This paragraph will discuss special case linear equations second order, when the coefficients of the equation are constant, that is, they are numbers. Such equations are called equations with constant coefficients. This type of equations finds particularly wide application.
1. Linear homogeneous differential equations
second order with constant coefficientsConsider the equation
in which the coefficients are constant. Assuming that dividing all terms of the equation by and denoting
let's write down given equation as
As is known, to find a general solution to a linear homogeneous second-order equation, it is enough to know its fundamental system of partial solutions. Let's show you how it is fundamental system partial solutions for a homogeneous linear differential equation with constant coefficients. We will look for a particular solution to this equation in the form
Differentiating this function twice and substituting expressions for into equation (59), we obtain
Since , then, reducing by we get the equation
From this equation, those values of k are determined for which the function will be a solution to equation (59).
The algebraic equation (61) for determining the coefficient k is called the characteristic equation of this differential equation (59).
The characteristic equation is an equation of the second degree and therefore has two roots. These roots can be either real distinct, real and equal, or complex conjugate.
Let us consider what form the fundamental system of particular solutions has in each of these cases.
1. Roots characteristic equation real and distinct: . In this case, using formula (60) we find two partial solutions:
These two particular solutions form a fundamental system of solutions on the entire numerical axis, since the Wronski determinant does not vanish anywhere:
Consequently, the general solution of the equation according to formula (48) has the form
2. The roots of the characteristic equation are equal: . In this case, both roots will be real. Using formula (60), we obtain only one particular solution
Let us show that the second particular solution, which together with the first forms a fundamental system, has the form
First of all, let's check that the function is a solution to equation (59). Really,
But, since there is a root of the characteristic equation (61). In addition, according to Vieta’s theorem, Therefore . Consequently, , i.e., the function is indeed a solution to equation (59).
Let us now show that the found partial solutions form a fundamental system of solutions. Really,
Thus, in this case the general solution of the homogeneous linear equation has the form
3. The roots of the characteristic equation are complex. As is known, the complex roots of a quadratic equation with real coefficients are conjugate complex numbers, i.e. they look like: . In this case, partial solutions of equation (59), according to formula (60), will have the form:
Using Euler's formulas (see Chapter XI, § 5, paragraph 3), expressions for can be written as:
These solutions are comprehensive. To get valid solutions, consider the new functions
They are linear combinations of solutions and, therefore, are themselves solutions to equation (59) (see § 3, item 2, Theorem 1).
It is easy to show that the Wronski determinant for these solutions is nonzero and, therefore, the solutions form a fundamental system of solutions.
Thus, the general solution of a homogeneous linear differential equation in the case of complex roots of the characteristic equation has the form
In conclusion, we present a table of formulas for the general solution of equation (59) depending on the type of roots of the characteristic equation.
Here we will apply the method of variation of Lagrange constants to solve linear inhomogeneous second-order differential equations. Detailed description this method for solving equations of arbitrary order is described on the page
Solution of linear inhomogeneous differential equations of higher orders by the Lagrange method >>>.
Example 1
Solve a second-order differential equation with constant coefficients using the method of variation of Lagrange constants:
(1)
Solution
First we solve the homogeneous differential equation:
(2)
This is a second order equation.
Solving the quadratic equation:
.
Multiple roots: . The fundamental system of solutions to equation (2) has the form:
(3)
.
From here we obtain a general solution to the homogeneous equation (2):
(4)
.
Varying the constants C 1
and C 2
. That is, we replace constants in (4) with functions:
.
We are looking for a solution to the original equation (1) in the form:
(5)
.
Finding the derivative:
.
Let's connect the functions and the equation:
(6)
.
Then
.
We find the second derivative:
.
Substitute into the original equation (1):
(1)
;
.
Since and satisfy homogeneous equation (2), the sum of the terms in each column of the last three rows gives zero and the previous equation takes the form:
(7)
.
Here .
Together with equation (6) we obtain a system of equations for determining the functions and:
(6)
:
(7)
.
Solving a system of equations
We solve the system of equations (6-7). Let's write down expressions for the functions and:
.
We find their derivatives:
;
.
We solve the system of equations (6-7) using the Cramer method. We calculate the determinant of the system matrix:
.
Using Cramer's formulas we find:
;
.
So, we found the derivatives of the functions:
;
.
Let's integrate (see Methods for integrating roots). Making a substitution
;
;
;
.
.
.
;
.
Answer
Example 2
Solve the differential equation by the method of variation of Lagrange constants:
(8)
Solution
Step 1. Solving the homogeneous equation
We solve the homogeneous differential equation:
(9)
We are looking for a solution in the form . We compose the characteristic equation:
This equation has complex roots:
.
The fundamental system of solutions corresponding to these roots has the form:
(10)
.
General solution of homogeneous equation (9):
(11)
.
Step 2. Variation of constants - replacing constants with functions
Now we vary the constants C 1
and C 2
. That is, we replace the constants in (11) with functions:
.
We are looking for a solution to the original equation (8) in the form:
(12)
.
Further, the solution progress is the same as in example 1. We arrive at the following system of equations for determining the functions and:
(13)
:
(14)
.
Here .
Solving a system of equations
Let's solve this system. Let's write down the expressions for the functions and :
.
From the table of derivatives we find:
;
.
We solve the system of equations (13-14) using the Cramer method. Determinant of the system matrix:
.
Using Cramer's formulas we find:
;
.
.
Since , the modulus sign under the logarithm sign can be omitted. Multiply the numerator and denominator by:
.
Then
.
General solution to the original equation:
.
Linear differential equation of the second order called an equation of the form
y"" + p(x)y" + q(x)y = f(x) ,
Where y is the function to be found, and p(x) , q(x) And f(x) - continuous functions on a certain interval ( a, b) .
If the right side of the equation is zero ( f(x) = 0), then the equation is called linear homogeneous equation . The practical part of this lesson will mainly be devoted to such equations. If the right side of the equation is not equal to zero ( f(x) ≠ 0), then the equation is called .
In the problems we are required to solve the equation for y"" :
y"" = −p(x)y" − q(x)y + f(x) .
Linear differential equations second order have a unique solution Cauchy problems .
Linear homogeneous differential equation of the second order and its solution
Consider a linear homogeneous differential equation of the second order:
y"" + p(x)y" + q(x)y = 0 .
If y1 (x) And y2 (x) are particular solutions of this equation, then the following statements are true:
1) y1 (x) + y 2 (x) - is also a solution to this equation;
2) Cy1 (x) , Where C- an arbitrary constant (constant), is also a solution to this equation.
From these two statements it follows that the function
C1 y 1 (x) + C 2 y 2 (x)
is also a solution to this equation.
A fair question arises: is this solution general solution of a linear homogeneous differential equation of the second order , that is, such a solution in which, for different values C1 And C2 Is it possible to get all possible solutions to the equation?
The answer to this question is: maybe, but under certain conditions. This condition on what properties particular solutions should have y1 (x) And y2 (x) .
And this condition is called the condition of linear independence of partial solutions.
Theorem. Function C1 y 1 (x) + C 2 y 2 (x) is a general solution to a linear homogeneous second order differential equation if the functions y1 (x) And y2 (x) linearly independent.
Definition. Functions y1 (x) And y2 (x) are called linearly independent if their ratio is a constant non-zero:
y1 (x)/y 2 (x) = k ; k = const ; k ≠ 0 .
However, determining by definition whether these functions are linearly independent is often very laborious. There is a way to establish linear independence using the Wronski determinant W(x) :
If the Wronski determinant is not equal to zero, then the solutions are linearly independent . If the Wronski determinant is zero, then the solutions are linearly dependent.
Example 1. Find the general solution of a linear homogeneous differential equation.
Solution. We integrate twice and, as is easy to see, in order for the difference between the second derivative of a function and the function itself to be equal to zero, the solutions must be associated with an exponential whose derivative is equal to itself. That is, the partial solutions are and .
Since the Wronski determinant
is not equal to zero, then these solutions are linearly independent. Therefore, the general solution to this equation can be written as
.
Linear homogeneous second order differential equations with constant coefficients: theory and practice
Linear homogeneous differential equation of the second order with constant coefficients called an equation of the form
y"" + py" + qy = 0 ,
Where p And q- constant values.
The fact that this is a second-order equation is indicated by the presence of the second derivative of the desired function, and its homogeneity is indicated by zero on the right side. The values already mentioned above are called constant coefficients.
To solve a linear homogeneous second order differential equation with constant coefficients , you must first solve the so-called characteristic equation of the form
k² + pq + q = 0 ,
which, as can be seen, is an ordinary quadratic equation.
Depending on the solution of the characteristic equation, three different options are possible solutions to a linear homogeneous second order differential equation with constant coefficients , which we will now analyze. For complete definiteness, we will assume that all particular solutions have been tested by the Wronski determinant and it is not equal to zero in all cases. Doubters, however, can check this themselves.
The roots of the characteristic equation are real and distinct
In other words, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form
.
Example 2. Solve a linear homogeneous differential equation
.
Example 3. Solve a linear homogeneous differential equation
.
Solution. The characteristic equation has the form , its roots and are real and distinct. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form
.
The roots of the characteristic equation are real and equal
That is, . In this case, the solution to a linear homogeneous second-order differential equation with constant coefficients has the form
.
Example 4. Solve a linear homogeneous differential equation
.
Solution. Characteristic equation 
has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form
Example 5. Solve a linear homogeneous differential equation
.
Solution. The characteristic equation has equal roots. The corresponding partial solutions of the equation are: and . The general solution of this differential equation has the form
Consider a linear homogeneous differential equation with constant coefficients:
(1)
.
Its solution can be obtained by following general method reduction of order.
However, it is easier to immediately obtain the fundamental system n linearly independent solutions and based on it create a general solution. In this case, the entire solution procedure is reduced to the following steps.
We are looking for a solution to equation (1) in the form . We get characteristic equation:
(2)
.
It has n roots. We solve equation (2) and find its roots. Then the characteristic equation (2) can be represented in the following form:
(3)
.
Each root corresponds to one of the linearly independent solutions of the fundamental system of solutions to equation (1). Then the general solution to the original equation (1) has the form:
(4)
.
Real roots
Let's consider real roots. Let the root be single. That is, the factor enters the characteristic equation (3) only once. Then this root corresponds to the solution
.
Let be a multiple root of multiplicity p. That is
. In this case, the multiplier is p times:
.
These multiple (equal) roots correspond to p linearly independent solutions of the original equation (1):
;
;
;
...;
.
Complex roots
Consider complex roots. Let us express the complex root in terms of the real and imaginary parts:
.
Since the coefficients of the original are real, then in addition to the root there is a complex conjugate root
.
Let the complex root be multiple. Then a pair of roots corresponds to two linearly independent solutions:
;
.
Let be a multiple complex root of multiplicity p. Then the complex conjugate value is also the root of the characteristic equation of multiplicity p and the multiplier enters p times:
.
This 2p roots correspond 2p linearly independent solutions:
;
;
;
...
;
;
;
;
...
.
After the fundamental system of linearly independent solutions has been found, we obtain the general solution.
Examples of problem solutions
Example 1
Solve the equation:
.
Solution
.
Let's transform it:
;
;
.
Let's look at the roots of this equation. We got four complex roots of multiplicity 2:
;
.
They correspond to four linearly independent solutions of the original equation:
;
;
;
.
We also have three real roots of multiple 3:
.
They correspond to three linearly independent solutions:
;
;
.
The general solution to the original equation has the form:
.
Answer
Example 2
Solve the equation
Solution
We are looking for a solution in the form . We compose the characteristic equation:
.
Solving a quadratic equation.
.
We got two complex roots:
.
They correspond to two linearly independent solutions:
.
General solution to the equation:
.
Fundamentals of solving linear inhomogeneous second order differential equations (LNDE-2) with constant coefficients (PC)
A 2nd order LDDE with constant coefficients $p$ and $q$ has the form $y""+p\cdot y"+q\cdot y=f\left(x\right)$, where $f\left(x \right)$ is a continuous function.
With regard to LNDU 2 with PC, the following two statements are true.
Let us assume that some function $U$ is an arbitrary partial solution of an inhomogeneous differential equation. Let us also assume that some function $Y$ is the general solution (GS) of the corresponding linear homogeneous differential equation (HLDE) $y""+p\cdot y"+q\cdot y=0$. Then the GR of LHDE-2 is equal to the sum of the indicated private and general solutions, that is, $y=U+Y$.
If the right-hand side of a 2nd order LMDE is a sum of functions, that is, $f\left(x\right)=f_(1) \left(x\right)+f_(2) \left(x\right)+. ..+f_(r) \left(x\right)$, then first we can find the PDs $U_(1) ,U_(2) ,...,U_(r)$ that correspond to each of the functions $f_( 1) \left(x\right),f_(2) \left(x\right),...,f_(r) \left(x\right)$, and after that write the CR LNDU-2 in the form $U=U_(1) +U_(2) +...+U_(r) $.
Solution of 2nd order LPDE with PC
It is obvious that the type of one or another PD $U$ of a given LNDU-2 depends on the specific form of its right-hand side $f\left(x\right)$. The simplest cases of searching for PD LNDU-2 are formulated in the form of the following four rules.
Rule #1.
The right side of LNDU-2 has the form $f\left(x\right)=P_(n) \left(x\right)$, where $P_(n) \left(x\right)=a_(0) \cdot x^(n) +a_(1) \cdot x^(n-1) +...+a_(n-1) \cdot x+a_(n) $, that is, it is called a polynomial of degree $n$. Then its PD $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) $, where $Q_(n) \left(x\right)$ is another polynomial of that the same degree as $P_(n) \left(x\right)$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 that are equal to zero. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the method of indefinite coefficients (UK).
Rule No. 2.
The right side of LNDU-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot P_(n) \left(x\right)$, where $P_(n) \left( x\right)$ is a polynomial of degree $n$. Then its PD $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) \cdot e^(\alpha \cdot x) $, where $Q_(n) \ left(x\right)$ is another polynomial of the same degree as $P_(n) \left(x\right)$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2 equal to $\alpha $. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the NC method.
Rule No. 3.
The right side of LNDU-2 has the form $f\left(x\right)=a\cdot \cos \left(\beta \cdot x\right)+b\cdot \sin \left(\beta \cdot x\right) $, where $a$, $b$ and $\beta$ are known numbers. Then its PD $U$ is sought in the form $U=\left(A\cdot \cos \left(\beta \cdot x\right)+B\cdot \sin \left(\beta \cdot x\right)\right )\cdot x^(r) $, where $A$ and $B$ are unknown coefficients, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2, equal to $i\cdot \beta $. The coefficients $A$ and $B$ are found using the non-destructive method.
Rule No. 4.
The right side of LNDU-2 has the form $f\left(x\right)=e^(\alpha \cdot x) \cdot \left$, where $P_(n) \left(x\right)$ is a polynomial of degree $ n$, and $P_(m) \left(x\right)$ is a polynomial of degree $m$. Then its PD $U$ is sought in the form $U=e^(\alpha \cdot x) \cdot \left\cdot x^(r) $, where $Q_(s) \left(x\right)$ and $ R_(s) \left(x\right)$ are polynomials of degree $s$, the number $s$ is the maximum of two numbers $n$ and $m$, and $r$ is the number of roots of the characteristic equation of the corresponding LODE-2, equal to $\alpha +i\cdot \beta $. The coefficients of the polynomials $Q_(s) \left(x\right)$ and $R_(s) \left(x\right)$ are found by the NC method.
The NDT method consists of using next rule. In order to find the unknown coefficients of the polynomial that are part of the partial solution of the inhomogeneous differential equation LNDU-2, it is necessary:
- substitute the PD $U$ written in general view, V left side LNDU-2;
- on the left side of LNDU-2, perform simplifications and group terms with the same powers $x$;
- in the resulting identity, equate the coefficients of terms with the same powers $x$ of the left and right sides;
- solve the resulting system of linear equations for unknown coefficients.
Example 1
Task: find OR LNDU-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. Find also PD , satisfying the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$.
We write down the corresponding LOD-2: $y""-3\cdot y"-18\cdot y=0$.
Characteristic equation: $k^(2) -3\cdot k-18=0$. The roots of the characteristic equation are: $k_(1) =-3$, $k_(2) =6$. These roots are valid and distinct. Thus, the OR of the corresponding LODE-2 has the form: $Y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) $.
The right side of this LNDU-2 has the form $\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. It is necessary to consider the coefficient of the exponent $\alpha =3$. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PD of this LNDU-2 has the form $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $.
We will search for the coefficients $A$, $B$ using the NC method.
We find the first derivative of the Czech Republic:
$U"=\left(A\cdot x+B\right)^((") ) \cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot \left( e^(3\cdot x) \right)^((") ) =$
$=A\cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot 3\cdot e^(3\cdot x) =\left(A+3\cdot A\ cdot x+3\cdot B\right)\cdot e^(3\cdot x) .$
We find the second derivative of the Czech Republic:
$U""=\left(A+3\cdot A\cdot x+3\cdot B\right)^((") ) \cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot \left(e^(3\cdot x) \right)^((") ) =$
$=3\cdot A\cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot 3\cdot e^(3\cdot x) =\left(6\cdot A+9\cdot A\cdot x+9\cdot B\right)\cdot e^(3\cdot x) .$
We substitute the functions $U""$, $U"$ and $U$ instead of $y""$, $y"$ and $y$ into the given NLDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x).$ Moreover, since the exponent $e^(3\cdot x)$ is included as a factor in all components, then its can be omitted. We get:
$6\cdot A+9\cdot A\cdot x+9\cdot B-3\cdot \left(A+3\cdot A\cdot x+3\cdot B\right)-18\cdot \left(A\ cdot x+B\right)=36\cdot x+12.$
We perform the actions on the left side of the resulting equality:
$-18\cdot A\cdot x+3\cdot A-18\cdot B=36\cdot x+12.$
We use the NDT method. We obtain a system of linear equations with two unknowns:
$-18\cdot A=36;$
$3\cdot A-18\cdot B=12.$
The solution to this system is: $A=-2$, $B=-1$.
PD $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $ for our problem looks like this: $U=\left(-2\cdot x-1\right) \cdot e^(3\cdot x) $.
The OR $y=Y+U$ for our problem looks like this: $y=C_(1) \cdot e^(-3\cdot x) +C_(2) \cdot e^(6\cdot x) +\ left(-2\cdot x-1\right)\cdot e^(3\cdot x) $.
In order to search for a PD that satisfies the given initial conditions, we find the derivative $y"$ of the OP:
$y"=-3\cdot C_(1) \cdot e^(-3\cdot x) +6\cdot C_(2) \cdot e^(6\cdot x) -2\cdot e^(3\ cdot x) +\left(-2\cdot x-1\right)\cdot 3\cdot e^(3\cdot x) .$
We substitute into $y$ and $y"$ the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$:
$6=C_(1) +C_(2) -1; $
$1=-3\cdot C_(1) +6\cdot C_(2) -2-3=-3\cdot C_(1) +6\cdot C_(2) -5.$
We received a system of equations:
$C_(1) +C_(2) =7;$
$-3\cdot C_(1) +6\cdot C_(2) =6.$
Let's solve it. We find $C_(1) $ using Cramer's formula, and $C_(2) $ we determine from the first equation:
$C_(1) =\frac(\left|\begin(array)(cc) (7) & (1) \\ (6) & (6) \end(array)\right|)(\left|\ begin(array)(cc) (1) & (1) \\ (-3) & (6) \end(array)\right|) =\frac(7\cdot 6-6\cdot 1)(1\ cdot 6-\left(-3\right)\cdot 1) =\frac(36)(9) =4; C_(2) =7-C_(1) =7-4=3.$
Thus, the PD of this differential equation has the form: $y=4\cdot e^(-3\cdot x) +3\cdot e^(6\cdot x) +\left(-2\cdot x-1\right )\cdot e^(3\cdot x) $.