Homogeneous differential equation with constant coefficients. Inhomogeneous Second Order Differential Equations
We have verified that, in the case when it is known common decision linear homogeneous equation, it is possible to find the general solution by the method of variation of arbitrary constants inhomogeneous equation. However, the question of how to find the general solution of the homogeneous equation remained open. In a particular case, when in the linear differential equation (3) all the coefficients p i(X)= a i - constants, it is solved quite simply, even without integration.
Consider a linear homogeneous differential equation with constant coefficients, i.e., equations of the form
y (n) + a 1 y (n – 1) + ... a n – 1 y " + a n y = 0, (14)
where a i- constants (i= 1, 2, ...,n).
As is known, for a linear homogeneous equation of the 1st order, the solution is a function of the form e kx . We will seek a solution to Eq. (14) in the form j (X) = e kx.
Let us substitute into equation (14) the function j (X) and its order derivatives m (1 £ m£ n)j (m) (X) = k m e kx. Get
(k n + a 1 k n – 1 +… and n – 1 k + a n)e kx = 0,
but e k x ¹ 0 for any X, That's why
k n + a 1 k n – 1 + ... a n – 1 k + a n = 0. (15)
Equation (15) is called characteristic equation, polynomial on the left side,- characteristic polynomial , its roots- characteristic roots differential equation (14).
Conclusion:
functionj (X) = e kx - solution of the linear homogeneous equation (14) if and only if the number k - root of the characteristic equation (15).
Thus, the process of solving the linear homogeneous equation (14) is reduced to solving the algebraic equation (15).
There are various cases of characteristic roots.
1.All roots of the characteristic equation are real and distinct.
In this case n different characteristic roots k 1 ,k 2 ,..., k n corresponds n different solutions of the homogeneous equation (14)
It can be shown that these solutions are linearly independent and therefore form fundamental system solutions. Thus, the general solution to the equation is the function
where With 1 , C 2 , ..., ~ n - arbitrary constants.
EXAMPLE 7. Find the general solution of the linear homogeneous equation:
a) at¢ ¢ (X) - 6at¢ (X) + 8at(X) = 0,b) at¢ ¢ ¢ (X) + 2at¢ ¢ (X) - 3at¢ (X) = 0.
Decision. Let's make a characteristic equation. To do this, we replace the order derivative m functions y(x) to the corresponding degree
k(at (m) (x) « k m),
while the function itself at(X) as the zeroth order derivative is replaced by k 0 = 1.
In case (a), the characteristic equation has the form k 2 - 6k + 8 = 0. The roots of this quadratic equation k 1 = 2,k 2 = 4. Since they are real and different, the general solution has the form j (X)= C 1 e 2X + From 2 e 4x.
For case (b), the characteristic equation is the third-degree equation k 3 + 2k 2 - 3k = 0. Find the roots of this equation:
k(k 2 + 2 k - 3)= 0 Þ k = 0i k 2 + 2 k - 3 = 0 Þ k = 0, (k - 1)(k + 3) = 0,
t . e . k 1 = 0, k 2 = 1, k 3 = - 3.
These characteristic roots correspond to the fundamental system of solutions of the differential equation:
j 1 (X)= e 0X = 1, j 2 (X) = e x, j 3 (X)= e - 3X .
The general solution, according to formula (9), is the function
j (X)= C 1 + C 2 e x + C 3 e - 3X .
II . All the roots of the characteristic equation are different, but some of them are complex.
All coefficients of the differential equation (14), and consequently, of its characteristic equation (15)- real numbers, then if c among the characteristic roots there is a complex root k 1 = a + ib, that is, its conjugate root k 2 = ` k 1 = a- ib.First root k 1 corresponds to the solution of the differential equation (14)
j 1 (X)= e (a+ib)X = e a x e ibx = e ax(cosbx + isinbx)
(we used the Euler formula e i x = cosx + isinx). Likewise, the root k 2 = a- ib corresponds decision
j 2 (X)= e (a - -ib)X = e a x e - ib x= e ax(cosbx - isinbx).
These solutions are complex. To obtain real solutions from them, we use the properties of solutions of a linear homogeneous equation (see 13.2). Functions
are real solutions of equation (14). Also, these solutions are linearly independent. Thus, the following conclusion can be drawn.
Rule 1.A pair of conjugate complex roots a± ib of the characteristic equation in the FSR of the linear homogeneous equation (14) corresponds to two real particular solutionsand .
EXAMPLE 8. Find the general solution of the equation:
a) at¢ ¢ (X) - 2at ¢ (X) + 5at(X) = 0 ;b) at¢ ¢ ¢ (X) - at¢ ¢ (X) + 4at ¢ (X) - 4at(X) = 0.
Decision. In the case of equation (a), the roots of the characteristic equation k 2 - 2k + 5 = 0 are two conjugate complex numbers
k 1, 2 = .
Therefore, according to rule 1, they correspond to two real linearly independent solutions: and , and the general solution of the equation is the function
j (X)= C 1 e x cos 2x + C 2 e x sin 2x.
In case (b), to find the roots of the characteristic equation k 3 - k 2 + 4k- 4 = 0, we factorize its left side:
k 2 (k - 1) + 4(k - 1) = 0 Þ (k - 1)(k 2 + 4) = 0 Þ (k - 1) = 0, (k 2 + 4) = 0.
Therefore, we have three characteristic roots: k 1 = 1,k2 , 3 = ± 2i. Cornu k 1 corresponds decision , and a pair of conjugate complex roots k 2, 3 = ± 2i = 0 ± 2i- two real solutions: and . We compose the general solution of the equation:
j (X)= C 1 e x + C 2 cos 2x + C 3 sin 2x.
III . Among the roots of the characteristic equation there are multiples.
Let be k 1 - real root of multiplicity m characteristic equation (15), i.e. among the roots there are m equal roots. Each of them corresponds to the same solution of differential equation (14) However, include m equal solutions in the FSR are impossible, since they constitute a linearly dependent system of functions.
It can be shown that in the case of a multiple root k 1 solutions of equation (14), in addition to the function, are the functions
The functions are linearly independent on the entire number axis, since , i.e., they can be included in the FSR.
Rule 2 real characteristic root k 1 multiplicities m in FSR corresponds m solutions:
If a k 1 - complex root of multiplicity m characteristic equation (15), then there is a conjugate root k 1 multiplicities m. By analogy, we get the following rule.
Rule 3. A pair of conjugate complex roots a± ib in the FSR corresponds to 2m real linearly independent solutions:
, , ..., ,
, , ..., .
EXAMPLE 9. Find the general solution of the equation:
a) at¢ ¢ ¢ (X) + 3at¢ ¢ (X) + 3at¢ (X)+ y ( X)= 0;b) IV(X) + 6at¢ ¢ (X) + 9at(X) = 0.
Decision. In case (a), the characteristic equation has the form
k 3 + 3 k 2 + 3 k + 1 = 0
(k + 1) 3 = 0,
i.e. k =- 1 - multiplicity root 3. Based on rule 2, we write the general solution:
j (X)= C 1 + C 2 x + C 3 x 2 .
The characteristic equation in case (b) is the equation
k 4 + 6k 2 + 9 = 0
or, otherwise,
(k 2 + 3) 2 = 0 Þ k 2 = - 3 Þ k 1, 2 = ± i .
We have a pair of conjugate complex roots, each of multiplicity 2. According to Rule 3, the general solution is written as
j (X)= C 1 + C 2 x + C 3 + C 4 x .
It follows from the above that for any linear homogeneous equation with constant coefficients, one can find a fundamental system of solutions and form a general solution. Therefore, the solution of the corresponding inhomogeneous equation for any continuous function f(x) on the right side can be found using the method of variation of arbitrary constants (see Section 5.3).
Example r10. Using the method of variation, find the general solution of the inhomogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = x e 2x .
Decision. First, we find the general solution of the corresponding homogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = 0. The roots of the characteristic equation k 2 - k- 6 = 0 are k 1 = 3,k 2 = - 2, a general solution of the homogeneous equation - function ` at ( X) = C 1 e 3X + C 2 e - 2X .
We will look for a solution to the inhomogeneous equation in the form
at( X) = With 1 (X)e 3X + C 2 (X)e 2X . (*)
Let's find the Vronsky determinant
W[e 3X , e 2X ] = .
Let us compose the system of equations (12) with respect to the derivatives of the unknown functions With ¢ 1 (X) and With¢ 2 (X):
Solving the system using Cramer's formulas, we obtain
Integrating, we find With 1 (X) and With 2 (X):
Substituting functions With 1 (X) and With 2 (X) into equality (*), we obtain the general solution of the equation at¢ ¢ (X) - at¢ (X) - 6at(X) = x e 2x :
In case when right part linear inhomogeneous equation with constant coefficients has a special form, a particular solution of the inhomogeneous equation can be found without resorting to the method of variation of arbitrary constants.
Consider the equation with constant coefficients
y (n) + a 1 y (n– 1) + ... a n – 1 y " + a n y = f (x), (16)
f( x) = eax(P n(x)cosbx + Rm(x)sinbx), (17)
where P n(x) and R m(x) - degree polynomials n and m respectively.
Private solution y*(X) of equation (16) is determined by the formula
at* (X) = x se ax(M r(x)cosbx + Nr(x)sinbx), (18)
where M r(x) and N r(x) - degree polynomials r = max(n, m) with indeterminate coefficients , a s equal to the multiplicity of the root k 0 = a + ib characteristic polynomial of equation (16), while it is assumed s= 0 if k 0 is not a characteristic root.
To formulate a particular solution using formula (18), we need to find four parameters - a, b, r and s. The first three are determined from the right side of the equation, with r- it is actually the highest x found on the right side. Parameter s is found by comparing the number k 0 = a + ib and the set of all (taking into account multiplicities) characteristic roots of equation (16) that are found in solving the corresponding homogeneous equation.
Let us consider particular cases of the form of function (17):
1) at a ¹ 0, b= 0f(x)= e ax P n(x);
2) when a= 0, b ¹ 0f(x)= P n(x) withosbx + Rm(x)sinbx;
3) when a = 0, b = 0f(x)=Pn(x).
Remark 1. If P n (x) º 0 or R m (x)º 0, then the right side of the equation f(x) = e ax P n (x)с osbx or f(x) = e ax R m (x)sinbx, i.e. contains only one of the functions - cosine or sine. But in the notation of a particular solution, they must both be present, since, according to formula (18), each of them is multiplied by a polynomial with indefinite coefficients of the same degree r = max(n, m).
Example 11. Determine the form of a particular solution of a linear homogeneous equation of the 4th order with constant coefficients, if the right side of the equation is known f(X) = e x(2xcos 3x +(x 2 + 1)sin 3x) and the roots of the characteristic equation:
a ) k 1 = k 2 = 1, k 3 = 3,k 4 = - 1;
b ) k 1, 2 = 1 ± 3i,k 3, 4 = ± 1;
in ) k 1, 2 = 1 ± 3i,k 3, 4 = 1 ± 3i.
Decision. On the right side, we find that in the particular solution at*(X), which is determined by formula (18), parameters: a= 1, b= 3, r= 2. They remain the same for all three cases, hence the number k 0 , which specifies the last parameter s formula (18) is equal to k 0 = 1+ 3i. In case (a) there is no number among the characteristic roots k 0 = 1 + 3i, means, s= 0, and the particular solution has the form
y*(X) = x 0 e x(M 2 (x)cos 3x + N 2 (x)sin 3x) =
= ex( (Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.
In case (b) the number k 0 = 1 + 3i occurs only once among the characteristic roots, which means that s= 1 and
y*(X) = x e x((Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.
For case (c) we have s= 2 and
y*(X) = x 2 e x((Ax 2 + Bx + C)cos 3x +(A 1 x 2 + B 1 x + C 1)sin 3x.
In example 11, in the notation of a particular solution, there are two polynomials of the 2nd degree with indefinite coefficients. To find a solution, you need to determine the numerical values of these coefficients. Let's formulate a general rule.
To determine the unknown coefficients of polynomials M r(x) and N r(x) equality (17) is differentiated the required number of times, the function is substituted y*(X) and its derivatives into equation (16). Comparing its left and right parts, we get the system algebraic equations to find coefficients.
Example 12. Find a solution to the equation at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x, having determined a particular solution of the inhomogeneous equation by the form of the right side.
Decision. The general solution of the inhomogeneous equation has the form
at( X) = ` at(X)+ y*(X),
where ` at ( X) - the general solution of the corresponding homogeneous equation, and y*(X) - a particular solution of an inhomogeneous equation.
First we solve the homogeneous equation at¢ ¢ (X) - at¢ (X) - 6at(X) = 0. Its characteristic equation k 2 - k- 6 = 0 has two roots k 1 = 3,k 2 = - 2, hence, ` at ( X) = C 1 e 3X + C 2 e - 2X .
We use formula (18) to determine the type of particular solution at*(X). Function f(x) = xe 2x is a special case (a) of formula (17), while a = 2,b= 0 and r= 1, i.e. k 0 = 2 + 0i = 2. Comparing with the characteristic roots, we conclude that s= 0. Substituting the values of all parameters into formula (18), we have y*(X) = (Ah + B)e 2X .
To find values BUT and AT, find the derivatives of the first and second orders of the function y*(X) = (Ah + B)e 2X :
y*¢ (X)= Ae 2X + 2(Ah + B)e 2X = (2Ah + A + 2B)e 2x,
y*¢ ¢ (X) = 2Ae 2X + 2(2Ah + A + 2B)e 2X = (4Ah + 4A+ 4B)e 2X .
After substituting the function y*(X) and its derivatives into the equation we have
(4Ah + 4A+ 4B)e 2X - (2Ah + A + 2B)e 2X - 6(Ah + B)e 2X =xe 2x Þ Þ A=- 1/4,B=- 3/16.
Thus, a particular solution of the inhomogeneous equation has the form
y*(X) = (- 1/4X- 3/16)e 2X ,
and the general solution - at ( X) = C 1 e 3X + C 2 e - 2X + (- 1/4X- 3/16)e 2X .
Remark 2.In the case when the Cauchy problem for an inhomogeneous equation is posed, one must first find a general solution to the equation
at( X) = ,
having determined all the numerical values of the coefficients in at*(X). Then use the initial conditions and, substituting them into the general solution (and not into y*(X)), find the values of the constants C i.
Example 13. Find a solution to the Cauchy problem:
at¢ ¢ (X) - at¢ (X) - 6at(X) = xe 2x ,y(0) = 0, y ¢ (X) = 0.
Decision. General solution of this equation
at(X) = C 1 e 3X + C 2 e - 2X + (- 1/4X- 3/16)e 2X
was found in Example 12. To find a particular solution that satisfies the initial conditions of the given Cauchy problem, we obtain the system of equations
Solving it, we have C 1 = 1/8, C 2 = 1/16. Therefore, the solution to the Cauchy problem is the function
at(X) = 1/8e 3X + 1/16e - 2X + (- 1/4X- 3/16)e 2X .
Remark 3(superposition principle). If in a linear equation L n[y(x)]= f(x), where f(x) = f 1 (x)+ f 2 (x) and y* 1 (x) - solution of the equation L n[y(x)]= f 1 (x), a y* 2 (x) - solution of the equation L n[y(x)]= f 2 (x), then the function y*(X)= y* 1 (x)+ y* 2 (x) is an solution of the equation L n[y(x)]= f(x).
EXAMPLE 14. Specify the type of general solution linear equation
at¢ ¢ (X) + 4at(X) = x + sinx.
Decision. General solution of the corresponding homogeneous equation
` at(x) = C 1 cos 2x + C 2 sin 2x,
since the characteristic equation k 2 + 4 = 0 has roots k 1, 2 = ± 2i.The right side of the equation does not correspond to formula (17), but if we introduce the notation f 1 (x) = x, f 2 (x) = sinx and use the principle of superposition , then a particular solution of the inhomogeneous equation can be found in the form y*(X)= y* 1 (x)+ y* 2 (x), where y* 1 (x) - solution of the equation at¢ ¢ (X) + 4at(X) = x, a y* 2 (x) - solution of the equation at¢ ¢ (X) + 4at(X) = sinx. By formula (18)
y* 1 (x) = Ax + B,y* 2 (x) = Ccosx + Dsinx.
Then a particular solution
y*(X) \u003d Ax + B + Ccosx + Dsinx,
hence the general solution has the form
at(X) = C 1 cos 2x + C 2 e - 2X + A x + B + Ccosx + Dsinx.
EXAMPLE 15. The electrical circuit consists of a series-connected current source with emf e(t) = E sinw t, inductance L and containers With, and
Fundamentals of solving linear inhomogeneous differential equations of the second order (LNDE-2) with constant coefficients (PC)
A second-order CLDE 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.
The following two statements are true with respect to the 2nd LNDE with PC.
Assume that some function $U$ is an arbitrary particular solution of an inhomogeneous differential equation. Let us also assume that some function $Y$ is a general solution (OR) of the corresponding linear homogeneous differential equation (LODE) $y""+p\cdot y"+q\cdot y=0$. Then the OR of LNDE-2 is equal to the sum of the indicated private and general solutions, i.e. $y=U+Y$.
If the right side of the 2nd order LIDE is the 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 you can find the PD $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 LNDE-2 PD as $U=U_(1) +U_(2) +...+U_(r) $.
Solution of 2nd order LNDE with PC
Obviously, the form of one or another PD $U$ of a given LNDE-2 depends on the specific form of its right-hand side $f\left(x\right)$. The simplest cases of searching for the PD of LNDE-2 are formulated as the following four rules.
Rule number 1.
The right side of LNDE-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 PR $U$ is sought in the form $U=Q_(n) \left(x\right)\cdot x^(r) $, where $Q_(n) \left(x\right)$ is another polynomial of the the same degree as $P_(n) \left(x\right)$, and $r$ is the number of zero roots of the characteristic equation of the corresponding LODE-2. The coefficients of the polynomial $Q_(n) \left(x\right)$ are found by the method uncertain coefficients(NC).
Rule number 2.
The right side of LNDE-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 NK method.
Rule number 3.
The right part of LNDE-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 searched for 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 by the NDT method.
Rule number 4.
The right side of LNDE-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 searched for 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 NK method.
The NK method consists in applying the following rule. In order to find the unknown coefficients of the polynomial, which are part of the particular solution of the inhomogeneous differential equation LNDE-2, it is necessary:
- substitute the PD $U$ written in general view, to the left side of LNDU-2;
- on the left side of LNDE-2, perform simplifications and group terms with equal degrees$x$;
- in the resulting identity, equate the coefficients of the 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 the OR LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x) $. Also find the PR , satisfying the initial conditions $y=6$ for $x=0$ and $y"=1$ for $x=0$.
Write the corresponding LODA-2: $y""-3\cdot y"-18\cdot y=0$.
Characteristic equation: $k^(2) -3\cdot k-18=0$. The roots of the characteristic equation: $k_(1) =-3$, $k_(2) =6$. These roots are real 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 part of this LNDE-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 of the exponent $\alpha =3$. This coefficient does not coincide with any of the roots of the characteristic equation. Therefore, the PR of this LNDE-2 has the form $U=\left(A\cdot x+B\right)\cdot e^(3\cdot x) $.
We will look for the coefficients $A$, $B$ using the NK method.
We find the first derivative of the CR:
$U"=\left(A\cdot x+B\right)^((") ) \cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot \left( e^(3\cdot x) \right)^((") ) =$
$=A\cdot e^(3\cdot x) +\left(A\cdot x+B\right)\cdot 3\cdot e^(3\cdot x) =\left(A+3\cdot A\ cdot x+3\cdot B\right)\cdot e^(3\cdot x) .$
We find the second derivative of the CR:
$U""=\left(A+3\cdot A\cdot x+3\cdot B\right)^((") ) \cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot \left(e^(3\cdot x) \right)^((") ) =$
$=3\cdot A\cdot e^(3\cdot x) +\left(A+3\cdot A\cdot x+3\cdot B\right)\cdot 3\cdot e^(3\cdot x) =\left(6\cdot A+9\cdot A\cdot x+9\cdot B\right)\cdot e^(3\cdot x) .$
We substitute the functions $U""$, $U"$ and $U$ instead of $y""$, $y"$ and $y$ into the given LNDE-2 $y""-3\cdot y"-18\cdot y=\left(36\cdot x+12\right)\cdot e^(3\cdot x).$ At the same time, since the exponent $e^(3\cdot x) $ is included as a factor in all components, then its can be omitted.
$6\cdot A+9\cdot A\cdot x+9\cdot B-3\cdot \left(A+3\cdot A\cdot x+3\cdot B\right)-18\cdot \left(A\ cdot x+B\right)=36\cdot x+12.$
We perform 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 NC method. We get 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: $A=-2$, $B=-1$.
The CR $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 $y"$ OR:
$y"=-3\cdot C_(1) \cdot e^(-3\cdot x) +6\cdot C_(2) \cdot e^(6\cdot x) -2\cdot e^(3\ cdot x) +\left(-2\cdot x-1\right)\cdot 3\cdot e^(3\cdot x) .$
We substitute in $y$ 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 got a system of equations:
$C_(1) +C_(2) =7;$
$-3\cdot C_(1) +6\cdot C_(2) =6.$
We solve it. We find $C_(1) $ using Cramer's formula, and $C_(2) $ is determined 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 is: $y=4\cdot e^(-3\cdot x) +3\cdot e^(6\cdot x) +\left(-2\cdot x-1\right )\cdot e^(3\cdot x) $.
Educational Institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
Guidelines
on the study of the topic "Linear differential equations of the second order" by students of the accounting department of the correspondence form of education (NISPO)
Gorki, 2013
Linear differential equations
second order with constantcoefficients
Linear homogeneous differential equations
Linear differential equation of the second order with constant coefficients is called an equation of the form
those. an equation that contains the desired function and its derivatives only to the first degree and does not contain their products. In this equation 
and 
are some numbers, and the function 
given on some interval 
.
If a 
on the interval 
, then equation (1) takes the form
,
(2)
and called linear homogeneous . Otherwise, equation (1) is called linear inhomogeneous .
Consider the complex function
,
(3)
where 
and 
are real functions. If function (3) is a complex solution of equation (2), then the real part 
, and the imaginary part 
solutions 
taken separately are solutions of the same homogeneous equation. Thus, every complete solution equation (2) generates two real solutions of this equation.
Solutions of a homogeneous linear equation have the following properties:
If a 
is a solution to equation (2), then the function 
, where With- an arbitrary constant, will also be a solution to equation (2);
If a 
and 
are solutions of equation (2), then the function 
will also be a solution to equation (2);
If a 
and 
are solutions of equation (2), then their linear combination 
will also be a solution to equation (2), where 
and 
are arbitrary constants.
Functions 
and 
called linearly dependent
on the interval 
if there are such numbers 
and 
, not zero at the same time that on this interval the equality
If equality (4) holds only when 
and 
, then the functions 
and 
called linearly independent
on the interval 
.
Example 1
. Functions 
and 
are linearly dependent, since 
along the whole number line. In this example 
.
Example 2
. Functions 
and 
are linearly independent on any interval, since the equality 
possible only if and 
, and 
.
Construction of a general solution of a linear homogeneous
equations
In order to find a general solution to equation (2), you need to find two of its linearly independent solutions 
and 
. Linear combination of these solutions 
, where 
and 
are arbitrary constants, and will give the general solution of a linear homogeneous equation.
Linearly independent solutions of Eq. (2) will be sought in the form
,
(5)
where 
- some number. Then 
,
. Let us substitute these expressions into equation (2):
or 
.
As 
, then 
. So the function 
will be a solution to equation (2) if 
will satisfy the equation
.
(6)
Equation (6) is called characteristic equation for equation (2). This equation is an algebraic quadratic equation.
Let be 
and 
are the roots of this equation. They can be either real and different, or complex, or real and equal. Let's consider these cases.
Let the roots 
and 
characteristic equations are real and distinct. Then the solutions of equation (2) will be the functions 
and 
. These solutions are linearly independent, since the equality 
can only be performed when 
, and 
. Therefore, the general solution of Eq. (2) has the form
,
where 
and 
are arbitrary constants.
Example 3
.
Decision
. The characteristic equation for this differential will be 
. Solving it quadratic equation, find its roots 
and 
. Functions 
and 
are solutions of the differential equation. The general solution of this equation has the form 
.
complex number 
is called an expression of the form 
, where 
and 
are real numbers, and 
is called the imaginary unit. If a 
, then the number 
is called purely imaginary. If 
, then the number 
is identified with a real number 
.
Number 
is called the real part of the complex number, and 
- the imaginary part. If two complex numbers differ from each other only in the sign of the imaginary part, then they are called conjugate: 
,
.
Example 4
. Solve a quadratic equation 
.
Decision
. Equation discriminant 
. Then. Likewise, 
. Thus, this quadratic equation has conjugate complex roots.
Let the roots of the characteristic equation be complex, i.e. 
,
, where 
. Solutions to equation (2) can be written as 
,
or 
,
. According to Euler's formulas
,
.
Then ,. As is known, if a complex function is a solution of a linear homogeneous equation, then the solutions of this equation are both the real and imaginary parts of this function. Thus, the solutions of equation (2) will be the functions 
and 
. Since equality
can only be performed if 
and 
, then these solutions are linearly independent. Therefore, the general solution of equation (2) has the form
where 
and 
are arbitrary constants.
Example 5
. Find the general solution of the differential equation 
.
Decision
. The equation 
is characteristic for the given differential. We solve it and get complex roots 
,
. Functions 
and 
are linearly independent solutions of the differential equation. The general solution of this equation has the form.
Let the roots of the characteristic equation be real and equal, i.e. 
. Then the solutions of equation (2) are the functions 
and 
. These solutions are linearly independent, since the expression can be identically equal to zero only when 
and 
. Therefore, the general solution of equation (2) has the form 
.
Example 6
. Find the general solution of the differential equation 
.
Decision
. Characteristic equation 
has equal roots 
. In this case, the linearly independent solutions of the differential equation are the functions 
and 
. The general solution has the form 
.
Inhomogeneous second-order linear differential equations with constant coefficients
and special right side
The general solution of the linear inhomogeneous equation (1) is equal to the sum of the general solution 
corresponding homogeneous equation and any particular solution 
inhomogeneous equation: 
.
In some cases, a particular solution of an inhomogeneous equation can be found quite simply by the form of the right side 
equations (1). Let's consider cases when it is possible.
those. the right side of the inhomogeneous equation is a polynomial of degree m. If a 
is not a root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form of a polynomial of degree m, i.e.
Odds 
are determined in the process of finding a particular solution.
If 
is the root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form
Example 7
. Find the general solution of the differential equation 
.
Decision
. The corresponding homogeneous equation for this equation is 
. Its characteristic equation 
has roots 
and 
. The general solution of the homogeneous equation has the form 
.
As 
is not a root of the characteristic equation, then we will seek a particular solution of the inhomogeneous equation in the form of a function 
. Find the derivatives of this function 
,
and substitute them into this equation:
or . Equate the coefficients at 
and free members: 
Solving this system, we get 
,
. Then a particular solution of the inhomogeneous equation has the form 
, and the general solution of this inhomogeneous equation will be the sum of the general solution of the corresponding homogeneous equation and the particular solution of the inhomogeneous one: 
.
Let the inhomogeneous equation have the form
If a 
is not a root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form. If 
is the root of the characteristic multiplicity equation k
(k=1 or k=2), then in this case the particular solution of the inhomogeneous equation will have the form .
Example 8
. Find the general solution of the differential equation 
.
Decision
. The characteristic equation for the corresponding homogeneous equation has the form 
. its roots 
,
. In this case, the general solution of the corresponding homogeneous equation is written as 
.
Since the number 3 is not the root of the characteristic equation, then a particular solution of the inhomogeneous equation should be sought in the form 
. Let's find derivatives of the first and second orders:,
Substitute into the differential equation: 
+
+,
+,.
Equate the coefficients at 
and free members:
From here 
,
. Then a particular solution of this equation has the form 
, and the general solution
.
Lagrange method of variation of arbitrary constants
The method of variation of arbitrary constants can be applied to any inhomogeneous linear equation with constant coefficients, regardless of the form of the right side. This method makes it possible to always find a general solution to an inhomogeneous equation if the general solution of the corresponding homogeneous equation is known.
Let be 
and 
are linearly independent solutions of Eq. (2). Then the general solution to this equation is 
, where 
and 
are arbitrary constants. The essence of the method of variation of arbitrary constants is that the general solution of equation (1) is sought in the form
where 
and 
- new unknown features to be found. Since there are two unknown functions, two equations containing these functions are needed to find them. These two equations make up the system
which is a linear algebraic system of equations with respect to 
and 
. Solving this system, we find 
and 
. Integrating both parts of the obtained equalities, we find
and 
.
Substituting these expressions into (9), we obtain the general solution of the inhomogeneous linear equation (1).
Example 9
. Find the general solution of the differential equation 
.
Decision.
The characteristic equation for the homogeneous equation corresponding to the given differential equation is 
. Its roots are complex 
,
. As 
and 
, then 
,
, and the general solution of the homogeneous equation has the form Then the general solution of this inhomogeneous equation will be sought in the form where 
and 
- unknown functions.
The system of equations for finding these unknown functions has the form
Solving this system, we find 
,
. Then
,
. Let us substitute the obtained expressions into the general solution formula:
This is the general solution of this differential equation obtained by the Lagrange method.
Questions for self-control of knowledge
Which differential equation is called a second-order linear differential equation with constant coefficients?
Which linear differential equation is called homogeneous, and which one is called non-homogeneous?
What are the properties of a linear homogeneous equation?
What equation is called characteristic for a linear differential equation and how is it obtained?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of different roots of the characteristic equation?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of equal roots of the characteristic equation?
In what form is the general solution of a linear homogeneous differential equation with constant coefficients written in the case of complex roots of the characteristic equation?
How is the general solution of a linear inhomogeneous equation written?
In what form is a particular solution of a linear inhomogeneous equation sought if the roots of the characteristic equation are different and not equal to zero, and the right side of the equation is a polynomial of degree m?
In what form is a particular solution of a linear inhomogeneous equation sought if there is one zero among the roots of the characteristic equation, and the right side of the equation is a polynomial of degree m?
What is the essence of the Lagrange method?
