1228 ORDINARY DIFFERENTIAL EQUATIONS 28. y xx + α(y x ) 2 = e βx f(y) + β y x . Solution: e αy dy F (y)+C 1 = C 2 + 1 β e βx ,whereF (y)= e αy f(y) dy. 29. y xx + f(y)(y x ) 2 + g(y) =0. The substitution w(y)=(y x ) 2 leads to a first-order linear equation: w y +2f (y)w+2g(y)=0. 30. y xx + f(y)(y x ) 2 – 1 2 y x = e x g(y). The substitution w(y)=e –x (y x ) 2 leads to a first-order linear equation: w y +2f(y)w = 2g(y). 31. y xx = xf(y)(y x ) 3 . Taking y to be the independent variable, we obtain a linear equation with respect to x = x(y): x yy =–f(y)x. 32. y xx = f(y)(y x ) 2 + g(x)y x . Dividing by y x , we obtain an exact differential equation. Its solution follows from the equation: ln |y x | = f(y) dy + g(x) dx + C. Solving the latter for y x , we arrive at a separable equation. In addition, y = C 1 is a singular solution, with C 1 being an arbitrary constant. 33. y xx = f(x)g(xy x – y). The substitution w = xy x – y leads to a first-order separable equation: w x = xf (x)g(w). 34. y xx = y x 2 f xy x y . The substitution w(x)=xy x /y leadsto a first-order separable equation: xw x =f(w)+w–w 2 . 35. gy xx + 1 2 g x y x = f(y)h y x √ g , g = g(x). The substitution w(y)=y x √ g leads to a first-order separable equation: ww y = f (y)h(w). 36. y xx = f y 2 x + ay . The substitution w(y)=(y x ) 2 +ay leads to a first-order separable equation: w y = 2f(w)+a. References for Chapter T5 Kamke, E., Differentialgleichungen: L ¨ osungsmethoden und L ¨ osungen, I, Gew ¨ ohnliche Differentialgleichungen, B. G. Teubner, Leipzig, 1977. Murphy, G. M., Ordinary Differential Equations and Their Solutions, D. Van Nostrand, New York, 1960. Polyanin, A. D. and Zaitsev, V. F., Handbook of Exact Solutions for Ordinary Differential Equations, 2nd Edition, Chapman & Hall/CRC Press, Boca Raton, 2003. Zaitsev, V. F. and Polyanin, A. D., Discrete-Group Methods for Integrating Equations of Nonlinear Mechanics, CRC Press, Boca Raton, 1994. Chapter T6 Systems of Ordinary Differential Equations T6.1. Linear Systems of Two Equations T6.1.1. Systems of First-Order Equations 1. x t = ax + by, y t = cx + dy. System of two constant-coefficient first-order linear homogeneous differential equations. Let us write out the characteristic equation λ 2 –(a + d)λ + ad – bc = 0 (1) and find its discriminant D =(a – d) 2 + 4bc.(2) 1 ◦ .Casead – bc ≠ 0. The origin of coordinates x = y = 0 is the only one stationary point; it is a node if D = 0; a node if D > 0 and ad – bc > 0; a saddle if D > 0 and ad – bc < 0; a focus if D < 0 and a + d ≠ 0; a center if D < 0 and a + d = 0. 1.1. Suppose D > 0. The characteristic equation (1) has two distinct real roots, λ 1 and λ 2 . The general solution of the original system of differential equations is expressed as x = C 1 be λ 1 t + C 2 be λ 2 t , y = C 1 (λ 1 – a)e λ 1 t + C 2 (λ 2 – a)e λ 2 t , where C 1 and C 2 are arbitrary constants. 1.2. Suppose D < 0. The characteristic equation (1) has two complex conjugate roots, λ 1,2 = σ iβ. The general solution of the original system of differential equations is given by x = be σt C 1 sin(βt)+C 2 cos(βt) , y = e σt [(σ – a)C 1 – βC 2 ]sin(βt)+[βC 1 +(σ – a)C 2 cos(βt) , where C 1 and C 2 are arbitrary constants. 1.3. Suppose D = 0 and a ≠ d. The characteristic equation (1) has two equal real roots, λ 1 = λ 2 . The general solution of the original system of differential equations is x = 2b C 1 + C 2 a – d + C 2 t exp a + d 2 t , y =[(d – a)C 1 + C 2 +(d – a)C 2 t]exp a + d 2 t , 1229 1230 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS where C 1 and C 2 are arbitrary constants. 1.4. Suppose a = d ≠ 0 and b = 0. Solution: x = C 1 e at , y =(cC 1 t + C 2 )e at . 1.5. Suppose a = d ≠ 0 and c = 0. Solution: x =(bC 1 t + C 2 )e at , y = C 1 e at . 2 ◦ .Casead – bc = 0 and a 2 + b 2 > 0. The whole of the line ax + by = 0 consists of singular points. The system in question may be rewritten in the form x t = ax + by, y t = k(ax + by). 2.1. Suppose a + bk ≠ 0. Solution: x = bC 1 + C 2 e (a+bk)t , y =–aC 1 + kC 2 e (a+bk)t . 2.2. Suppose a + bk = 0. Solution: x = C 1 (bkt – 1)+bC 2 t, y = k 2 bC 1 t +(bk 2 t + 1)C 2 . 2. x t = a 1 x + b 1 y + c 1 , y t = a 2 x + b 2 y + c 2 . The general solution of this system is given by the sum of its any particular solution and the general solution of the corresponding homogeneous system (see system T6.1.1.1). 1 ◦ . Suppose a 1 b 2 – a 2 b 1 ≠ 0. A particular solution: x = x 0 , y = y 0 , where the constants x 0 and y 0 are determined by solving the linear algebraic system of equations a 1 x 0 + b 1 y 0 + c 1 = 0, a 2 x 0 + b 2 y 0 + c 2 = 0. 2 ◦ . Suppose a 1 b 2 – a 2 b 1 = 0 and a 2 1 + b 2 1 > 0. Then the original system can be rewritten as x t = ax + by + c 1 , y t = k(ax + by)+c 2 . 2.1. If σ = a + bk ≠ 0, the original system has a particular solution of the form x = bσ –1 (c 1 k – c 2 )t – σ –2 (ac 1 + bc 2 ), y = kx +(c 2 – c 1 k)t. 2.2. If σ = a + bk = 0, the original system has a particular solution of the form x = 1 2 b(c 2 – c 1 k)t 2 + c 1 t, y = kx +(c 2 – c 1 k)t. 3. x t = f(t)x + g(t)y, y t = g(t)x + f(t)y. Solution: x = e F (C 1 e G + C 2 e –G ), y = e F (C 1 e G – C 2 e –G ), where C 1 and C 2 are arbitrary constants, and F = f(t) dt, G = g(t) dt. T6.1. LINEAR SYSTEMS OF TWO EQUATIONS 1231 4. x t = f(t)x + g(t)y, y t =–g(t)x + f(t)y. Solution: x = F (C 1 cos G + C 2 sin G), y = F (–C 1 sin G + C 2 cos G), where C 1 and C 2 are arbitrary constants, and F =exp f(t) dt , G = g(t) dt. 5. x t = f(t)x + g(t)y, y t = ag(t)x + [f (t) + bg(t)]y. The transformation x =exp f(t) dt u, y =exp f(t) dt v, τ = g(t) dt leads to a system of constant coefficient linear differential equations of the form T6.1.1.1: u τ = v, v τ = au + bv. 6. x t = f(t)x + g(t)y, y t = a[f (t) + ah(t)]x + a[g(t) – h(t)]y. Let us multiply the first equation by –a and add it to the second equation to obtain y t – ax t =–ah(t)(y – ax). By setting U = y – ax and then integrating, one obtains y – ax = C 1 exp –a h(t) dt ,(∗) where C 1 is an arbitrary constant. On solving (∗)fory and on substituting the resulting expression into the first equation of the system, one arrives at a first-order linear differential equation for x. 7. x t = f(t)x + g(t)y, y t = h(t)x + p(t)y. 1 ◦ . Let us express y from the first equation and substitute into the second one to obtain a second-order linear equation: gx tt –(fg+ gp + g t )x t +(fgp– g 2 h + fg t – f t g)x = 0.(1) This equation is easy to integrate if, for example, the following conditions are met: 1) fgp– g 2 h + fg t – f t g = 0; 2) fgp– g 2 h + fg t – f t g = ag, fg + gp + g t = bg. In the first case, equation (1) has a particular solution u = C = const. In the second case, it is a constant-coefficient equation. A considerable number of other solvable cases of equation (1) can be found in the handbooks by Kamke (1977) and Polyanin and Zaitsev (2003). 1232 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS 2 ◦ . Suppose a particular solution of the system in question is known, x = x 0 (t), y = y 0 (t). Then the general solution can be written out in the form x(t)=C 1 x 0 (t)+C 2 x 0 (t) g(t)F (t)P (t) x 2 0 (t) dt, y(t)=C 1 y 0 (t)+C 2 F (t)P (t) x 0 (t) + y 0 (t) g(t)F (t)P (t) x 2 0 (t) dt , where C 1 and C 2 are arbitrary constants, and F (t)=exp f(t) dt , P(t)=exp p(t) dt . T6.1.2. Systems of Second-Order Equations 1. x tt = ax + by, y tt = cx + dy. System of two constant-coefficient second-order linear homogeneous differential equations. The characteristic equation has the form λ 4 –(a + d)λ 2 + ad – bc = 0. 1 ◦ .Casead – bc ≠ 0. 1.1. Suppose (a – d) 2 + 4bc ≠ 0. The characteristic equation has four distinct roots λ 1 , , λ 4 . The general solution of the system in question is written as x = C 1 be λ 1 t + C 2 be λ 2 t + C 3 be λ 3 t + C 4 be λ 4 t , y = C 1 (λ 2 1 – a)e λ 1 t + C 2 (λ 2 2 – a)e λ 2 t + C 3 (λ 2 3 – a)e λ 3 t + C 4 (λ 2 4 – a)e λ 4 t , where C 1 , , C 4 are arbitrary constants. 1.2. Solution with (a – d) 2 + 4bc = 0 and a ≠ d: x = 2C 1 bt + 2bk a – d e kt/2 + 2C 2 bt – 2bk a – d e –kt/2 + 2bC 3 te kt/2 + 2bC 4 te –kt/2 , y = C 1 (d – a)te kt/2 + C 2 (d – a)te –kt/2 + C 3 [(d – a)t + 2k]e kt/2 + C 4 [(d – a)t – 2k]e –kt/2 , where C 1 , , C 4 are arbitrary constants and k = √ 2(a + d). 1.3. Solution with a = d ≠ 0 and b = 0: x = 2 √ aC 1 e √ at + 2 √ aC 2 e – √ at , y = cC 1 te √ at – cC 2 te – √ at + C 3 e √ at + C 4 e – √ at . 1.4. Solution with a = d ≠ 0 and c = 0: x = bC 1 te √ at – bC 2 te – √ at + C 3 e √ at + C 4 e – √ at , y = 2 √ aC 1 e √ at + 2 √ aC 2 e – √ at . T6.1. LINEAR SYSTEMS OF TWO EQUATIONS 1233 2 ◦ .Casead – bc = 0 and a 2 + b 2 > 0. The original system can be rewritten in the form x tt = ax + by, y tt = k(ax + by). 2.1. Solution with a + bk ≠ 0: x = C 1 exp t √ a + bk + C 2 exp –t √ a + bk + C 3 bt + C 4 b, y = C 1 k exp t √ a + bk + C 2 k exp –t √ a + bk – C 3 at – C 4 a. 2.2. Solution with a + bk = 0: x = C 1 bt 3 + C 2 bt 2 + C 3 t + C 4 , y = kx + 6C 1 t + 2C 2 . 2. x tt = a 1 x + b 1 y + c 1 , y tt = a 2 x + b 2 y + c 2 . The general solution of this system is expressed as the sum of its any particular solution and the general solution of the corresponding homogeneous system (see system T6.1.2.1). 1 ◦ . Suppose a 1 b 2 – a 2 b 1 ≠ 0. A particular solution: x = x 0 , y = y 0 , where the constants x 0 and y 0 are determined by solving the linear algebraic system of equations a 1 x 0 + b 1 y 0 + c 1 = 0, a 2 x 0 + b 2 y 0 + c 2 = 0. 2 ◦ . Suppose a 1 b 2 – a 2 b 1 = 0 and a 2 1 + b 2 1 > 0. Then the system can be rewritten as x tt = ax + by + c 1 , y tt = k(ax + by)+c 2 . 2.1. If σ = a + bk ≠ 0, the original system has a particular solution x = 1 2 bσ –1 (c 1 k – c 2 )t 2 – σ –2 (ac 1 + bc 2 ), y = kx + 1 2 (c 2 – c 1 k)t 2 . 2.2. If σ = a + bk = 0, the system has a particular solution x = 1 24 b(c 2 – c 1 k)t 4 + 1 2 c 1 t 2 , y = kx + 1 2 (c 2 – c 1 k)t 2 . 3. x tt – ay t + bx =0, y tt + ax t + by =0. This system is used to describe the horizontal motion of a pendulum taking into account the rotation of the earth. Solution with a 2 + 4b > 0: x = C 1 cos(αt)+C 2 sin(αt)+C 3 cos(βt)+C 4 sin(βt), y =–C 1 sin(αt)+C 2 cos(αt)–C 3 sin(βt)+C 4 cos(βt), where C 1 , , C 4 are arbitrary constants and α = 1 2 a + 1 2 a 2 + 4b, β = 1 2 a – 1 2 a 2 + 4b. 1234 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS 4. x tt + a 1 x t + b 1 y t + c 1 x + d 1 y = k 1 e iωt , y tt + a 2 x t + b 2 y t + c 2 x + d 2 y = k 2 e iωt . Systems of this type often arise in oscillation theory (e.g., oscillations of a ship and a ship gyroscope). The general solution of this constant-coefficient linear nonhomogeneous system of differential equations is expressed as the sum of its any particular solution and the general solution of the corresponding homogeneous system (with k 1 = k 2 = 0). 1 ◦ . A particular solution is sought by the method of undetermined coefficients in the form x = A ∗ e iωt , y = B ∗ e iωt . On substituting these expressions into the system of differential equations in question, one arrives at a linear nonhomogeneous system of algebraic equations for the coefficients A ∗ and B ∗ . 2 ◦ . The general solution of a homogeneous system of differential equations is determined by a linear combination of its linearly independent particular solutions, which are sought using the method of undetermined coefficients in the form of exponential functions, x = Ae λt , y = Be λt . On substituting these expressions into the system and on collecting the coefficients of the unknowns A and B, one obtains (λ 2 + a 1 λ + c 1 )A +(b 1 λ + d 1 )B = 0, (a 2 λ + c 2 )A +(λ 2 + b 2 λ + d 2 )B = 0. For a nontrivial solution to exist, the determinant of this system must vanish. This require- ment results in the characteristic equation (λ 2 + a 1 λ + c 1 )(λ 2 + b 2 λ + d 2 )–(b 1 λ + d 1 )(a 2 λ + c 2 )=0, which is used to determine λ. If the roots of this equation, k 1 , , k 4 , are all distinct, then the general solution of the original system of differential equations has the form x =–C 1 (b 1 λ 1 + d 1 )e λ 1 t – C 2 (b 1 λ 2 + d 1 )e λ 2 t – C 3 (b 1 λ 1 + d 1 )e λ 3 t – C 4 (b 1 λ 4 + d 1 )e λ 4 t , y = C 1 (λ 2 1 + a 1 λ 1 + c 1 )e λ 1 t + C 2 (λ 2 2 + a 1 λ 2 + c 1 )e λ 2 t + C 3 (λ 2 3 + a 1 λ 3 + c 1 )e λ 3 t + C 4 (λ 2 4 + a 1 λ 4 + c 1 )e λ 4 t , where C 1 , , C 4 are arbitrary constants. 5. x tt = a(ty t – y), y tt = b(tx t – x). The transformation u = tx t – x, v = ty t – y (1) leads to a fi rst-order system: u t = atv, v t = btu. The general solution of this system is expressed as with ab > 0: u(t)=C 1 a exp 1 2 √ ab t 2 + C 2 a exp – 1 2 √ ab t 2 , v(t)=C 1 √ ab exp 1 2 √ ab t 2 – C 2 √ ab exp – 1 2 √ ab t 2 ; with ab < 0: u(t)=C 1 a cos 1 2 √ |ab| t 2 + C 2 a sin 1 2 √ |ab| t 2 , v(t)=–C 1 √ |ab| sin 1 2 √ |ab| t 2 + C 2 √ |ab| cos 1 2 √ |ab| t 2 , (2) . number of other solvable cases of equation (1) can be found in the handbooks by Kamke (1977) and Polyanin and Zaitsev (2003). 1232 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS 2 ◦ . Suppose a particular. M., Ordinary Differential Equations and Their Solutions, D. Van Nostrand, New York, 1960. Polyanin, A. D. and Zaitsev, V. F., Handbook of Exact Solutions for Ordinary Differential Equations,. k 2 e iωt . Systems of this type often arise in oscillation theory (e.g., oscillations of a ship and a ship gyroscope). The general solution of this constant-coefficient linear nonhomogeneous system of differential