so that we end up with a system of first-order differential equations. We will also consider such systems and learn how to solve them.

Why are we only considering first order? Because you need to start with something simple, and it is simply impossible to describe everything related to differential equations in one article.

Separable equations

These are perhaps the simplest first order differential equations. These include examples that can be written as follows: y"=f(x)*f(y). To solve this equation, we need a formula for representing the derivative as a ratio of differentials: y"=dy/dx. Using it we get the following equation: dy/dx=f(x)*f(y). Now we can turn to the method for solving standard examples: we will divide the variables into parts, that is, we will move everything with the variable y to the part where dy is located, and do the same with the variable x. We obtain an equation of the form: dy/f(y)=f(x)dx, which is solved by taking integrals of both sides. Don't forget about the constant that needs to be set after taking the integral.

The solution to any “diffure” is a function of the dependence of x on y (in our case) or, if a numerical condition is present, then the answer in the form of a number. Let's look at the whole solution process using a specific example:

Let's move the variables in different directions:

Now let's take the integrals. All of them can be found in a special table of integrals. And we get:

ln(y) = -2*cos(x) + C

If required, we can express "y" as a function of "x". Now we can say that our differential equation is solved if the condition is not specified. A condition can be specified, for example, y(n/2)=e. Then we simply substitute the values of these variables into the solution and find the value of the constant. In our example it is 1.

Homogeneous differential equations of the first order

Now let's move on to the more difficult part. Homogeneous differential equations of the first order can be written in general form as follows: y"=z(x,y). It should be noted that the right-hand function of two variables is homogeneous, and it cannot be divided into two dependences: z on x and z on y. Check , whether the equation is homogeneous or not is quite simple: we make the replacement x=k*x and y=k*y. Now we cancel all k. If all these letters are canceled, then the equation is homogeneous and you can safely start solving it. Looking ahead , let's say: the principle of solving these examples is also very simple.

We need to make a replacement: y=t(x)*x, where t is a certain function that also depends on x. Then we can express the derivative: y"=t"(x)*x+t. Substituting all this into our original equation and simplifying it, we get an example with separable variables t and x. We solve it and get the dependence t(x). When we received it, we simply substitute y=t(x)*x into our previous replacement. Then we get the dependence of y on x.

To make it clearer, let's look at an example: x*y"=y-x*e y/x .

When checking with replacement, everything is reduced. This means that the equation is truly homogeneous. Now we make another replacement that we talked about: y=t(x)*x and y"=t"(x)*x+t(x). After simplification, we obtain the following equation: t"(x)*x=-e t. We solve the resulting example with separated variables and get: e -t =ln(C*x). All we have to do is replace t with y/x (after all, if y =t*x, then t=y/x), and we get the answer: e -y/x =ln(x*C).

Linear differential equations of the first order

It's time to look at another broad topic. We will analyze first-order inhomogeneous differential equations. How are they different from the previous two? Let's figure it out. Linear differential equations of the first order in general form can be written as follows: y" + g(x)*y=z(x). It is worth clarifying that z(x) and g(x) can be constant quantities.

And now an example: y" - y*x=x 2 .

There are two solutions, and we will look at both in order. The first is the method of varying arbitrary constants.

In order to solve the equation in this way, you must first equate the right side to zero and solve the resulting equation, which, after transferring the parts, will take the form:

ln|y|=x 2 /2 + C;

y=e x2/2 *y C =C 1 *e x2/2 .

Now we need to replace the constant C 1 with the function v(x), which we have to find.

Let's replace the derivative:

y"=v"*e x2/2 -x*v*e x2/2 .

And substitute these expressions into the original equation:

v"*e x2/2 - x*v*e x2/2 + x*v*e x2/2 = x 2 .

You can see that on the left side two terms cancel. If in some example this did not happen, then you did something wrong. Let's continue:

v"*e x2/2 = x 2 .

Now we solve the usual equation in which we need to separate the variables:

dv/dx=x 2 /e x2/2 ;

dv = x 2 *e - x2/2 dx.

To extract the integral, we will have to apply integration by parts here. However, this is not the topic of our article. If you are interested, you can learn how to perform such actions yourself. It is not difficult, and with sufficient skill and care it does not take much time.

Let's turn to the second method of solving inhomogeneous equations: Bernoulli's method. Which approach is faster and easier is up to you to decide.

So, when solving an equation using this method, we need to make a substitution: y=k*n. Here k and n are some x-dependent functions. Then the derivative will look like this: y"=k"*n+k*n". We substitute both replacements into the equation:

k"*n+k*n"+x*k*n=x 2 .

Grouping:

k"*n+k*(n"+x*n)=x 2 .

Now we need to equate to zero what is in parentheses. Now, if we combine the two resulting equations, we get a system of first-order differential equations that needs to be solved:

We solve the first equality as an ordinary equation. To do this you need to separate the variables:

We take the integral and get: ln(n)=x 2 /2. Then, if we express n:

Now we substitute the resulting equality into the second equation of the system:

k"*e x2/2 =x 2 .

And transforming, we get the same equality as in the first method:

dk=x 2 /e x2/2 .

We will also not discuss further actions. It is worth saying that at first solving first-order differential equations causes significant difficulties. However, as you delve deeper into the topic, it starts to work out better and better.

Where are differential equations used?

Differential equations are used very actively in physics, since almost all the basic laws are written in differential form, and the formulas that we see are solutions to these equations. In chemistry they are used for the same reason: fundamental laws are derived with their help. In biology, differential equations are used to model the behavior of systems, such as predator and prey. They can also be used to create reproduction models of, say, a colony of microorganisms.

How can differential equations help you in life?

The answer to this question is simple: not at all. If you are not a scientist or engineer, then they are unlikely to be useful to you. However, for general development it will not hurt to know what a differential equation is and how it is solved. And then the son or daughter’s question is “what is a differential equation?” won't confuse you. Well, if you are a scientist or engineer, then you yourself understand the importance of this topic in any science. But the most important thing is that now the question “how to solve a first-order differential equation?” you can always give an answer. Agree, it’s always nice when you understand something that people are even afraid to understand.

Main problems in studying

The main problem in understanding this topic is poor skill in integrating and differentiating functions. If you are not good at derivatives and integrals, then it is probably worth studying more, mastering different methods of integration and differentiation, and only then starting to study the material that was described in the article.

Some people are surprised when they learn that dx can be carried over, because previously (at school) it was stated that the fraction dy/dx is indivisible. Here you need to read the literature on the derivative and understand that it is a ratio of infinitesimal quantities that can be manipulated when solving equations.

Many people do not immediately realize that solving first-order differential equations is often a function or an integral that cannot be taken, and this misconception gives them a lot of trouble.

What else can you study for a better understanding?

It is best to begin further immersion in the world of differential calculus with specialized textbooks, for example, on mathematical analysis for students of non-mathematical specialties. Then you can move on to more specialized literature.

It is worth saying that, in addition to differential equations, there are also integral equations, so you will always have something to strive for and something to study.

Conclusion

We hope that after reading this article you have an idea of what differential equations are and how to solve them correctly.

In any case, mathematics will be useful to us in life in some way. It develops logic and attention, without which every person is without hands.