Handbook of mathematics for engineers and scienteists part 180 pps

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