514 ORDINARY DIFFERENTIAL EQUATIONS Example 1. The eigenvalue problem for the equation y  xx + λ(1 + x 2 ) –2 y = 0 with the boundary conditions (12.3.7.8) admits an exact analytic solution and has eigenvalues λ 1 = 15, λ 2 = 63, , λ n = 16n 2 – 1. According to the Rayleigh–Ritz principle, formula (12.2.5.6) for z =sin(πx) yields the approximate value λ 0 1 = 15.33728. The solution of the Cauchy problem (12.3.7.9)–(12.3.7.10) with f(x)=1, g(x)=λ(1 + x 2 ) –2 , h(x)=0 yields x 0 = 0.983848, 1 – x 0 = 0.016152, y 2 = 0.024585, y  x (x 0 )=–0.70622822. The first iteration for the first eigenvalue is determined by (12.3.7.11) and results in the value λ 1 1 = 14.99245 with the relative error Δλ/λ 1 1 = 5×10 –4 . The second iteration results in λ 2 1 = 14.999986 with the relative error Δλ/λ 2 1 < 10 –6 . Example 2. Consider the eigenvalue problem for the equation ( √ 1 + xy  x )  x + λy = 0 with the boundary conditions (12.3.7.8). The Rayleigh–Ritz principle yields λ 0 1 = 11.995576. The next two iterations result in the values λ 1 1 = 11.898578 and λ 2 1 = 11.898458. For the relative error we have Δλ/λ 2 1 < 10 –5 .  For more details about finite-difference methods and other numerical methods, see, for instance, the books by Lambert (1973), Keller (1976), Schiesser (1993), and Zwillinger (1997). 12.4. Linear Equations of Arbitrary Order 12.4.1. Linear Equations with Constant Coefficients 12.4.1-1. Homogeneous linear equations. An nth-order homogeneous linear equation with constant coefficients has the general form y (n) x + a n–1 y (n–1) x + ···+ a 1 y  x + a 0 y = 0.(12.4.1.1) The general solution of this equation is determined by the roots of the characteristic equation P (λ)=0,whereP (λ)=λ n + a n–1 λ n–1 + ···+ a 1 λ + a 0 .(12.4.1.2) The following cases are possible: 1 ◦ . All roots λ 1 , λ 2 , , λ n of the characteristic equation (12.4.1.2) are real and distinct. Then the general solution of the homogeneous linear differential equation (12.4.1.1) has the form y = C 1 exp(λ 1 x)+C 2 exp(λ 2 x)+···+ C n exp(λ n x). 2 ◦ .Therearem equal real roots λ 1 = λ 2 = ···= λ m (m ≤ n), and the other roots are real and distinct. In this case, the general solution is given by y =exp(λ 1 x)(C 1 + C 2 x + ···+ C m x m–1 ) + C m+1 exp(λ m+1 x)+C m+2 exp(λ m+2 x)+···+ C n exp(λ n x). 3 ◦ .Therearem equal complex conjugate roots λ = α iβ (2m ≤ n), and the other roots are real and distinct. In this case, the general solution is y =exp(αx)cos(βx)(A 1 + A 2 x + ···+ A m x m–1 ) +exp(αx)sin(βx)(B 1 + B 2 x + ···+ B m x m–1 ) + C 2m+1 exp(λ 2m+1 x)+C 2m+2 exp(λ 2m+2 x)+···+ C n exp(λ n x), where A 1 , , A m , B 1 , , B m , C 2m+1 , , C n are arbitrary constants. 12.4. LINEAR EQUATIONS OF ARBITRARY ORDER 515 4 ◦ . In the general case, where there are r different roots λ 1 , λ 2 , , λ r of multiplicities m 1 , m 2 , , m r , respectively, the left-hand side of the characteristic equation (12.4.1.2) can be represented as the product P (λ)=(λ – λ 1 ) m 1 (λ – λ 2 ) m 2 (λ – λ r ) m r , where m 1 + m 2 + ···+ m r = n. The general solution of the original equation is given by the formula y = r  k=1 exp(λ k x)(C k,0 + C k,1 x + ···+ C k,m k –1 x m k –1 ), where C k,l are arbitrary constants. If the characteristic equation (12.4.1.2) has complex conjugate roots, then in the above solution, one should extract the real part on the basis of the relation exp(α iβ)=e α (cos β i sin β). Example 1. Find the general solution of the linear third-order equation y  + ay  – y  – ay = 0. Its characteristic equation is λ 3 + aλ 2 – λ – a = 0, or, in factorized form, (λ + a)(λ – 1)(λ + 1)=0. Depending on the value of the parameter a, three cases are possible. 1. Case a ≠ 1. There are three different roots, λ 1 =–a, λ 2 =–1,andλ 3 = 1. The general solution of the differential equation is expressed as y = C 1 e –ax + C 2 e –x + C 3 e x . 2. Case a = 1. There is a double root, λ 1 = λ 2 =–1, and a simple root, λ 3 = 1. The general solution of the differential equation has the form y =(C 1 + C 2 x)e –x + C 3 e x . 3. Case a =–1. There is a double root, λ 1 = λ 2 = 1, and a simple root, λ 3 =–1. The general solution of the differential equation is expressed as y =(C 1 + C 2 x)e x + C 3 e –x . Example 2. Consider the linear fourth-order equation y  xxxx – y = 0. Its characteristic equation, λ 4 – 1 = 0, has four distinct roots, two real and two pure imaginary, λ 1 = 1, λ 2 =–1, λ 3 = i, λ 4 =–i. Therefore the general solution of the equation in question has the form (see Item 3 ◦ ) y = C 1 e x + C 2 e –x + C 3 sin x + C 4 cos x. 12.4.1-2. Nonhomogeneous linear equations. Forms of particular solutions. 1 ◦ .Annth-order nonhomogeneous linear equation with constant coefficients has the gen- eral form y (n) x + a n–1 y (n–1) x + ···+ a 1 y  x + a 0 y = f (x). (12.4.1.3) The general solution of this equation is the sum of the general solution of the corre- sponding homogeneous equation with f(x) ≡ 0 (see Paragraph 12.4.1-1) and any particular solution of the nonhomogeneous equation (12.4.1.3). If all the roots λ 1 , λ 2 , , λ n of the characteristic equation (12.4.1.2) are different, equation (12.4.1.3) has the general solution: y = n  ν=1 C ν e λ ν x + n  ν=1 e λ ν x P  λ (λ ν )  f(x)e –λ ν x dx (12.4.1.4) (for complex roots, the real part should be taken). In the general case, if the characteristic equation (12.4.1.2) has multiple roots, the solution to equation (12.4.1.3) can be constructed using formula (12.4.2.5). 516 ORDINARY DIFFERENTIAL EQUATIONS TABLE 12.4 Forms of particular solutions of the constant coefficient nonhomogeneous linear equation y (n) x + a n–1 y (n–1) x + ···+ a 1 y  x + a 0 y = f(x) that correspond to some special forms of the function f(x) Form of the function f (x) Roots of the characteristic equation λ n + a n–1 λ n–1 + ···+ a 1 λ + a 0 = 0 Form of a particular solution y = y(x) Zero is not a root of the characteristic equation (i.e., a 0 ≠ 0)  P m (x) P m (x) Zero is a root of the characteristic equation (multiplicity r) x r  P m (x) α is not a root of the characteristic equation  P m (x)e αx P m (x)e αx (α is a real constant) α is a root of the characteristic equation (multiplicity r) x r  P m (x)e αx iβ is not a root of the characteristic equation  P ν (x)cosβx +  Q ν (x)sinβx P m (x)cosβx + Q n (x)sinβx iβ is a root of the characteristic equation (multiplicity r) x r [  P ν (x)cosβx +  Q ν (x)sinβx] α + iβ is not a root of the characteristic equation [  P ν (x)cosβx +  Q ν (x)sinβx]e αx [P m (x)cosβx + Q n (x)sinβx]e αx α + iβ is a root of the characteristic equation (multiplicity r) x r [  P ν (x)cosβx +  Q ν (x)sinβx]e αx Notation: P m and Q n are polynomials of degrees m and n with given coefficients;  P m ,  P ν ,and  Q ν are polynomials of degrees m and ν whose coefficients are determined by substituting the particular solution into the basic equation; ν =max(m, n); and α and β are real numbers, i 2 =–1. 2 ◦ . Table 12.4 lists the forms of particular solutions corresponding to some special forms of functions on the right-hand side of the linear nonhomogeneous equation. 3 ◦ . Consider the Cauchy problem for equation (12.4.1.3) subject to the homogeneous initial conditions y(0)=y  x (0)= = y (n–1) x (0)=0.(12.4.1.5) Let y(x) be the solution of problem (12.4.1.3), (12.4.1.5) for arbitrary f(x)andletu(x)be the solution of the auxiliary, simpler problem (12.4.1.3), (12.4.1.5) with f (x) ≡ 1,sothat u(x)=y(x)| f(x)≡1 . Then the formula y(x)=  x 0 f(t)u  x (x – t) dt holds. It is called the Duhamel integral. 12.4.1-3. Solution of the Cauchy problem using the Laplace transform. Consider the Cauchy problem for equation (12.4.1.3) with arbitrary initial conditions y(0)=y 0 , y  x (0)=y 1 , , y (n–1) x (0)=y n–1 ,(12.4.1.6) where y 0 , y 1 , , y n–1 are given constants. 12.4. LINEAR EQUATIONS OF ARBITRARY ORDER 517 Problem (12.4.1.3), (12.4.1.6) can be solved using the Laplace transform based on the formulas (for details, see Section 11.2) y(p)=L  y(x)  ,  f(p)=L  f(x)  ,whereL  f(x)  ≡  ∞ 0 e –px f(x) dx. To this end, let us multiply equation (12.4.1.3) by e –px and then integrate with respect to x from zero to infinity. Taking into account the differentiation rule L  y (n) x (x)  = p n y(p)– n  k=1 p n–k y (k–1) x (+0) and the initial conditions (12.4.1.6), we arrive at a linear algebraic equation for the trans- form y(p): P (p)y(p)–Q(p)=  f(p), (12.4.1.7) where P (p)=p n + a n–1 p n–1 + ···+ a 1 p + a 0 , Q(p)=b n–1 p n–1 + ···+ b 1 p + b 0 , b k = y n–k–1 + a n–1 y n–k–2 + ···+ a k+2 y 1 + a k+1 y 0 , k = 0, 1, , n – 1. The polynomial P (p) coincides with the characteristic polynomial (12.4.1.2) at λ = p. The solution of equation (12.4.1.7) is given by the formula y(p)=  f(p)+Q(p)  P (p) .(12.4.1.8) On applying the Laplace inversion formula (see in Section 11.2) to (12.4.1.8), we obtain a solution to problem (12.4.1.3), (12.4.1.6) in the form y(x)= 1 2πi  c+i∞ c–i∞  f(p)+Q(p)  P (p) e px dp.(12.4.1.9) Since the transform y(p) (12.4.1.8) is a rational function, the inverse Laplace transform (12.4.1.9) can be obtained using the formulas from Paragraph 11.2.2-2 or the tables of Section T3.2. Remark. In practice, the solution method for the Cauchy problem based on the Laplace transform leads to the solution faster than the direct application of general formulas like (12.4.1.4), where one has to determine the coefficients C 1 , , C n . Example 3. Consider the following Cauchy problem for a homogeneous fourth-order equation: y  xxxx + a 4 y = 0; y(0)=y  x (0)=y  xxx (0)=0, y  xx (0)=b. Using the Laplace transform reduces this problem to a linear algebraic equation for y(p): (p 4 + a 4 )y(p)– bp = 0. It follows that y(p)= bp p 4 + a 4 . In order to invert this expression, let us use the table of inverse Laplace transforms T3.2.2 (see row 52) and take into account that a constant multiplier can be taken outside the transform operator to obtain the solution to the original Cauchy problem in the form y(x)= b a 2 sin  ax √ 2  sinh  ax √ 2  . 518 ORDINARY DIFFERENTIAL EQUATIONS 12.4.2. Linear Equations with Variable Coefficients 12.4.2-1. Homogeneous linear equations. Structure of the general solution. The general solution of the nth-order homogeneous linear differential equation f n (x)y (n) x + f n–1 (x)y (n–1) x + ···+ f 1 (x)y  x + f 0 (x)y = 0 (12.4.2.1) has the form y = C 1 y 1 (x)+C 2 y 2 (x)+···+ C n y n (x). (12.4.2.2) Here, y 1 (x), y 2 (x), , y n (x) is a fundamental system of solutions (the y k are linearly independent particular solutions, y k 0); C 1 , C 2 , , C n are arbitrary constants. 12.4.2-2. Utilization of particular solutions for reducing the order of the equation. 1 ◦ .Lety 1 =y 1 (x) be a nontrivial particular solution of equation (12.4.2.1). The substitution y = y 1 (x)  z(x) dx results in a linear equation of order n – 1 for the function z(x). 2 ◦ .Lety 1 = y 1 (x)andy 2 = y 2 (x) be two nontrivial linearly independent solutions of equation (12.4.2.1). The substitution y = y 1  y 2 wdx– y 2  y 1 wdx results in a linear equation of order n – 2 for w(x). 3 ◦ . Suppose that m linearly independent solutions y 1 (x), y 2 (x), , y m (x) of equation (12.4.2.1) are known. Then one can reduce the order of the equation to n –m by successive application of the following procedure. The substitution y = y m (x)  z(x) dx leads to an equation of order n – 1 for the function z(x) with known linearly independent solutions: z 1 =  y 1 y m   x , z 2 =  y 2 y m   x , , z m–1 =  y m–1 y m   x . The substitution z = z m–1 (x)  w(x) dx yields an equation of order n – 2. Repeating this procedure m times, we arrive at a homogeneous linear equation of order n – m. 12.4.2-3. Wronskian determinant and Liouville formula. The Wronskian determinant (or simply, Wronskian) is the function defined as W (x)=        y 1 (x) ··· y n (x) y  1 (x) ··· y  n (x) ··· ··· ··· y (n–1) 1 (x) ··· y (n–1) n (x)        ,(12.4.2.3) where y 1 (x), , y n (x) is a fundamental system of solutions of the homogeneous equa- tion (12.4.2.1); y (m) k (x)= d m y k dx m , m = 1, , n – 1; k = 1, , n. The following Liouville formula holds: W (x)=W (x 0 )exp  –  x x 0 f n–1 (t) f n (t) dt  . 12.4. LINEAR EQUATIONS OF ARBITRARY ORDER 519 12.4.2-4. Nonhomogeneous linear equations. Construction of the general solution. 1 ◦ . The general nonhomogeneous nth-order linear differential equation has the form f n (x)y (n) x + f n–1 (x)y (n–1) x + ···+ f 1 (x)y  x + f 0 (x)y = g(x). (12.4.2.4) The general solution of the nonhomogeneous equation (12.4.2.4) can be represented as the sum of its particular solution and the general solution of the corresponding homogeneous equation (12.4.2.1). 2 ◦ .Lety 1 (x), , y n (x) be a fundamental system of solutions of the homogeneous equa- tion (12.4.2.1), and let W (x) be the Wronskian determinant (12.4.2.3). Then the general solution of the nonhomogeneous linear equation (12.4.2.4) can be represented as y = n  ν=1 C ν y ν (x)+ n  ν=1 y ν (x)  W ν (x) dx f n (x)W (x) ,(12.4.2.5) where W ν (x) is the determinant obtained by replacing the νth column of the matrix (12.4.2.3) by the column vector with the elements 0, 0, ,0,g. 3 ◦ .Lety(x, σ) be the solution to the Cauchy problem for the homogeneous equation (12.4.2.1) with nonhomogeneous initial conditions at x = σ: y(σ)=y  x (σ)=···= y (n–2) x (σ)=0, y (n–1) x (σ)=1, where σ is an arbitrary parameter. Then a particular solution of the nonhomogeneous linear equation (12.4.2.4) with homogeneous boundary conditions y(x 0 )=y  x (x 0 )=···= y (n–1) x (x 0 )=0 isgivenbytheCauchy formula ¯y(x)=  x x 0 y(x, σ) g(σ) f n (σ) dσ. 4 ◦ . Superposition principle. The solution of a nonhomogeneous linear equation L[y]= m  k=1 g k (x), L[y] ≡ f n (x)y (n) x + f n–1 (x)y (n–1) x + ···+ f 1 (x)y  x + f 0 (x)y is determined by adding together the solutions, y = m  k=1 y k , of m (simpler) equations, L[y k ]=g k (x), k = 1, 2, , m, corresponding to respective nonhomogeneous terms in the original equation. 520 ORDINARY DIFFERENTIAL EQUATIONS 12.4.2-5. Euler equation. 1 ◦ . The nonhomogeneous Euler equation has the form x n y (n) x + a n–1 x n–1 y (n–1) x + ···+ a 1 xy  x + a 0 y = f (x). The substitution x = be t (b ≠ 0) leads to a constant coefficient linear equation of the form (12.4.1.3). 2 ◦ . Particular solutions of the homogeneous Euler equation [with f(x) ≡ 0] are sought in the form y = x k .Ifallk are real and distinct, its general solution is expressed as y(x)=C 1 |x| k 1 + C 2 |x| k 2 + ···+ C n |x| k n . Remark. To a pair of complex conjugate values k = α iβ there corresponds a pair of particular solutions: y = |x| α sin(β|x|)andy = |x| α cos(β|x|). 12.4.2-6. Solution of equations using the Laplace transform. Laplace equation. 1 ◦ . Some classes of equations (12.4.2.1) or (12.4.2.4) with polynomial coefficients f k (x)= s k  m=0 a km x m may be solved using the Laplace transform (see Paragraph 12.4.1-3 and Section 11.2). To this end, one uses the following formula for the Laplace transform of the product of a power function and a derivative of the unknown function: L  x m y (n) x (x)  =(–1) m d m dp m  p n y(p)– n  k=1 p n–k y (k–1) x (+0)  .(12.4.2.6) The right-hand side contains initial data y (m) x (+0), m = 0, 1, , n – 1 (specified in the Cauchy problem). As a result, one arrives at a linear ordinary differential equation, with respect to p, for the transform y(p); the order of this equation is equal to max 1≤k≤n {s k },the highest degree of the polynomials that determine the equation coefficients. In some cases, the equation for y(p) turns out to be simpler than the initial equation for y(x) and can be solved in closed form. The desired function y(x) is found by inverting the transform y(p) using the formulas from Paragraph 11.2.2-2 of the tables from Section T3.2. 2 ◦ . Consider the Laplace equation (a n + b n x)y (n) x +(a n–1 + b n–1 x)y (n–1) x + ···+(a 1 + b 1 x)y  x +(a 0 + b 0 x)y = 0,(12.4.2.7) whose coefficients are linear functions of the independent variable x. The application of the Laplace transform, in view of formulas (12.4.2.6), brings it to a linear first-order ordinary differential equation for the transform y(p). . =max(m, n); and α and β are real numbers, i 2 =–1. 2 ◦ . Table 12.4 lists the forms of particular solutions corresponding to some special forms of functions on the right-hand side of the linear. r) x r [  P ν (x)cosβx +  Q ν (x)sinβx]e αx Notation: P m and Q n are polynomials of degrees m and n with given coefficients;  P m ,  P ν ,and  Q ν are polynomials of degrees m and ν whose coefficients are determined by substituting the particular. function f(x) Form of the function f (x) Roots of the characteristic equation λ n + a n–1 λ n–1 + ···+ a 1 λ + a 0 = 0 Form of a particular solution y = y(x) Zero is not a root of the characteristic