Fundamental system of solutions to a homogeneous system of differential equations. Linear differential equations with constant coefficients
LDE of the nth order - ur-e, linear with respect to the unknown function and its derivatives and has the form
a 0 (x)y (n) +a 1 (x)y (n-1) +…+a n-1 (x)y'+a n (x)y=φ(x)|: a 0 (x )
φ(x)≠0- LNOU
y (n) +p 1 (x)y (n -1) +…+p n -1 (x)y’+p n (x)y=g(x)- (1) ur-e in the given form
*if y 1 is a solution to the LOU, then C y 1, where C is an arbitrary constant is also a solution to this equation.
*The sum of y 1 + y 2 solutions of the LOE is a solution of the same level.
1 0 Linear combination with arbitrary solution constants y 1 , y 2 ,…, y m LOU is a solution of the same equation.
*if LOU (1) with real coefficients p i (x)∈R has comprehensive solution y(x)=u(x)+iv(x), then the real part of this solution Rey=u(x) and its imaginary part Imy=v(x) are separately solutions of the same equation.
The functions y 1 (x), y 2 (x),…, y n (x) are called linearly dependent on some interval (a,b), if there are constants a1,a2,…,an≠0 such that for all x of the interval (a,b) the identity a 1 y 1 (x)+a 2 y 2 (x)+…+a n -1 (x)y'+ is true a n y n (x)=0. If the functions are linearly dependent, then at least one of them is a linear combination of the others.
If the identity is valid only for a1=a2=…=an=0, then the functions y 1 (x), y 2 (x),…, y n (x) are called linearly independent on the interval (a,b).
*if functions y 1 (x), y 2 (x),…, y n (x) linearly dependent on the interval (a,b), then the determinant (Vronsky Island)
W(x)=W= 
=0 on this interval.
Condition linear independence private solutions:
* if linearly independent functions y 1 (x), y 2 (x),…, y n (x) are solutions of LOE (1) with coefficients p i (x) continuous on the interval (a,b), then compiled for them the Wronski determinant does not = 0 at any point in the interval (a,b).
The general solution of LOU (1) with coefficients p i (x) continuous on (a,b) (i=1,2,...,n) is a linear combination y oo = n linearly independent on the same interval of partial solutions y i with arbitrary constant coefficients.
1 0 the maximum number of linearly independent solutions of the LOU is equal to its order.
FSR- any n independent partial solving LOUs of the nth order.
*y on =y oo +y chn
Structure of the general solution of a linear inhomogeneous differential equation. Method of variation of arbitrary constants for finding a particular solution to a linear inhomogeneous differential equation of the nth order.
LPDEs are solved by the method of varying arbitrary constants. First is common decision 
homogeneous equation 
, having the same left side as the original inhomogeneous equation. Then the solution to the equation is found in the form, i.e. It is assumed that the constants C are the f-mi of the independent variable x. In this case, the functions C 1 (x) and C 2 (x) can be obtained as a solution to the system
U he = u oo + u chn
the maximum number of solutions to an equation is equal to its order.
common decision
44*. Linear homogeneous differential equation with constant coefficients. Characteristic polynomial and characteristic equation. Construction fundamental system solutions in case simple roots characteristic polynomial (real and complex).
Equation of the form y"+p(x)y=f(x), where p(x), f(x) are continuous functions on the interval a If f(x)= 0, then the equation is called homogeneous. If in LO ur-ii y (n) +p 1 (x)y (n -1) +…+p n-1 (x)y’+p n (x)y=0 All coefficients pi are constant, then its partial solutions can be found in the form y=e kx, where k is a constant. Substituting in ur (k n +p 1 k n -1 +….+p n-1 k+ p n) e kx =0 Reducing by e kx we get the so-called Characteristic level k n +p 1 k n -1 +….+p n -1 k+ p n =0 This equation of the nth degree determines those values of k at which y= e kx is a solution to the original differential equation with constant coefficients. 1.k 1 , k 2 ,…,k n – real and different FSR: e k 1 x , e k 2 x ,…, e knx 2. k 1 = k 2 =…=k m =k ~ , k ~ - m - multiple root of ur-i, and all other n- m roots are different FSR: e k ~ x ,x e k ~ x ,…, x m -1 e k ~ x , e km +1 x , e k n x Linear differential equations of the second order The second order differential equation has the form . Definition. A general solution to a second-order equation is a function that, for any value, is a solution to this equation. Definition. A linear homogeneous equation of the second order is called an equation. If the coefficients are constant, i.e. do not depend on , then this equation is called an equation with constant coefficients and is written as follows: . The equation Definition. The equation that is obtained from a linear homogeneous equation by replacing the function by one, and and by the corresponding powers, is called a characteristic equation. It is known that a quadratic equation has a solution depending on the discriminant: Example 4. Solve the equation. Solution. The discriminant of this quadratic equation is therefore . We will show how to find the general solution of a homogeneous second-order linear equation using the form of the roots of the characteristic equation. If are the real roots of the characteristic equation, then If the roots of the characteristic equation are the same, i.e. , then the general solution of the differential equation is sought using the formula If the characteristic equation has complex roots, then. Example 5. Find the general solution of the equation. Solution. Let's create a characteristic equation for this differential equation: . Its roots are valid and different. Therefore the general solution Fundamental system of solutions to a linear homogeneous differential equation. A theorem on the structure of the general solution of solutions to a linear homogeneous differential equation. In this section we will prove that the basis of the linear space of partial solutions of a homogeneous equation can be any set of n
its linearly independent solutions. To understand what it is fundamental decision system you can watch a video tutorial for the same example by clicking. Now let's move on to the actual description of all the necessary work. This will help you understand the essence of this issue in more detail. Let's take for example the following system of linear equations: Let's find the solution to this linear system of equations. To begin with, we you need to write out the coefficient matrix of the system. see also Solving linear differential equations online Let's call a linear differential equation (*) an equation with constant coefficients if the coefficients in this equation are constant, that is, a i (x)=const. Then the corresponding homogeneous equation L(y)=0 will have the form Example. For the equation y""-3y" + 2y=0, the roots of the characteristic equation r 2 - 3r + 2 = 0 are equal to r 1 = 1, r 2 = 2 (the roots were found through the service for finding the discriminant). Consequently, the fundamental system of solutions consists of the functions y 1 = e x, y 2 = e 2 x, and the general solution is written as y = C 1 e x + C 2 e 2 x. We will continue to polish our technology elementary transformations on homogeneous system of linear equations. The answer suggests itself. A system of linear equations is homogeneous if the free term everyone equation of the system is zero. For example: It is absolutely clear that a homogeneous system is always consistent, that is, it always has a solution. And, first of all, what catches your eye is the so-called trivial solution Example 1 Solution: to solve a homogeneous system it is necessary to write system matrix and with the help of elementary transformations bring it to a stepwise form. Please note that here there is no need to write down the vertical bar and the zero column of free terms - after all, no matter what you do with zeros, they will remain zeros: (1) The first line was added to the second line, multiplied by –2. The first line was added to the third line, multiplied by –3. (2) The second line was added to the third line, multiplied by –1. Dividing the third line by 3 doesn't make much sense. As a result of elementary transformations, an equivalent homogeneous system is obtained Answer: Let us formulate an obvious criterion: a homogeneous system of linear equations has just a trivial solution, If system matrix rank(in this case 3) is equal to the number of variables (in this case – 3 pieces). Let's warm up and tune our radio to the wave of elementary transformations: Example 2 Solve a homogeneous system of linear equations To finally consolidate the algorithm, let’s analyze the final task: Example 7 Solve a homogeneous system, write the answer in vector form. Solution: let’s write down the matrix of the system and, using elementary transformations, bring it to a stepwise form: (1) The sign of the first line has been changed. Once again I draw attention to a technique that has been encountered many times, which allows you to significantly simplify the next action. (1) The first line was added to the 2nd and 3rd lines. The first line, multiplied by 2, was added to the 4th line. (3) The last three lines are proportional, two of them have been removed. As a result, a standard step matrix is obtained, and the solution continues along the knurled track: – basic variables; Let us express the basic variables in terms of free variables. From the 2nd equation: – substitute into the 1st equation: So the general solution is: Since in the example under consideration there are three free variables, the fundamental system contains three vectors. Let's substitute a triple of values For a triple of values And finally for the three Answer: , Where Those wishing to avoid fractional values may consider triplets Speaking of fractions. Let's look at the matrix obtained in the problem Second solution: The idea is to try choose other basis variables. Let's look at the matrix and notice two ones in the third column. So why not have a zero at the top? Let's carry out one more elementary transformation: 
we will call it a linear inhomogeneous equation.
, i.e. if , then the roots and are distinct real numbers. If, then. If, i.e. , then will be an imaginary number, and the roots and will be complex numbers. In this case, we agree to denote .
.
or .
.
Def. 14.5.5.1. fundamental system of solutions. Fundamental system of solutions linear homogeneous differential equation n
-th order is 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
) of a linear homogeneous differential equation is a linear combination of functions from the fundamental system of solutions to this equation:
y
(x
) = C
1 y
1 (x
) + C
2 y
2 (x
) + …+ C n y n
(x
).
Document. Let y
1 (x
), y
2 (x
), …, y n
(x
) is a fundamental system of solutions to a linear homogeneous differential equation. It is required to prove that any particular solution y
what ( x
) of this equation is contained in the formula y
(x
) = C
1 y
1 (x
) + C
2 y
2 (x
) + …+ C n y n
(x
) for a certain set of constants C
1 , C
2 , …, Cn
. Let's take any point, calculate the numbers at this point and find the constants C
1 , C
2 , …, Cn
as a solution to a linear inhomogeneous system of algebraic equations 
Such a solution exists and is unique, since the determinant of this system is equal to . Consider the linear combination y
(x
) = C
1 y
1 (x
) + C
2 y
2 (x
) + …+ C n y n
(x
) functions from the fundamental system of solutions with these values of the constants C
1 , C
2 , …, Cn
and compare it with the function y
what ( x
). Functions y
(x
) And y
what ( x
) satisfy the same equation and the same initial conditions at the point x
0, therefore, due to the uniqueness of the solution to the Cauchy problem, they coincide: y
what ( x
) = C
1 y
1 (x
) + C
2 y
2 (x
) + … + C n y n
(x
). The theorem is proven.
From this theorem it follows that the dimension of the linear space of partial solutions of a homogeneous equation with continuous coefficients does not exceed n
. It remains to prove that this dimension is not less than n
.
Theorem 14.5.5.1.2 on the existence of a fundamental system of solutions to a linear homogeneous differential equation. Any linear homogeneous differential equation n
th order with continuous coefficients has a fundamental system of solutions, i.e. system from n
linearly independent solutions.
Document. Let's take any numerical determinant n
-th order, not equal to zeroHow to find the fundamental system of solutions to a linear equation?
Let's transform this matrix to a triangular one. We rewrite the first line without changes. And all the elements that are under $a_(11)$ must be made zeros. To make a zero in place of the element $a_(21)$, you need to subtract the first from the second line, and write the difference in the second line. To make a zero in place of the element $a_(31)$, you need to subtract the first from the third line and write the difference in the third line. To make a zero in place of the element $a_(41)$, you need to subtract the first multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(31)$, you need to subtract the first multiplied by 2 from the fifth line and write the difference in the fifth line. 
We rewrite the first and second lines without changes. And all the elements that are under $a_(22)$ must be made zeros. To make a zero in place of the element $a_(32)$, you need to subtract the second one multiplied by 2 from the third line and write the difference in the third line. To make a zero in place of the element $a_(42)$, you need to subtract the second multiplied by 2 from the fourth line and write the difference in the fourth line. To make a zero in place of the element $a_(52)$, you need to subtract the second multiplied by 3 from the fifth line and write the difference in the fifth line. 
We see that the last three lines are the same, so if you subtract the third from the fourth and fifth, they will become zero. 
According to this matrix write a new system of equations.
We see that we have only three linearly independent equations, and five unknowns, so the fundamental system of solutions will consist of two vectors. So we we need to move the last two unknowns to the right.
Now, we begin to express those unknowns that are on the left side through those that are on the right side. We start with the last equation, first we express $x_3$, then we substitute the resulting result into the second equation and express $x_2$, and then into the first equation and here we express $x_1$. Thus, we expressed all the unknowns that are on the left side through the unknowns that are on the right side. 
Then instead of $x_4$ and $x_5$, we can substitute any numbers and find $x_1$, $x_2$ and $x_3$. Each five of these numbers will be the roots of our original system of equations. To find the vectors that are included in FSR we need to substitute 1 instead of $x_4$, and substitute 0 instead of $x_5$, find $x_1$, $x_2$ and $x_3$, and then vice versa $x_4=0$ and $x_5=1$.
Finding a fundamental system of solutions in the general case is a rather difficult task. However, there is a class of equations for which this problem can be solved quite easily. We are now starting to study this class.
(*)
. (6)
We will look for a solution to equation (6) in the form y = erx. Then y" = r e rx , y"" = r 2 e rx ,…, y (n) = r n e rx . Substituting into (6), we get
Since e rx does not vanish anywhere, then 
. (7)
Equation (7) is called the characteristic equation of a linear homogeneous differential equation with constant coefficients.
Thus, we have proven the following theorem. Theorem. The function y = e rx is a solution to a linear homogeneous differential equation with constant coefficients (6) if and only if r is the root of the characteristic equation (7).
The following cases are possible.
1. All roots of the characteristic polynomial are real and distinct. Let us denote them r 1 ,r 2 ,…,r n . Then we get n different solutions
y 1 = e r1x , y 2 = e r2x ,…, y n = e rnx (8)
equation (6). Let us prove that the resulting system of solutions is linearly independent. Let us consider its Wronsky determinant 
.
The factor e (r 1+ r 2+..+ rn) x on the right side of W(e r 1 x, e r 2 x,…, e rnx) does not vanish anywhere. Therefore, it remains to show that the second factor (determinant) is not equal to zero. Let's assume that 
Then the rows of this determinant are linearly dependent, i.e. there are numbers α 1, α 2, ..., α n such that
Thus, we found that r i , i = 1,2,..,n are n different roots of a polynomial of (n-1) degree, which is impossible. Consequently, the determinant on the right side W(e r 1 x , e r 2 x ,…, e rnx) is not equal to zero and the system of functions (8) forms a fundamental system of solutions to equation (6) in the case when the roots of the characteristic equation are different.
2. Among the roots of the characteristic equation there are multiples. Suppose that r 1 has multiplicity α, and all others are different. Let us first consider the case r 1 = 0. Then the characteristic equation has the form
since otherwise it would not be a root of the multiplicity α. Therefore, the differential equation has the form
that is, it does not contain derivatives of order below α. This equation is satisfied by all functions whose derivatives of order α and higher are equal to zero. In particular, these are all polynomials of degree not higher than α-1, for example,
1, x, x 2, …, x α-1. (9)
Let us show that this system is linearly independent. Having compiled the Wronski determinant of this system of functions, we obtain 
.
This is a triangular determinant with non-zero elements on the main diagonal. Therefore, it is different from zero, which proves the linear independence of the system of functions (9). Note that in one of the examples in the previous paragraph we proved the linear independence of the system of functions (9) in a different way. Let now the root of the characteristic equation of multiplicity α be the number r 1 ≠0. Let us make the replacement y = z r 1 x = z exp(r 1 x) in equation (6) L(y) = 0. Then 
and so on. Substituting the obtained values of the derivatives into the original equation, we again obtain a linear homogeneous equation with constant coefficients
(0)
with characteristic equation 
. (1)
Note that if k is the root of the characteristic equation (1), then z = e kx is a solution to equation (0), and y = y r 1 x = e (k + r 1) x is a solution to equation (6). Then r=k+r 1 is the root of the characteristic equation (7). On the other hand, equation (6) can be obtained from equation (0) by the reverse substitution z = ye - r 1 x and therefore each root of the characteristic equation (7) corresponds to the root k = r - r 1 of the characteristic equation (1). Thus, a one-to-one correspondence has been established between the roots of the characteristic equations (7) and (1), and different roots of one equation correspond to different roots of the other. Since r = r 1 is the root of the multiplicity α of equation (7), then equation (1) has k=0 as the root of the multiplicity α. According to what was proven earlier, equation (0) has α linearly independent solutions
which correspond to α linearly independent solutions
(2)
equation (7). By adding the resulting system of solutions (2) to the n-α solutions corresponding to the remaining roots of the characteristic equation, we obtain a fundamental system of solutions for a linear homogeneous differential equation with constant coefficients in the case of multiple roots.
Example.
For the equation y"""-4y""+4y" = 0, the characteristic equation r 3 -4r 2 + 4r = 0 has roots r=0 of multiple 1 and r=2 of multiple 2, since r 3 -4r 2 + 4r = r(r-2) 2, therefore the fundamental system of solutions to the original equation is the system of functions y 1 = 1, y 2 = e 2 x, y 3 = xe 2 x, and the general solution has the form y = C 1 + C 2 e 2 x + C 3 xe 2 x .
3. Among the roots of the characteristic equation there are complex roots. You can consider complex solutions, but for equations with real coefficients this is not very convenient. Let us find real solutions corresponding to complex roots. Since we are considering an equation with real coefficients, then for each complex root r j = a+bi of multiplicity α of the characteristic equation, its complex conjugate number r k = a-bi is also a root of multiplicity α of this equation. The pairs of solutions corresponding to these roots are the functions and , l=0,1,.., α-1. Instead of these solutions, consider their linear combinations 3. For the equation y (4) + 8y"" + 16y =0, the characteristic equation r 4 +8r 2 +16=0 has r 1 = 2i, r 2 = -2i of multiplicity 2, since r 4 +8r 2 +16= (r 2 + 4) 2, therefore the fundamental system of solutions to the original equation is the system of functions y 1 = cos2x, y 2 = sin2x, y 3 = xcos2x, y 4 = xsin2x, and the general solution has the form y = C 1 cos2x+ C 2 sin2x+ C 3 xcos2x+ C 4 xsin2x.
Based on the first paragraphs, the material may seem boring and mediocre, but this impression is deceptive. In addition to further development of techniques, there will be a lot of new information, so please try not to neglect the examples in this article.What is a homogeneous system of linear equations?
. Trivial, for those who do not understand the meaning of the adjective at all, means without a show-off. Not academically, of course, but intelligibly =) ...Why beat around the bush, let's find out if this system has any other solutions:
, and, using the inverse of the Gaussian method, it is easy to verify that the solution is unique.
– free variables.
into the general solution and obtain a vector whose coordinates satisfy each equation of the homogeneous system. And again, I repeat that it is highly advisable to check each received vector - it will not take much time, but it will completely protect you from errors.
find the vector 
we get the third vector: 
and get an answer in equivalent form:
and let us ask ourselves: is it possible to simplify the further solution? After all, here we first expressed the basic variable through fractions, then through fractions the basic variable, and, I must say, this process was not the simplest and not the most pleasant.
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