T5.2. SECOND-ORDER LINEAR EQUATIONS 1221 8. y xx + (a –2q cos 2x)y =0. Mathieu equation. 1 ◦ . Given numbers a and q, there exists a general solution y(x) and a characteristic index μ such that y(x + π)=e 2πμ y(x). For small values of q, an approximate value of μ can be found from the equation cosh(πμ)=1 + 2 sin 2 1 2 π √ a + πq 2 (1 – a) √ a sin π √ a + O(q 4 ). If y 1 (x) is the solution of the Mathieu equation satisfying the initial conditions y 1 (0)=1 and y 1 (0)=0, the characteristic index can be determined from the relation cosh(2πμ)=y 1 (π). The solution y 1 (x), and hence μ, can be determined with any degree of accuracy by means of numerical or approximate methods. The general solution differs depending on the value of y 1 (π) and can be expressed in terms of two auxiliary periodical functions ϕ 1 (x)andϕ 2 (x)(seeTableT5.3). TABLE T5.3 The general solution of the Mathieu equation T5.2.2.8 expressed in terms of auxiliary periodical functions ϕ 1 (x)andϕ 2 (x) Constraint General solution y = y(x) Period of ϕ 1 and ϕ 2 Index y 1 (π)>1 C 1 e 2μx ϕ 1 (x)+C 2 e –2μx ϕ 2 (x) π μ is a real number y 1 (π)<–1 C 1 e 2ρx ϕ 1 (x)+C 2 e –2ρx ϕ 2 (x) 2π μ = ρ+ 1 2 i, i 2 =–1, ρ is the real part of μ |y 1 (π)| < 1 (C 1 cos νx+ C 2 sin νx)ϕ 1 (x) +(C 1 cos νx– C 2 sin νx)ϕ 2 (x) π μ = iν is a pure imaginary number, cos(2πν)=y 1 (π) y 1 (π)= 1 C 1 ϕ 1 (x)+C 2 xϕ 2 (x) π μ = 0 2 ◦ . In applications, of major interest are periodical solutions of the Mathieu equation that exist for certain values of the parameters a and q (those values of a are referred to as eigenvalues). The most important periodical solutions of the Mathieu equation have the form ce 2n (x, q)= ∞ m=0 A 2n 2m cos(2mx), ce 2n+1 (x, q)= ∞ m=0 A 2n+1 2m+1 cos (2m + 1)x ; se 2n (x, q)= ∞ m=0 B 2n 2m sin(2mx), se 2n+1 (x, q)= ∞ m=0 B 2n+1 2m+1 sin (2m + 1)x ; where A i j and B i j are constants determined by recurrence relations. The Mathieu functions ce 2n (x, q)andse 2n (x, q) are discussed in Section 18.16 in more detail. 1222 ORDINARY DIFFERENTIAL EQUATIONS 9. y xx + a tan xy x + by =0. 1 ◦ . The substitution ξ =sinx leads to a linear equation of the form T5.2.1.21: (ξ 2 –1)y ξξ + (1 – a)ξy ξ – by = 0. 2 ◦ . Solution for a =–2: y cos x = C 1 sin(kx)+C 2 cos(kx)ifb + 1 = k 2 > 0, C 1 sinh(kx)+C 2 cosh(kx)ifb + 1 =–k 2 < 0. 3 ◦ . Solution for a = 2 and b = 3: y = C 1 cos 3 x + C 2 sin x (1 + 2 cos 2 x). T5.2.3. Equations Involving Arbitrary Functions Notation: f = f(x) and g = g(x) are arbitrary functions; a , b ,and λ are arbitrary parameters. 1. y xx + fy x + a(f – a)y =0. Particular solution: y 0 = e –ax . 2. y xx + xfy x – fy =0. Particular solution: y 0 = x. 3. xy xx + (xf + a)y x + (a –1)fy =0. Particular solution: y 0 = x 1–a . 4. xy xx + [(ax +1)f + ax –1]y x + a 2 xfy =0. Particular solution: y 0 =(ax + 1)e –ax . 5. xy xx + [(ax 2 + bx)f +2]y x + bfy =0. Particular solution: y 0 = a + b/x. 6. x 2 y xx + xfy x + a(f – a –1)y =0. Particular solution: y 0 = x –a . 7. y xx + (f + ae λx )y x + ae λx (f + λ)y =0. Particular solution: y 0 =exp – a λ e λx . 8. y xx – (f 2 + f x )y =0. Particular solution: y 0 =exp fdx . 9. y xx +2fy x + (f 2 + f x )y =0. Solution: y =(C 2 x + C 1 )exp – fdx . T5.3. SECOND-ORDER NONLINEAR EQUATIONS 1223 10. y xx + (1–a)fy x – a(f 2 + f x )y =0. Particular solution: y 0 =exp a fdx . 11. y xx + fy x + (fg – g 2 + g x )y =0. Particular solution: y 0 =exp – gdx . 12. fy xx – af x y x – bf 2a+1 y =0. Solution: y = C 1 e u + C 2 e –u ,whereu = √ b f a dx. 13. f 2 y xx + f(f x + a)y x + by =0. The substitution ξ = f –1 dx leads to a constant coefficient linear equation: y ξξ +ay ξ +by = 0. 14. y xx – f x y x + a 2 e 2f y =0. Solution: y = C 1 sin a e f dx + C 2 cos a e f dx . 15. y xx – f x y x – a 2 e 2f y =0. Solution: y = C 1 exp a e f dx + C 2 exp –a e f dx . T5.3. Second-Order Nonlinear Equations T5.3.1. Equations of the Form y xx = f(x, y) 1. y xx = f(y). Autonomous equation. Solution: C 1 + 2 f(y) dy –1/2 dy = C 2 x. Particular solutions: y = A k ,whereA k are roots of the algebraic (transcendental) equation f(A k )=0. 2. y xx = Ax n y m . Emden–Fowler equation. 1 ◦ . With m ≠ 1, the Emden–Fowler equation has a particular solution: y = λx n+2 1–m ,whereλ = (n + 2)(n + m + 1) A(m – 1) 2 1 m–1 . 2 ◦ . The transformation z = x n+2 y m–1 , w = xy x /y leads to a first-order (Abel) equation: z[(m – 1)w + n + 2]w z =–w 2 + w + Az. 3 ◦ . The transformation y = w/t, x = 1/t leads to the Emden–Fowler equation with the independent variable raised to a different power: w tt = At –n–m–3 w m . 4 ◦ . The books by Zaitsev and Polyanin (1994) and Polyanin and Zaitsev (2003) present 28 solvable cases of the Emden–Fowler equation (corresponding to some pairs of n and m). 1224 ORDINARY DIFFERENTIAL EQUATIONS 3. y xx + f(x)y = ay –3 . Ermakov’s equation. Let w = w(x) be a nontrivial solution of the second-order linear equation w xx + f(x)w = 0. The transformation ξ = dx w 2 , z = y w leads to an autonomous equation of the form T5.3.1: z ξξ = az –3 . Solution: C 1 y 2 = aw 2 + w 2 C 2 + C 1 dx w 2 2 . Further on, f , g , h ,and ψ are arbitrary composite functions of their arguments indicated in parentheses after the function name (the arguments can depend on x , y , y x ). 4. y xx = f(ay + bx + c). The substitution w = ay + bx + c leads to an equation of the form T5.3.1.1: w xx = af(w). 5. y xx = f(y + ax 2 + bx + c). The substitution w=y+ax 2 +bx+c leadstoanequationoftheformT5.3.1.1: w xx =f(w)+2a. 6. y xx = x –1 f(yx –1 ). Homogeneous equation. The transformation t =–ln|x|, z = y/x leads to an autonomous equation: z tt – z t = f (z). 7. y xx = x –3 f(yx –1 ). The transformation ξ =1/x, w =y/x leads to the equation of the form T5.3.1.1: w ξξ =f(w). 8. y xx = x –3/2 f(yx –1/2 ). Having set w = yx –1/2 , we obtain d dx (xw x ) 2 = 1 2 ww x + 2f(w)w x . Integrating the latter equation, we arrive at a separable equation. Solution: C 1 + 1 4 w 2 + 2 f(w) dw –1/2 dw = C 2 ln x. 9. y xx = x k–2 f(x –k y). Generalized homogeneous equation. The transformation z = x –k y, w = xy x /y leads to a first-order equation: z(w – k)w z = z –1 f(z)+w – w 2 . 10. y xx = yx –2 f(x n y m ). Generalized homogeneous equation. The transformation z = x n y m , w = xy x /y leads to a first-order equation: z(mw + n)w z = f (z)+w – w 2 . 11. y xx = y –3 f y √ ax 2 + bx + c . Setting u(x)=y(ax 2 + bx + c) –1/2 and integrating the equation, we obtain a first-order separable equation: (ax 2 + bx + c) 2 (u x ) 2 =( 1 4 b 2 – ac)u 2 + 2 u –3 f(u) du + C 1 . T5.3. SECOND-ORDER NONLINEAR EQUATIONS 1225 12. y xx = e –ax f(e ax y). The transformation z = e ax y, w = y x /y leads to a first-order equation: z(w + a)w z = z –1 f(z)–w 2 . 13. y xx = yf(e ax y m ). The transformation z = e ax y m , w = y x /y leads to a first-order equation: z(mw + a)w z = f(z)–w 2 . 14. y xx = x –2 f(x n e ay ). The transformation z = x n e ay , w = xy x leads to a first-order equation: z(aw + n)w z = f(z)+w. 15. y xx = ψ xx ψ y + ψ –3 f y ψ , ψ = ψ(x). The transformation ξ = dx ψ 2 , w = y ψ leadstoanequationoftheformT5.3.1.1: w ξξ =f(w). Solution: C 1 + 2 f(w) dw –1/2 dw = C 2 dx ψ 2 (x) . T5.3.2. Equations of the Form f (x, y)y xx = g(x, y, y x ) 1. y xx – y x = f(y). Autonomous equation. The substitution w(y)=y x leads to a first-order equation. For solvable equations of the form in question, see the book by Polyanin and Zaitsev (2003). 2. y xx + f(y)y x + g(y) =0. Lienard equation. The substitution w(y)=y x leads to a first-order equation. For solvable equations of the form in question, see the book by Polyanin and Zaitsev (2003). 3. y xx + [ay + f (x)]y x + f x (x)y =0. Integrating yields a Riccati equation: y x + f (x)y + 1 2 ay 2 = C. 4. y xx + [2ay + f(x)]y x + af(x)y 2 = g(x). On setting u = y x + ay 2 , we obtain a first-order linear equation: u x + f (x)u = g(x). 5. y xx = ay x + e 2ax f(y). Solution: C 1 + 2 f(y) dy –1/2 dy = C 2 1 a e ax . 6. y xx = f(y)y x . Solution: dy F (y)+C 1 = C 2 + x,whereF (y)= f(y) dy. 1226 ORDINARY DIFFERENTIAL EQUATIONS 7. y xx = e αx f(y) + α y x . The substitution w(y)=e –αx y x leads to a first-order separable equation: w y = f(y). Solution: dy F (y)+C 1 = C 2 + 1 α e αx ,whereF (y)= f(y) dy. 8. xy xx = ny x + x 2n+1 f(y). 1 ◦ . Solution for n ≠ –1: C 1 + 2 f(y) dy –1/2 dy = x n+1 n + 1 + C 2 . 2 ◦ . Solution for n =–1: C 1 + 2 f(y) dy –1/2 dy = ln |x| + C 2 . 9. xy xx = f(y)y x . The substitution w(y)=xy x /y leads to a first-order linear equation: yw y =–w + 1 + f(y). 10. xy xx = x k f(y) + k –1 y x . Solution: dy F (y)+C 1 = C 2 + 1 k x k ,whereF (y)= f(y) dy. 11. x 2 y xx + xy x = f(y). The substitution x = e t leads to an autonomous equation of the form T5.3.1.1: y tt = f(y). 12. (ax 2 + b)y xx + axy x + f(y) =0. The substitution ξ = dx √ ax 2 + b leads to an autonomous equation of the form T5.3.1.1: y ξξ + f(y)=0. 13. y xx = f(y)y x + g(x). Integrating yields a first-order equation: y x = f(y) dy + g(x) dx + C. 14. xy xx + (n +1)y x = x n–1 f(yx n ). The transformation ξ = x n , w = yx n leads to an autonomous equation of the form T5.3.1.1: n 2 w ξξ = f (w). 15. gy xx + 1 2 g x y x = f(y), g = g(x). Integrating yields a first-order separable equation: g(x)(y x ) 2 = 2 f(y) dy + C 1 . Solution for g(x) ≥ 0: C 1 + 2 f(y) dy –1/2 dy = C 2 dx √ g(x) . T5.3. SECOND-ORDER NONLINEAR EQUATIONS 1227 16. y xx =–ay x + e ax f(ye ax ). The transformation ξ = e ax , w = ye ax leads to the equation w ξξ = a –2 f(w), which is of the form of T5.3.1.1. 17. xy xx = f(x n e ay )y x . The transformation z = x n e ay , w = xy x leads to the following first-order separable equation: z(aw + n)w z =[f(z)+1]w. 18. x 2 y xx + xy x = f(x n e ay ). The transformation z = x n e ay , w = xy x leads to the following first-order separable equation: z(aw + n)w z = f (z). 19. yy xx + (y x ) 2 + f(x)yy x + g(x) =0. The substitution u = y 2 leads to a linear equation, u xx + f(x)u x + 2g(x)=0, which can be reduced by the change of variable w(x)=u x to a first-order linear equation. 20. yy xx – (y x ) 2 + f(x)yy x + g(x)y 2 =0. The substitution u = y x /y leads to a first-order linear equation: u x + f (x)u + g(x)=0. 21. yy xx – n(y x ) 2 + f(x)y 2 + ay 4n–2 =0. 1 ◦ .Forn = 1, this is an equation of the form T5.3.2.22. 2 ◦ .Forn ≠ 1, the substitution w = y 1–n leads to Ermakov’s equation T5.3.1.5: w xx + (1 – n)f(x)w + a(1 – n)w –3 = 0. 22. yy xx – n(y x ) 2 + f(x)y 2 + g(x)y n+1 =0. The substitution w = y 1–n leads to a nonhomogeneous linear equation: w xx +(1–n)f(x)w+ (1 – n)g(x)=0. 23. yy xx + a(y x ) 2 + f(x)yy x + g(x)y 2 =0. The substitution w = y a+1 leads to a linear equation: w xx + f(x)w x +(a + 1)g(x)w = 0. 24. yy xx = f(x)(y x ) 2 . The substitution w(x)=xy x /y leads to aBernoulli equation T5.1.4: xw x =w+[f(x)–1]w 2 . 25. y xx – a(y x ) 2 + f(x)e ay + g(x) =0. The substitution w = e –ay leads to a nonhomogeneous linear equation: w xx – ag(x)w = af(x). 26. y xx – a(y x ) 2 + be 4ay + f(x) =0. The substitution w = e –ay leads to Ermakov’s equation T5.3.1.5: w xx – af (x)w = abw –3 . 27. y xx + a(y x ) 2 – 1 2 y x = e x f(y). The substitution w(y)=e –x (y x ) 2 leads to a first-order linear equation: w y + 2aw = 2f (y). . = 0 2 ◦ . In applications, of major interest are periodical solutions of the Mathieu equation that exist for certain values of the parameters a and q (those values of a are referred to as eigenvalues) determined with any degree of accuracy by means of numerical or approximate methods. The general solution differs depending on the value of y 1 (π) and can be expressed in terms of two auxiliary periodical. equation. 1 ◦ . Given numbers a and q, there exists a general solution y(x) and a characteristic index μ such that y(x + π)=e 2πμ y(x). For small values of q, an approximate value of μ can be found from