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Equations and systems of equations of the first degree

Two numbers or some expressions connected by the sign "=" form equality. If the given numbers or expressions are equal for any values ​​of the letters, then such equality is called identity.

For example, when it is stated that for any a valid:

a + 1 = 1 + a, here equality is an identity.

Equation is called an equality containing unknown numbers marked with letters. These letters are called unknown. There can be more than one unknown in an equation.

For example, in equation 2 X + at = 7X– 3 two unknowns: X and at.

The expression on the left side of the equation (2 X + at) is called the left side of the equation, and the expression on the right side of the equation (7 X– 3) is called its right side.

The value of the unknown at which the equation becomes an identity is called decision or root equations.

For example, if in equation 3 X+ 7=13 instead of unknown X substitute the number 2, we get the identity. Therefore, the value X= 2 satisfies the given equation and the number 2 is the solution or root of the given equation.

The two equations are called equivalent(or equivalent), if all solutions of the first equation are solutions of the second and vice versa, all solutions of the second equation are solutions of the first. To equivalent equations also include equations that have no solutions.

For example Equations 2 X– 5 = 11 and 7 X+ 6 = 62 are equivalent since they have the same root X= 8; equations X + 2 = X+ 5 and 2 X + 7 = 2X are equivalent because both have no solutions.

Properties of equivalent equations

1. To both sides of the equation, you can add any expression that makes sense for all allowed values unknown; the resulting equation will be equivalent to the given one.

Example. Equation 2 X– 1 = 7 has a root X= 4. Adding 5 to both sides, we get equation 2 X– 1 + 5 = 7 + 5 or 2 X+ 4 = 12 which has the same root X = 4.

2. If both parts of the equation have the same terms, then they can be omitted.

Example. Equation 9 x + 5X = 18 + 5X has one root X= 2. Omitting in both parts 5 X, we get equation 9 X= 18 which has the same root X = 2.

3. Any term of the equation can be transferred from one part of the equation to another by changing its sign to the opposite.

Example. Equation 7 X - 11 = 3 has one root X= 2. If we transfer 11 to the right side with opposite sign, we get equation 7 X= 3 + 11 which has the same solution X = 2.

4. Both parts of the equation can be multiplied by any expression (number) that makes sense and is non-zero for all admissible values ​​of the unknown, the resulting equation will be equivalent to this one.

Example. Equation 2 X - 15 = 10 – 3X has a root X= 5. Multiplying both sides by 3, we get the equation 3(2 X - 15) = 3(10 – 3X) or 6 X – 45 =30 – 9X, which has the same root X = 5.

5. The signs of all terms of the equation can be reversed (this is equivalent to multiplying both parts by (-1)).

Example. Equation - 3 x + 7 = - 8 after multiplying both parts by (-1) will take the form 3 X - 7 = 8. The first and second equations have a single root X = 5.

6. Both sides of the equation can be divided by the same number other than zero (that is, not equal to zero).

Example..gif" width="49 height=25" height="25">.gif" width="131" height="28"> is equivalent to this one because it has the same two roots: and https:/ /pandia.ru/text/78/105/images/image006_96.gif" width="125" height="48 src="> after multiplying both parts by 14, it will look like this:

https://pandia.ru/text/78/105/images/image009_71.gif" width="77 height=20" height="20">, where arbitrary numbers, X- unknown, called first degree equation with one unknown(or linear equation with one unknown).

Example. 2 X + 3 = 7 – 0,5X ; 0,3X = 0.

A first degree equation with one unknown always has one solution; a linear equation may not have solutions () or have them infinite set(https://pandia.ru/text/78/105/images/image013_59.gif" width="344 height=48" height="48">.

Decision. Multiply all terms in the equation by the least common multiple of the denominators, which is 12.

https://pandia.ru/text/78/105/images/image015_49.gif" width="183 height=24" height="24">.gif" width="371" height="20 src="> .

We group in one part (left) the terms containing the unknown, and in the other part (right) - the free terms:

https://pandia.ru/text/78/105/images/image019_34.gif" width="104" height="20">. Dividing both parts by (-22), we get X = 7.

Systems of two equations of the first degree with two unknowns

An equation like https://pandia.ru/text/78/105/images/image021_34.gif" width="87" height="24 src="> is called first degree equation with two unknowns x and at. If they find common solutions to two or more equations, then they say that these equations form a system, they are usually written one under the other and combined with a curly bracket, for example.

Each pair of unknowns that simultaneously satisfies both equations of the system is called system solution. Solve the system- this means finding all solutions of this system or showing that it does not have them. The two systems of equations are called equivalent (equivalent), if all solutions of one of them are solutions of the other, and vice versa, all solutions of the other are solutions of the first.

For example, the solution to the system is a pair of numbers X= 4 and at= 3. These numbers are also the only solution systems . Therefore, these systems of equations are equivalent.

Ways to solve systems of equations

1. Way algebraic addition. If the coefficients for some unknown in both equations are equal in absolute value, then by adding both equations (or subtracting one from the other), you can get an equation with one unknown. By solving this equation, one unknown is determined, and by substituting it into one of the equations of the system, the second unknown is found.

Examples: Solve systems of equations: 1) .

Here the coefficients at at are equal in absolute value but opposite in sign. To get an equation with one unknown equation we add the systems term by term:

Received value X= 4 we substitute into some equation of the system, for example, into the first one, and find the value at: .

Answer: X = 4; at = 3.

2) https://pandia.ru/text/78/105/images/image029_23.gif" width="112" height="57 src=">.gif" width="220" height="87 src=" >

https://pandia.ru/text/78/105/images/image033_21.gif" width="103" height="47 src=">.

2. Substitution method. From any equation of the system, we express one of the unknowns in terms of the rest, and then we substitute the value of this unknown into the remaining equations. Consider this method with specific examples:

1) Let's solve the system of equations. Let us express one of the unknowns from the first equation, for example X: https://pandia.ru/text/78/105/images/image036_18.gif" width="483" height="24 src=">

Substitute at= 1 into the expression for X, we get .

Answer: https://pandia.ru/text/78/105/images/image039_18.gif" width="99" height="55 src=">. In this case, it is convenient to express at from the second equation:

https://pandia.ru/text/78/105/images/image041_16.gif" width="660" height="24">Substitute the value X= 5 into the expression for at, we get https://pandia.ru/text/78/105/images/image043_15.gif" width="96" height="24 src=">.

3) Let's solve the system of equations https://pandia.ru/text/78/105/images/image045_12.gif" width="205" height="48">. Substituting this value into the second equation, we get an equation with one unknown at: https://pandia.ru/text/78/105/images/image049_11.gif" width="128" height="48">

Answer: https://pandia.ru/text/78/105/images/image051_12.gif" width="95" height="108 src="> .

Let's rewrite the system as: . We replace the unknowns by setting , we get linear system ..gif" width="11 height=17" height="17"> into the second equation, we get an equation with one unknown:

Substituting the value v into the expression for t, we get: https://pandia.ru/text/78/105/images/image060_9.gif" width="92 height=51" height="51"> we find .

Answer: https://pandia.ru/text/78/105/images/image062_9.gif" width="120" height="57">, where are coefficients for unknowns, https://pandia.ru/text/ 78/105/images/image065_10.gif" width="67" height="52 src=">, then the system has the only thing decision.

B) If https://pandia.ru/text/78/105/images/image067_9.gif" width="105" height="52 src=">, then the system has infinite set solutions.

Example..gif" width="47" height="48 src=">), so the system has a unique solution.

Really, .

https://pandia.ru/text/78/105/images/image073_7.gif" width="115" height="48 src=">.

Example..gif" width="91 height=48" height="48"> or after reduction , hence the system has no solutions.

Example..gif" width="116 height=48" height="48"> or after shortening , so the system has an infinite number of solutions.

Equations Containing Modulus

When solving equations containing a module, the concept of a module is used real number. module (absolute value ) real number a the number itself is called if and opposite number (– a), if https://pandia.ru/text/78/105/images/image082_7.gif" width="20" height="28">.

So, https://pandia.ru/text/78/105/images/image084_8.gif" width="44" height="28 src=">, since the number 3 > 0; , since the number is 5< 0, поэтому ; , as (); , as .

Module properties:

1) https://pandia.ru/text/78/105/images/image093_7.gif" width="72" height="28 src=">

3) https://pandia.ru/text/78/105/images/image095_8.gif" width="123" height="56 src=">

5) https://pandia.ru/text/78/105/images/image097_7.gif" width="73" height="28 src=">.

Given that the expression under the module can take two values ​​https://pandia.ru/text/78/105/images/image099_8.gif" width="68" height="20 src=">, then given equation reduces to solving two equations: and or and ..gif" width="52" height="20 src=">. Let's do a check by substituting each value X into the condition: if https://pandia.ru/text/78/105/images/image106_5.gif" width="165" height="28 src=">..gif" width="144" height="28 src=">.

Answer: https://pandia.ru/text/78/105/images/image104_6.gif" width="49" height="20 src=">.

Example..gif" width="408" height="55">

Answer: https://pandia.ru/text/78/105/images/image111_6.gif" width="41" height="20 src=">.

Example..gif" width="137" height="20"> and . Set aside the resulting values X on the numerical axis, breaking it into intervals:

If https://pandia.ru/text/78/105/images/image117_5.gif" width="144" height="24">, because in this interval, both expressions are under the module sign less than zero, and, removing the module, we must change the sign of the expression to the opposite. Let's solve the resulting equation:

Gif" width="75 height=24" height="24">. The boundary value can be included in both the first and second span, just as the value can be included in both the second and third. In the second interval, our equation will take the form: - this expression does not make sense, i.e., on this interval, the equation of solutions does not have solutions under the modulus sign, we equate them to zero. We find the roots of all expressions, with

Next spacing https://pandia.ru/text/78/105/images/image124_6.gif" width="225" height="20">..gif" width="52" height="20 src="> .gif" width="125" height="25">, where a, b, c are arbitrary numbers ( a≠ 0), and x is a variable called square. To solve this equation, you need to calculate the discriminant D = b 2 – 4ac. If a D> 0, then the quadratic equation has two solutions (roots): and .

If a D= 0, the quadratic equation obviously has two identical solutions(multiples of the root).

If a D< 0, квадратное уравнение не имеет real roots.

If one of the coefficients b or c zero, then the quadratic equation can be solved without calculating the discriminant:

1) https://pandia.ru/text/78/105/images/image131_5.gif" width="28" height="18 src="> x(ax+ b)=0

2)ax 2 + c = 0 ax 2 = – c; if https://pandia.ru/text/78/105/images/image135_3.gif" width="101" height="52">.

There are dependencies between the coefficients and roots of the quadratic equation, known as formulas or Vieta's theorem:

Bisquare equations are equations of the form https://pandia.ru/text/78/105/images/image138_4.gif" width="53" height="29">, then from the original equation we get a quadratic equation, from which we find at, and then X, according to the formula .

Example. solve the equation . We bring the expressions in both parts of the equality to common denominator..gif" width="212" height="29 src=">. We solve the resulting quadratic equation: , in this equation a= 1, b= –2,c= -15, then the discriminant is equal to: D = b 2 – 4ac= 64. Equation roots: , ..gif" width="130 height=25" height="25">. We make the replacement. Then the equation becomes is a quadratic equation, where a= 1, b= – 4,c= 3, its discriminant is: D = b 2 – 4ac = 16 – 12 = 4.

The roots of the quadratic equation are equal, respectively: and .

Roots of the original equation , , , ..gif" width="78" height="51">, where PN(x) and Pm(x) are polynomials of degrees n and m respectively. A fraction is zero if the numerator is zero and the denominator is not, but such a polynomial equation is mainly obtained only after lengthy transformations, transitions from one equation to another. In the process of solving, therefore, each equation is replaced by some new one, and the new one may have new roots. To follow these changes in the roots, to prevent the loss of roots and to be able to reject the extra ones is the task right decision equations.

It is clear that the best way- each time replace one equation with an equivalent one, then the roots of the last equation will be the roots of the original one. However, such perfect path difficult to implement in practice. As a rule, the equation is replaced by its consequence, which is not necessarily equivalent to it at all, while all the roots of the first equation are the roots of the second, i.e., the loss of roots does not occur, but extraneous ones may appear (or may not appear). In the case when at least once in the process of transformations the equation was replaced by an unequal one, we need mandatory check obtained roots.

So, if the solution was carried out without an analysis of the equivalence and sources of extraneous roots, the check is obligatory part solutions. Without verification, the solution will not be considered complete, even if extraneous roots did not appear. When they appeared and were not discarded, then this decision is simply wrong.

Here are some properties of a polynomial:

The root of the polynomial call the value x, for which the polynomial is equal to zero. Any polynomial of degree n has exactly n roots. If the polynomial equation is written as , then , where x 1, x 2,…, xn are the roots of the equation.

Any polynomial has even degree with real coefficients there is at least one real root, and in general it always has an odd number of real roots. A polynomial of even degree may not have real roots, and when they do, their number is even.

A polynomial under any circumstances can be decomposed into linear factors and square trinomials with negative discriminant. If we know its root x 1, then PN(x) = (x -x 1) Pn- 1(x).

If a PN(x) = 0 is an equation of even degree, then in addition to the method of factoring it into factors, you can try to introduce a change of variable, with the help of which the degree of the equation will decrease.

Example. Solve the equation:

This equation of the third (odd) degree means that it is impossible to introduce an auxiliary variable that will lower the degree of the equation. It must be solved by factoring the left side, for which we first open the brackets, and then write it in standard form.

We get: x 3 + 5x – 6 = 0.

This is the reduced equation (coefficient at the highest degree equal to one), so we are looking for its roots among the factors of the free term - 6. These are the numbers ±1, ±2, ±3, ±6. Substituting x= 1 into the equation, we see that x= 1 is its root, so the polynomial x 3 + 5x–6 = 0 divided by ( x- 1) no residue. Let's do this division:

x 3 + 5x –6 = 0 x- 1

x 3 – x 2 x 2+x + 6

x 2 + 5x- 6

x 2– x

https://pandia.ru/text/78/105/images/image167_4.gif"> 6 x- 6

https://pandia.ru/text/78/105/images/image168_4.gif" width="50"> 6 x- 6

So x 3 + 5x –6 = 0; (x- 1)(x 2+ x + 6) = 0

The first equation gives the root x= 1, which is already selected, and in the second equation D< 0, it does not have real solutions. Since the ODZ of this equation , it is possible not to check.

Example..gif" width="52" height="21 src=">. If you multiply the first factor with the third, and the second with the fourth, then these products will have the same parts, which depend on x: (x 2 + 4x – 5)(x 2 + 4x – = 0.

Let be x 2 + 4x = y, then we write the equation in the form ( y – 5)(y- 21) – 297 = 0.

This quadratic equation has solutions: y 1 = 32, y 2 = - 6 ..gif" width="140" height="61 src=">; ODZ: x ≠ – 9.

If we reduce this equation to a common denominator, a polynomial of the fourth degree will appear in the numerator. So, it is allowed to change the variable, which will lower the degree of the equation. Therefore, it is not necessary to immediately reduce this equation to a common denominator. Here you can see that on the left is the sum of the squares. So, you can add it to full square sums or differences. In fact, subtract and add twice the product of the bases of these squares: https://pandia.ru/text/78/105/images/image179_3.gif" width="80" height="59 src=">, then y 2 + 18y– 40 = 0. According to the Vieta theorem y 1 = 2; y 2 = – 20. https://pandia.ru/text/78/105/images/image183_4.gif" width="108 height=32" height="32">, and in the second D< 0. Эти корни удовлетворяют ОДЗ

Answer: https://pandia.ru/text/78/105/images/image185_4.gif" width="191 height=51" height="51">.gif" width="73 height=48" height=" 48"> .gif" width="132" height="50 src=">.

We get a quadratic equation a(y 2 https://pandia.ru/text/78/105/images/image192_3.gif" width="213" height="31">.

Irrational equations

irrational called an equation in which the variable is contained under the sign of the radical (root ) or under the sign of elevation to fractional degree()..gif" width="120" height="32"> and have the same domain of definition of the unknown. When squaring the first and second equations, we get the same equation . The solutions of this equation are the solutions of both irrational equations.

1. Substitution method: from any equation of the system we express one unknown in terms of another and substitute it into the second equation of the system.

Task. Solve the system of equations:

Decision. From the first equation of the system, we express at through X and substitute into the second equation of the system. Let's get the system equivalent to the original.

After bringing such terms, the system will take the form:

From the second equation we find: . Substituting this value into the equation at = 2 - 2X, we get at= 3. Therefore, the solution of this system is a pair of numbers .

2. Algebraic addition method: by adding two equations, get an equation with one variable.

Task. Solve the system equation:

Decision. Multiplying both sides of the second equation by 2, we get the system equivalent to the original. Adding the two equations of this system, we arrive at the system

After reducing similar terms, this system will take the form: From the second equation we find . Substituting this value into Equation 3 X + 4at= 5, we get , where . Therefore, the solution of this system is a pair of numbers .

3. Method for introducing new variables: we are looking for some repeated expressions in the system, which we will denote by new variables, thereby simplifying the form of the system.

Task. Solve the system of equations:

Decision. Let's write down this system otherwise:

Let be x + y = u, hu = v. Then we get the system

Let's solve it by the substitution method. From the first equation of the system, we express u through v and substitute into the second equation of the system. Let's get the system those.

From the second equation of the system we find v 1 = 2, v 2 = 3.

Substituting these values ​​into the equation u = 5 - v, we get u 1 = 3,
u 2 = 2. Then we have two systems

Solving the first system, we get two pairs of numbers (1; 2), (2; 1). The second system has no solutions.

Exercises for independent work

1. Solve systems of equations using the substitution method.

Systems of equations received wide application in the economic sector mathematical modeling various processes. For example, when solving problems of production management and planning, logistics routes (transport problem) or equipment placement.

Equation systems are used not only in the field of mathematics, but also in physics, chemistry and biology, when solving problems of finding the population size.

system linear equations name two or more equations with several variables for which it is necessary to find common decision. Such a sequence of numbers for which all equations become true equalities or prove that the sequence does not exist.

Linear Equation

Equations of the form ax+by=c are called linear. The designations x, y are the unknowns, the value of which must be found, b, a are the coefficients of the variables, c is the free term of the equation.
Solving the equation by plotting its graph will look like a straight line, all points of which are the solution of the polynomial.

Types of systems of linear equations

The simplest are examples of systems of linear equations with two variables X and Y.

F1(x, y) = 0 and F2(x, y) = 0, where F1,2 are functions and (x, y) are function variables.

Solve a system of equations - it means to find such values ​​(x, y) for which the system becomes a true equality, or to establish that there are no suitable values ​​of x and y.

A pair of values ​​(x, y), written as point coordinates, is called a solution to a system of linear equations.

If the systems have one common solution or there is no solution, they are called equivalent.

Homogeneous systems of linear equations are systems right part which is equal to zero. If the right part after the "equal" sign has a value or is expressed by a function, such a system is not homogeneous.

The number of variables can be much more than two, then we should talk about an example of a system of linear equations with three variables or more.

Faced with systems, schoolchildren assume that the number of equations must necessarily coincide with the number of unknowns, but this is not so. The number of equations in the system does not depend on the variables, there can be an arbitrarily large number of them.

Simple and complex methods for solving systems of equations

There is no common analytical method solutions of similar systems, all methods are based on numerical solutions. The school mathematics course describes in detail such methods as permutation, algebraic addition, substitution, as well as the graphical and matrix method, the solution by the Gauss method.

The main task in teaching methods of solving is to teach how to correctly analyze the system and find optimal algorithm solutions for each example. The main thing is not to memorize a system of rules and actions for each method, but to understand the principles of applying a particular method.

Solving examples of systems of linear equations of the 7th class of the program secondary school quite simple and explained in great detail. In any textbook on mathematics, this section is given enough attention. The solution of examples of systems of linear equations by the method of Gauss and Cramer is studied in more detail in the first courses of higher educational institutions.

Solution of systems by the substitution method

The actions of the substitution method are aimed at expressing the value of one variable through the second. The expression is substituted into the remaining equation, then it is reduced to a single variable form. The action is repeated depending on the number of unknowns in the system

Let's give an example of a system of linear equations of the 7th class by the substitution method:

As can be seen from the example, the variable x was expressed through F(X) = 7 + Y. The resulting expression, substituted into the 2nd equation of the system in place of X, helped to obtain one variable Y in the 2nd equation. Decision this example does not cause difficulties and allows you to get the Y value. The last step is to check the received values.

It is not always possible to solve an example of a system of linear equations by substitution. The equations can be complex and the expression of the variable in terms of the second unknown will be too cumbersome for further calculations. When there are more than 3 unknowns in the system, the substitution solution is also impractical.

Solution of an example of a system of linear inhomogeneous equations:

Solution using algebraic addition

When searching for a solution to systems by the addition method, term-by-term addition and multiplication of equations by various numbers. ultimate goal mathematical operations is an equation with one variable.

For applications this method it takes practice and observation. It is not easy to solve a system of linear equations using the addition method with the number of variables 3 or more. Algebraic addition is useful when the equations contain fractions and decimal numbers.

Solution action algorithm:

1. Multiply both sides of the equation by some number. As a result arithmetic operation one of the coefficients of the variable must become equal to 1.
2. Add the resulting expression term by term and find one of the unknowns.
3. Substitute the resulting value into the 2nd equation of the system to find the remaining variable.

Solution method by introducing a new variable

A new variable can be introduced if the system needs to find a solution for no more than two equations, the number of unknowns should also be no more than two.

The method is used to simplify one of the equations by introducing a new variable. The new equation is solved with respect to the entered unknown, and the resulting value is used to determine the original variable.

The example shows that by introducing a new variable t, it was possible to reduce the 1st equation of the system to the standard square trinomial. You can solve a polynomial by finding the discriminant.

It is necessary to find the value of the discriminant by well-known formula: D = b2 - 4*a*c, where D is the desired discriminant, b, a, c are the multipliers of the polynomial. AT given example a=1, b=16, c=39, hence D=100. If the discriminant is greater than zero, then there are two solutions: t = -b±√D / 2*a, if the discriminant is less than zero, then there is only one solution: x= -b / 2*a.

The solution for the resulting systems is found by the addition method.

A visual method for solving systems

Suitable for systems with 3 equations. The method is to build on coordinate axis graphs of each equation included in the system. The coordinates of the points of intersection of the curves will be the general solution of the system.

The graphic method has a number of nuances. Consider several examples of solving systems of linear equations in a visual way.

As can be seen from the example, two points were constructed for each line, the values ​​of the variable x were chosen arbitrarily: 0 and 3. Based on the values ​​of x, the values ​​for y were found: 3 and 0. Points with coordinates (0, 3) and (3, 0) were marked on the graph and connected by a line.

The steps must be repeated for the second equation. The point of intersection of the lines is the solution of the system.

AT next example required to find graphic solution systems of linear equations: 0.5x-y+2=0 and 0.5x-y-1=0.

As can be seen from the example, the system has no solution, because the graphs are parallel and do not intersect along their entire length.

The systems from Examples 2 and 3 are similar, but when constructed, it becomes obvious that their solutions are different. It should be remembered that it is not always possible to say whether the system has a solution or not, it is always necessary to build a graph.

Matrix and its varieties

Matrices are used for abbreviation systems of linear equations. A table is called a matrix. special kind filled with numbers. n*m has n - rows and m - columns.

A matrix is ​​square when the number of columns and rows is equal. A matrix-vector is a single-column matrix with an infinitely possible number of rows. Matrix with units along one of the diagonals and others zero elements called singular.

An inverse matrix is ​​such a matrix, when multiplied by which the original one turns into a unit one, such a matrix exists only for the original square one.

Rules for transforming a system of equations into a matrix

With regard to systems of equations, the coefficients and free members of the equations are written as numbers of the matrix, one equation is one row of the matrix.

A matrix row is called non-zero if at least one element of the row is not equal to zero. Therefore, if in any of the equations the number of variables differs, then it is necessary to enter zero in place of the missing unknown.

The columns of the matrix must strictly correspond to the variables. This means that the coefficients of the variable x can only be written in one column, for example the first, the coefficient of the unknown y - only in the second.

When multiplying a matrix, all matrix elements are sequentially multiplied by a number.

Options for finding the inverse matrix

The formula for finding the inverse matrix is ​​quite simple: K -1 = 1 / |K|, where K -1 - inverse matrix, and |K| - matrix determinant. |K| must not be equal to zero, then the system has a solution.

The determinant is easily calculated for a two-by-two matrix, it is only necessary to multiply the elements diagonally by each other. For the "three by three" option, there is a formula |K|=a 1 b 2 c 3 + a 1 b 3 c 2 + a 3 b 1 c 2 + a 2 b 3 c 1 + a 2 b 1 c 3 + a 3 b 2 c 1 . You can use the formula, or you can remember that you need to take one element from each row and each column so that the column and row numbers of the elements do not repeat in the product.

Solution of examples of systems of linear equations by the matrix method

The matrix method of finding a solution makes it possible to reduce cumbersome notations when solving systems with large quantity variables and equations.

In the example, a nm are the coefficients of the equations, the matrix is ​​a vector x n are the variables, and b n are the free terms.

Solution of systems by the Gauss method

AT higher mathematics the Gauss method is studied together with the Cramer method, and the process of finding a solution to systems is called the Gauss-Cramer solution method. These methods are used to find system variables with a lot of linear equations.

The Gaussian method is very similar to substitution and algebraic addition solutions, but is more systematic. In the school course, the Gaussian solution is used for systems of 3 and 4 equations. The purpose of the method is to bring the system to the form of an inverted trapezoid. way algebraic transformations and substitutions is the value of one variable in one of the equations of the system. The second equation is an expression with 2 unknowns, and 3 and 4 - with 3 and 4 variables, respectively.

After bringing the system to the described form, the further solution is reduced to the sequential substitution of known variables into the equations of the system.

AT school textbooks for grade 7, an example of a solution by the Gauss method is described as follows:

As can be seen from the example, at step (3) two equations were obtained 3x 3 -2x 4 =11 and 3x 3 +2x 4 =7. The solution of any of the equations will allow you to find out one of the variables x n.

Theorem 5, which is mentioned in the text, states that if one of the equations of the system is replaced by an equivalent one, then the resulting system will also be equivalent to the original one.

The Gauss method is difficult for students to understand high school, but is one of the most interesting ways to develop the ingenuity of children enrolled in the program in-depth study in math and physics classes.

For ease of recording calculations, it is customary to do the following:

Equation coefficients and free terms are written in the form of a matrix, where each row of the matrix corresponds to one of the equations of the system. separates the left side of the equation from the right side. Roman numerals denote the numbers of equations in the system.

First, they write down the matrix with which to work, then all the actions carried out with one of the rows. The resulting matrix is ​​written after the "arrow" sign and continue to perform the necessary algebraic actions until the result is achieved.

As a result, a matrix should be obtained in which one of the diagonals is 1, and all other coefficients are equal to zero, that is, the matrix is ​​reduced to a single form. We must not forget to make calculations with the numbers of both sides of the equation.

This notation is less cumbersome and allows you not to be distracted by listing numerous unknowns.

The free application of any method of solution will require care and a certain amount of experience. Not all methods are applied. Some ways of finding solutions are more preferable in a particular area of ​​human activity, while others exist for the purpose of learning.

I. Ordinary differential equations

1.1. Basic concepts and definitions

A differential equation is an equation that relates an independent variable x, the desired function y and its derivatives or differentials.

Symbolically differential equation is written like this:

F(x,y,y")=0, F(x,y,y")=0, F(x,y,y",y",.., y (n))=0

A differential equation is called ordinary if the desired function depends on one independent variable.

By solving the differential equation is called such a function that turns this equation into an identity.

The order of the differential equation is the order of the highest derivative in this equation

Examples.

1. Consider the first order differential equation

The solution to this equation is the function y = 5 ln x. Indeed, by substituting y" into the equation, we get - an identity.

And this means that the function y = 5 ln x– is the solution of this differential equation.

2. Consider the second order differential equation y" - 5y" + 6y = 0. The function is the solution to this equation.

Really, .

Substituting these expressions into the equation, we get: , - identity.

And this means that the function is the solution of this differential equation.

Integration of differential equations is the process of finding solutions to differential equations.

General solution of the differential equation is called a function of the form , which includes as many independent arbitrary constants as the order of the equation.

Partial solution of the differential equation is called the solution obtained from the general solution for different numerical values ​​of arbitrary constants. The values ​​of arbitrary constants are found at certain initial values ​​of the argument and function.

The graph of a particular solution of a differential equation is called integral curve.

Examples

1. Find a particular solution to a first-order differential equation

xdx + ydy = 0, if y= 4 at x = 3.

Decision. Integrating both sides of the equation, we get

Comment. An arbitrary constant C obtained as a result of integration can be represented in any form convenient for further transformations. In this case, taking into account the canonical equation of the circle, it is convenient to represent an arbitrary constant С in the form .

is the general solution of the differential equation.

A particular solution of an equation that satisfies the initial conditions y = 4 at x = 3 is found from the general by substituting the initial conditions into the general solution: 3 2 + 4 2 = C 2 ; C=5.

Substituting C=5 into the general solution, we get x2+y2 = 5 2 .

This is a particular solution of the differential equation obtained from the general solution under given initial conditions.

2. Find the general solution of the differential equation

The solution of this equation is any function of the form , where C is an arbitrary constant. Indeed, substituting into the equations, we obtain: , .

Therefore, this differential equation has an infinite number of solutions, since for different values ​​of the constant C, the equality determines various solutions equations.

For example, by direct substitution, one can verify that the functions are solutions of the equation .

A problem in which it is required to find a particular solution to the equation y" = f(x, y) satisfying the initial condition y(x0) = y0, is called the Cauchy problem.

Equation solution y" = f(x, y), satisfying the initial condition, y(x0) = y0, is called a solution to the Cauchy problem.

The solution of the Cauchy problem has a simple geometric meaning. Indeed, according to these definitions, to solve the Cauchy problem y" = f(x, y) given that y(x0) = y0, means to find the integral curve of the equation y" = f(x, y) which goes through given point M0 (x0,y 0).

II. First order differential equations

2.1. Basic concepts

A first-order differential equation is an equation of the form F(x,y,y") = 0.

The first order differential equation includes the first derivative and does not include higher order derivatives.

The equation y" = f(x, y) is called a first-order equation solved with respect to the derivative.

A general solution of a first-order differential equation is a function of the form , which contains one arbitrary constant.

Example. Consider a first order differential equation.

The solution to this equation is the function .

Indeed, replacing in this equation with its value, we obtain

i.e 3x=3x

Therefore, the function is a general solution of the equation for any constant C.

Find a particular solution of this equation that satisfies the initial condition y(1)=1 Substituting initial conditions x=1, y=1 into the general solution of the equation , we obtain whence C=0.

Thus, we obtain a particular solution from the general one by substituting into this equation, the resulting value C=0 is a private decision.

2.2. Differential equations with separable variables

A differential equation with separable variables is an equation of the form: y"=f(x)g(y) or through differentials , where f(x) and g(y) are given functions.

For those y, for which , the equation y"=f(x)g(y) is equivalent to the equation in which the variable y is present only on the left side, and the variable x is present only on the right side. They say, "in the equation y"=f(x)g(y separating the variables.

Type equation is called a separated variable equation.

After integrating both parts of the equation on x, we get G(y) = F(x) + C is the general solution of the equation, where G(y) and F(x) are some antiderivatives, respectively, of functions and f(x), C arbitrary constant.

Algorithm for solving a first-order differential equation with separable variables

Example 1

solve the equation y" = xy

Decision. Derivative of a function y" replace with

we separate the variables

Let's integrate both parts of the equality:

Example 2

2yy" = 1- 3x 2, if y 0 = 3 at x0 = 1

This is a separated variable equation. Let's represent it in differentials. To do this, we rewrite this equation in the form From here

Integrating both parts of the last equality, we find

Substituting initial values x 0 = 1, y 0 = 3 find With 9=1-1+C, i.e. C = 9.

Therefore, the desired partial integral will be or

Example 3

Write an equation for a curve passing through a point M(2;-3) and having a tangent with a slope

Decision. According to the condition

This is a separable variable equation. Dividing the variables, we get:

Integrating both parts of the equation, we get:

Using the initial conditions, x=2 and y=-3 find C:

Therefore, the desired equation has the form

2.3. Linear differential equations of the first order

A first-order linear differential equation is an equation of the form y" = f(x)y + g(x)

where f(x) and g(x)- some given functions.

If a g(x)=0 then the linear differential equation is called homogeneous and has the form: y" = f(x)y

If then the equation y" = f(x)y + g(x) called heterogeneous.

General solution of a linear homogeneous differential equation y" = f(x)y given by the formula: where With is an arbitrary constant.

In particular, if C \u003d 0, then the solution is y=0 If linear homogeneous equation has the form y" = ky where k is some constant, then its general solution has the form: .

General solution of a linear inhomogeneous differential equation y" = f(x)y + g(x) given by the formula ,

those. is equal to the sum of the general solution of the corresponding linear homogeneous equation and the particular solution of this equation.

For a linear inhomogeneous equation of the form y" = kx + b,

where k and b- some numbers and a particular solution will be a constant function . Therefore, the general solution has the form .

Example. solve the equation y" + 2y +3 = 0

Decision. We represent the equation in the form y" = -2y - 3 where k=-2, b=-3 The general solution is given by the formula .

Therefore, where C is an arbitrary constant.

2.4. Solution of linear differential equations of the first order by the Bernoulli method

Finding a General Solution to a First-Order Linear Differential Equation y" = f(x)y + g(x) reduces to solving two differential equations with separated variables using the substitution y=uv, where u and v- unknown functions from x. This solution method is called the Bernoulli method.

Algorithm for solving a first-order linear differential equation

y" = f(x)y + g(x)

1. Enter a substitution y=uv.

2. Differentiate this equality y"=u"v + uv"

3. Substitute y and y" into this equation: u"v + uv" =f(x)uv + g(x) or u"v + uv" + f(x)uv = g(x).

4. Group the terms of the equation so that u take it out of brackets:

5. From the bracket, equating it to zero, find the function

This is a separable equation:

Divide the variables and get:

Where . .

6. Substitute the received value v into the equation (from item 4):

and find the function This is a separable equation:

7. Write the general solution in the form: , i.e. .

Example 1

Find a particular solution to the equation y" = -2y +3 = 0 if y=1 at x=0

Decision. Let's solve it with substitution y=uv,.y"=u"v + uv"

Substituting y and y" into this equation, we get

Grouping the second and third terms on the left side of the equation, we take out the common factor u out of brackets

We equate the expression in brackets to zero and, having solved the resulting equation, we find the function v = v(x)

We got an equation with separated variables. We integrate both parts of this equation: Find the function v:

Substitute the resulting value v into the equation We get:

This is a separated variable equation. We integrate both parts of the equation: Let's find the function u = u(x,c) Let's find a general solution: Let us find a particular solution of the equation that satisfies the initial conditions y=1 at x=0:

III. Higher order differential equations

3.1. Basic concepts and definitions

A second-order differential equation is an equation containing derivatives not higher than the second order. In the general case, the second-order differential equation is written as: F(x,y,y",y") = 0

The general solution of a second-order differential equation is a function of the form , which includes two arbitrary constants C1 and C2.

A particular solution of a second-order differential equation is a solution obtained from the general one for some values ​​of arbitrary constants C1 and C2.

3.2. Linear homogeneous differential equations of the second order with constant ratios.

Linear homogeneous differential equation of the second order with constant coefficients is called an equation of the form y" + py" + qy = 0, where p and q are constant values.

Algorithm for solving second-order homogeneous differential equations with constant coefficients

1. Write the differential equation in the form: y" + py" + qy = 0.

2. Compose its characteristic equation, denoting y" through r2, y" through r, y in 1: r2 + pr +q = 0