Examples of solutions of second-order differential equations by the Lagrange method. Linear differential equations of the second order. The informal system of solutions solve the system of second-order equations
In the theory of systems of linear equations and in some other matters it is convenient to use the concept of determinant, or determinant.
Consider any four numbers recorded in the form of a square table (matrix) for two in lines and two in columns. The determinant or determinant composed of the numbers of this table is called the number indicated as follows:
Such a determinant is called the second order determinant, since a table of two rows and two columns is taken to compile it. The numbers from which the determinant is compiled is called its elements, while it is said that the elements are the main diagonal of the determinant, and the elements are its side diagonal. It can be seen that the determinant is equal to the difference in the works of the elements standing on its main and side diagonals.
Example 1. Calculate the following second order determinants:
Decision, a) by definition we have
With the help of determinants, equality can be equality (66.6), (66.7) and (66.8) rewrite, changing their parts, so:
Note that the determinants are very simply compiled by system coefficients (66.2).
Indeed, the determinant is made up of coefficients at unknown in this system. It is called the main determinant of the system (66.2). We call the determinants for unknown x and y, respectively. It is possible to formulate the following rule of their compilation: the determinant for each of the unknowns is obtained from the main determinant if the coefficient column in this unknown is to be replaced by a column of free members (taken from the right parts of the system equations).
Example 2. System (66.12) to solve with the help of determinants.
Decision. We compile and calculate the main determinant of this system:
Now it is replaced by the column of coefficients at x (first column) with free members. We obtain the determinant for x:
Similarly, we find
From here by formulas (66.11) we get
We came to the already known solution (1, -1).
We now conduct a study of the system of linear equations (66.2). To do this, we will return to equality (66.9) and (66.10) and we will distinguish between two cases:
Let, then, as already noted, formulas (66.11) give the only solution of the system (66.2). So, if the main determinant of the system is zero, the system has a single solution determined by formulas (66.11); Such a system is called a certain.
2) Let now. Depending on the values \u200b\u200bwe will distinguish between two cases.
a) at least one of the determinants differ from zero; Then the system (66.2) has no solutions. Indeed, let, for example,. Equality (66.9) cannot be satisfied with any value as this equality is obtained as a consequence of the system (66.2), the system has no solutions. Such a system is called incomplete.
b) both determinants are zero; Equality (66.9) and (66.10) are satisfied identical and to study the system (66.2) are not used.
We prove that if at least one of the coefficients at system unknown in the system (66.2) differ from zero, the system has an infinite set of solutions. To make sure that, let's say, for example, that. From relations
and from the recording of the second equation of the system (66.2), substituting the expressions of the coefficients
we find that it differs from the first equation only a multiplier that is, essentially coincides with it (equivalent to him). The system (66.2) is reduced to one by the first equation and defines countless solutions (such a system is called uncertain). It is possible, in principle, such an extreme case as equality zero of all coefficients at unknowns (it can meet in the study of systems with lettering coefficients). Such a system
all determinants are zero: however, it is incomplete when or.
We will summarize the study of the system of linear equations (66.2). There are three types of such systems:
1) If, the system is defined, has a single solution (66.11).
2) If, but then the system is incomprehensible, the solutions do not have.
3) If at least one of the coefficients at unknown is zero), then the system is indefinite, has an infinite set of solutions (reduces to one equation).
Equality zero determinant,
means the proportionality of elements in its lines (and back):
Because of this, signs that distinguish linear systems of different types (defined, indefinite, incomplete) can be formulated in terms of proportions between system coefficients (without attracting determinants).
The condition is replaced so the requirement of proportionality (disproportionateness) of coefficients at unknown:
In the case, not only coefficients are proportional to unknown, but also free members:
(These proportions are obtained, for example, from (67.6)). If, for example, before, then from (66.6) we see that - free members are not proportional to the coefficients at unknown. So:
1) if the coefficients are not proportional to unknown:
that system is defined.
2) if the coefficients are proportional to unknown, and the free members are not proportional to them:
that system is incomplete.
3) if coefficients are proportional to unknown and free members:
then the system is uncertain.
The study of the systems of linear equations with two unknown permits a simple geometric interpretation. Any linear equation of the form (38.4) determines the direct line on the coordinate plane. System equations (66.2) can therefore interpret as the equations of two direct on the plane, and the task of solving the system is as a task of finding the intersection point of these direct.
It is clear that three cases are possible: 1) the data are two straight lines intersect (Fig. 61, a); This case corresponds to a specific system; 2) the data are two straight parallel (Fig. 61, b); This case corresponds to the incomplete system;
3) the direct data coincide (Fig. 61, c); This case matches an uncertain system: each point "twice specified" direct will be solving the system.
Example 3. Explore linear systems:
Decision, a) make up and calculate the main determinant of this system.
Definition. Determinant of the second order
(*)
; 
; 
Theoretically, the following three cases are possible.
1. If, then the system (*) has a single solution that can be found according to formulas, which are called crawler formulas: ,.
2. If, and (then), the system (*) has no solutions.
3. If and (then), the system (*) has an infinite set of solutions (namely, each solution of one system equation is and solving another equation).
Comment. The determinant is called the main determinant of the system (*). The system can be solved according to the crimera formulas only under the condition. Otherwise, you need to use other methods, such as Gauss method.
The determinant of the third order. Solution of the system of three linear equations with three variables according to Cramer formulas
Definition. Third-order determinant The number is called and is calculated as follows:
Let a system of equations of type be given 
(*)
We introduce the following determinants to consideration:
- main determinant of the system (*);
; 
; 
.
When solving the system, the following cases are possible.
1. If, the system (*) has a single solution that can be found by formulas, which are called craver formulas: 
.
2. If, it is impossible to solve the system (1) by the Cramer method.
Note 1. In the case of the system may not have solutions or have infinite set solutions. For a more detailed study and finding a general solution system, you can use, for example, the Gauss method.
Solution of the system of three linear equations with three variables
By Gauss
The essence of the Gauss method will consider on a specific example.
Example. Solve the system of equations: 
(*)
Direct move. This system is shown to the triangular species in the method of algebraic addition.
At the first stage, we exclude from the second and third equations of the system, containing a variable. It is better to use in both cases the same equation (we will take the first).
We get:
 |  |
The first equation of the system will rewrite unchanged, and the second and third equations are replaced by the obtained equations.
The system will take the form: 
At the second stage, the term, containing a variable, eliminate from the third equation. We use the second equation for this.
The first two equations of the system will be represented unchanged, and the third equation is replaced by the resulting equation.
We get a triangular type system: 
Return. We consistently find unknown, starting from the third equation.
From the third equation of the system we find the value of the variable: .
Substitting the value in the second equation of the system, we get where to find the value of the variable: 
.
Substituting the found values \u200b\u200band in the first equation of the system, we get where we find the value of the variable: 
.
Answer: .
22. Solution of linear inequality
| Examples | |
| 1. If, then. | |
| 2. If, then. | |
| 3. If, then. | |
| 4. If, then the inequality does not have solutions. | Inequalities and do not have solutions. |
23. Linear inequality
| When solving inequality, the following cases are possible: | Examples |
| 1. If, then. | |
| 2. If, then. | |
| 3. If,, the inequality does not have solutions. | Inequality is not solutions. |
| 4. If, then. |
24. Solution of linear inequalities with one variable
The system of inequalities - These are two or more inequalities for which general solutions are searched.
By solving the system of inequality It is called the general solution of all inequalities in the system.
Theoretically possible cases even for a system of two inequalities are very much, so consider the main cases for the system of two simple inequalities.
Example 1.. Solve the system of inequalities:
Answer: .
Example 2.. Solve the system of inequalities:
I will depict the solutions of inequalities graphically.
Answer: .
Example 3..
Solve the system of inequalities:
I will depict the solutions of inequalities graphically.
Answer: .
Example 4.Solve the system of inequalities:
I will depict the solutions of inequalities graphically.
Answer: The system has no solutions.
25. Decision of incomplete square equations,
Square equation Called the view equation 
.
Square equation is called incompleteif at least one of the coefficients or is zero.
Each of the incomplete equations can be solved by the general formula. But it is more convenient to use private methods.
Case 1.
Its left part can be decomposed on the factors :. It is known that the work is zero if and only if at least one of the multipliers is zero. We get: or where due to the condition it follows that.
Output:the equation always has two valid roots ,.
Example 1. Solve equation.
Decision: or , .
Case 2. If, the equation takes a view.
Then. Since, then.
If, this equation has no valid roots (as).
If, the equation has two valid roots.
Example 2. Solve equation.
Decision:. Since, this equation does not have valid roots.
Example 3. Solve equation.
Decision: .
Case 3. If, the equation takes the form.
Since, then, or, therefore the equation has two equal root.
Example 4. Solve equation.
Decision: .
26. Solution of the reduced square equation 
The specified square equation is called a square equation 
, whose senior coefficient.
To find his roots, highlight a full square with a variable x.. We get:
.
The number is called the discriminant of the given square equation. The number of valid roots of the equation depends on the discriminant sign.
If, the equation does not have valid roots, since.
If, then 
, 
,
that is, the equation has two valid roots. 
and 
.
Comment. Formula 
It is especially convenient to use if the P coefficient is an even number.
Example. Solve equation 
.
Decision.Since, then, therefore 
.
Then 
, 
.
Answer: , .
27. Vieta formulas for the given square equation
provided two valid roots and .
Then ,
Thus, the theorem is proved, which is called the Vieta theorem.
Theorem. If the roots of the given square equation , then equality are just.
These equalities are called Vieta Formulas.
Comment. Vieta Formulas are valid and if the equation It has integrated conjugate roots.
Example. In the previous paragraph showed that the equation It has roots. Then.
Since, then , .
28. Solution of the square equation 
Since, by determining the square equation, it can be divided into action of the equation. We obtain a given square equation 
, in which , . Then his roots can be found by the formula .
We get:
The number is called the discriminant of a square equation (and the discriminant of square three decar). Discriminant shows how many valid roots have this equation.
If, then the equation It has two unequal valid root 
and 
().
If, then the equation It has two equal valid root.
If, then the equation has no valid roots.
Comment. In this case, the equation has two complex conjugate roots.
and 
.
Example 1. Solve equation 
.
Decision. Since, (then), then
Since, then 
.
Then 
, 
.
Answer: ,.
Example 2. Solve equation 
.
Decision. Since, ,, then.
Since this equation has no valid roots.
29. Solution of square inequalities
, 
, 
, 
with positive discriminant
establishment to the system of two linear inequalities
The discriminant of square three decar is a number.
Roots of square three decrees are called the roots of the equation .
and , Moreover, it means).
Then it can be decomposed on linear multipliers :.
Since, it is possible to divide on a both parts of each of the inequalities under consideration (if, the sign of inequality (that is, the sign\u003e or<) сохранится, если , то знак неравенства поменяется на противоположный). В результате получится неравенство одного из видов: 
, 
, 
, 
. Consider the solution of these inequalities.
1) the work of two factors is positively, if both multipliers are positive or both negative multiplier, so 
if or.
Solutions of both systems are decisions of this square inequality.
As , so then ).
Since, then (then).
Answer: inequality
It has many solutions that can be written as or or in the form.
3) The product of two factors is negative, if one of the multipliers is positive, and the other is negative. therefore if or.
Since, then.
This inequality system has no solutions, since the number X cannot be simultaneously less than less of two numbers and, and more than more.
Answer: inequality
2) Similarly, we get that inequality It has many solutions that can be written in the form or in the form.
Example. Solve inequality 
.
Decision. Find the roots of square three declections, that is, the roots of the equation 
: ,
, 
.
Decuting the left part of this inequality by the formula, we get inequality 
.
Since, by dividing both parts of the last inequality by 3, we get an equivalent inequality 
.
The work of two factors is negative if one of the multipliers is positive, and the other is negative. Therefore, solutions of the last inequality are solutions to each of the inequality systems if or. Then or
The graphical solution of the systems is presented in the figures (for the first system drawing on the left, for the second right). It can be seen that the second solutions system does not have, therefore, only solutions of the first system are solutions to this inequality.
Answer:
30. Solution of square inequalities
, , ,
using a chart of a quadratic function
Comment.We can assume that in all these inequalities. Otherwise, multiplying both parts of inequality on and changing the sign of inequality to the opposite, we get the inequality of one of the specified four species, equivalent to this.
Then a graph of the function 
There will be a parabola, whose branch is directed up. The location of this parabola relative to the abscissa axis depends on the sign of the discriminant square three declections. 3 cases are possible.
Fig. 1 Fig. 2 Fig. 3.
Case 1. If, the square three decrease has two valid roots and , and .
Then Parabola cross the abscissa axis at points with abscissions and. For strict inequalities 
and 
numbers and are depicted in unwitted circles (as in Fig. 1). For non-strict inequality 
and numbers and are depicted with painted circles. In this case: and does not have valid roots. Then Parabola is not common points with the abscissa axis (see Fig. 3). In this case,: x break the axis of the abscissa by 3 intervals (see Fig. 1). and
Linear second-order differential equations
The second order differential equation is viewed.
Definition. The general solution of the second order equation is such a function, which for any values \u200b\u200band is the solution of this equation.
Definition. The linear homogeneous equation of the second order is called the equation. If coefficients and constant, i.e. It does not depend on, this equation is called the equation with constant coefficients and write it like this :.
The equation will be called a linear inhomogeneous equation.
Definition.The equation that is obtained from a linear homogeneous equation by replacing the function by one, and the corresponding degrees, is called the characteristic equation.
It is known that the square equation has a solution depending on the discriminant:, i.e. If, then roots and - valid different numbers. If, then. If, i.e. , It will be an imaginary number, and roots and - complex numbers. In this case, we agree to signify.
Example 4.Solve equation.
Decision. Discriminant of this square equation, therefore.
We show, as in appearance of the roots of the characteristic equation to find the general solution of the homogeneous linear equation of the second order.
If - the valid roots of the characteristic equation, then.
If the roots of the characteristic equation are the same, i.e. , the general solution of the differential equation is searched by the formula or.
If the characteristic equation has integrated roots, then.
Example 5. Find a general solution to the equation.
Decision.We comprise a characteristic equation for this differential equation :. His roots are valid and different. Therefore, a general solution.
Fundamental system of solutions of a linear homogeneous differential equation. The theorem on the structure of the overall solution of solutions of a linear homogeneous differential equation. In this section, we will prove that the basis of the linear space of private solutions of a homogeneous equation can be any set of n.
His linear independent solutions.
ORD. 14.5.5.1. Fundamental system solutions. Fundamental system solutions linear homogeneous differential equation n.
-o order called any linearly independent system y.
1 (x.
), y.
2 (x.
), …, y N.
(x.
) his n.
Private solutions.
Theorem 14.5.5.1.1 On the structure of the general solution of a linear homogeneous differential equation. Common decision y.
(x.
) A linear homogeneous differential equation is a linear combination of functions from the fundamental system of solutions of this equation:
y.
(x.
) = C.
1 y.
1 (x.
) + C.
2 Y.
2 (x.
) + …+ C n y n
(x.
).
Dock. Let be y.
1 (x.
), y.
2 (x.
), …, y N.
(x.
) - a fundamental system of solutions of a linear homogeneous differential equation. It is required to prove that any particular decision y.
Cho ( x.
) This equation is contained in the formula y.
(x.
) = C.
1 y.
1 (x.
) + C.
2 Y.
2 (x.
) + …+ C n y n
(x.
) with some set of permanent C.
1 , C.
2 , …, C N.
. Take any point, calculate the number at this point and find constant C.
1 , C.
2 , …, C N.
as a solution of a linear inhomogeneous system of algebraic equations
Such a solution exists and the only one, since the determinant of this system is equal. Consider a linear combination y.
(x.
) = C.
1 y.
1 (x.
) + C.
2 Y.
2 (x.
) + …+ C n y n
(x.
) functions from the fundamental solution system with these values \u200b\u200bof constant C.
1 , C.
2 , …, C N.
and compare it with the function y.
Cho ( x.
). Functions y.
(x.
) I. y.
Cho ( x.
) satisfy one equation and the same initial conditions at the point x.
0, therefore, by the uniqueness of the solution of the Cauchy problem, they coincide: y.
Cho ( x.
) = C.
1 y.
1 (x.
) + C.
2 Y.
2 (x.
) + … + C n y n
(x.
). Theorem is proved.
From this theorem it follows that the dimension of the linear space of private solutions of a homogeneous equation with continuous coefficients does not exceed n.
. It remains to prove that this dimension is no less n.
.
Theorem 14.5.5.1.2 on the existence of a fundamental system of solutions of linear homogeneous differential equation. Any linear homogeneous differential equation n.
-o order with continuous coefficients has a fundamental solutions system, i.e. system is n.
Linely independent solutions.
Dock. Take any numeric determinant n.
-o order not equal zero
Let a square table of four numbers A 1, A 2, B 1, B 2:
The number A 1 B 2 - A 2 B 1 is called the second order determinant, corresponding to the table (1). This definite is indicated by the symbol, respectively, we have:
Numbers A 1, A 2, B 1, B 2 are called the elements of the determinant. It is said that elements A 1, B 2 lie on the main diagonal of the determinant, and 2, b 1 - on the side. Thus, the second order determinant is equal to the difference between the works of elements lying on the main and side diagonals. For example,
Consider the system of two equations
with two unknown x, y. (Coefficients A 1, B 1, A 2, B 2 and free members of HXI H2 Suppose data.) We introduce notation
The determinant δ, compiled from coefficients at unknown system (3), is called the determinant of this system. The determinant Δ X is obtained by replacing the elements of the first column of the determinant δ by free members of the system (3); The determinant Δ Y is obtained from the determinant Δ by replacing the elements of its second column with free members of the system (3).
If δ ≠ 0, then the system (3) has a single solution; It is determined by formulas
x \u003d δ x / δ, y \u003d δ y / δ (5)
If δ \u003d 0 and at the same time, at least one of the determinants Δ x, Δ y is different from zero, then the system (3) does not have solutions at all (as they say, the equations of this system are incompatible).
If Δ \u003d 0, but also Δ x \u003d δ y \u003d 0, then the system (3) has infinitely many solutions (in this case, one of the equations of the system is a consequence of another).
Let in the equations of the system (3) H 1 \u003d H 2 \u003d 0; Then the system (3) will look at:
a 1 x + b 1 y \u003d 0, a 2 x + b 2 y \u003d 0. (6)
The system of equations of the form (6) is called homogeneous; It always has a zero solution: x \u003d 0, y \u003d 0. If Δ ≠ o, then this solution is the only one if δ \u003d 0, then the system (6), besides zero, has infinitely many other solutions.
1204. Calculate the determinants:
1205. Solve equations:
1206. Solve inequalities:
1207. Find all solutions of each of the following systems of equations:
1208. To determine at what values \u200b\u200ba and b system of Equations of the SK - AU \u003d 1, 6X + 4U \u003d B 1) has a single solution; 2) has no solutions; 3) has infinitely many solutions.
1209. Determine with what value A system of homogeneous equations 13x + 2ow \u003d 0, 5x + AU \u003d 0 has a nonzero solution.
Here we apply the method of variation of permanent Lagrange to solve linear inhomogeneous second-order differential equations. A detailed description of this method to solve the random order equations is set out on the page.
Solution of linear inhomogeneous differential equations of higher orders of Lagrange method \u003e\u003e\u003e.
Example 1.
Solve the second-order differential equation with constant coefficients by variation of permanent Lagrange:
(1)
Decision
Initially, we solve a homogeneous differential equation:
(2)
This is the second order equation.
We solve the square equation:
.
Roots multiples :. The fundamental system of solutions of equation (2) has the form:
(3)
.
From here we get a general solution of a homogeneous equation (2):
(4)
.
Various constant C. 1
and C. 2
. That is, we will replace in (4) permanent and functions:
.
We are looking for the solution of the initial equation (1) in the form:
(5)
.
Find a derivative:
.
We connect functions and equation:
(6)
.
Then
.
We find the second derivative:
.
We substitute in the initial equation (1):
(1)
;
.
Since and satisfy a homogeneous equation (2), the sum of members in each column of the last three lines gives zero and the previous equation acquires the form:
(7)
.
Here .
Together with equation (6), we obtain a system of equations for determining functions and:
(6)
:
(7)
.
Solving system of equations
We solve the system of equations (6-7). We write out expressions for functions and:
.
We find their derivatives:
;
.
We solve the system of equations (6-7) by the Cramer method. Calculate the system matrix determinant:
.
By crawler formulas, we find:
;
.
So, we found derived functions:
;
.
We integrate (see root integration methods). Making substitution
;
;
;
.
.
.
;
.
Answer
Example 2.
Solve the differential equation by variation of permanent Lagrange:
(8)
Decision
Step 1. Solution of a homogeneous equation
We solve a homogeneous differential equation:
(9)
We are looking for a decision in the form. We compile a characteristic equation:
This equation has integrated roots:
.
The fundamental system of solutions corresponding to these roots has the form:
(10)
.
General solution of a homogeneous equation (9):
(11)
.
Step 2. Variation of Permanent - Replacing Permanent Functions
Now vary constant C 1
and C. 2
. That is, replace in (11) permanent functions:
.
We are looking for the solution of the initial equation (8) as:
(12)
.
Further, the decision of the solution is the same as in Example 1. We arrive at the following system of equations for determining functions and:
(13)
:
(14)
.
Here .
Solving system of equations
We solve this system. We repel the expressions of functions and:
.
From the table of derivatives we find:
;
.
We solve the system of equations (13-14) by the Cramer method. System matrix determinant:
.
By crawler formulas, we find:
;
.
.
Because, the module sign under the logarithm sign can be omitted. Multiply the numerator and denominator on:
.
Then
.
General solution of the original equation:
.
