6.1-2. Equations of the Form b a G(···) dt = F (x) 12. 1 0 y(t)y(xt) dt = A,0≤ x ≤ 1. This is a special case of equation 6.2.2 with f(t)=1,a = 0, and b =1. 1 ◦ . Solutions: y 1 (x)= √ A, y 3 (x)= √ A (3x – 2), y 5 (x)= √ A (10x 2 – 12x + 3), y 2 (x)=– √ A, y 4 (x)=– √ A (3x – 2), y 6 (x)=– √ A (10x 2 – 12x + 3). 2 ◦ . The integral equation has some other solutions; for example, y 7 (x)= √ A C (2C +1)x C – C – 1 , y 9 (x)= √ A (ln x + 1), y 8 (x)=– √ A C (2C +1)x C – C – 1 , y 10 (x)=– √ A (ln x + 1), where C is an arbitrary constant. 3 ◦ . See 6.2.2 for some other solutions. 13. 1 0 y(t)y(xt β ) dt = A, β >0. 1 ◦ . Solutions: y 1 (x)= √ A, y 3 (x)= √ B (β +2)x – β – 1 , y 2 (x)=– √ A, y 4 (x)=– √ B (β +2)x – β – 1 , where B = 2A β(β +1) . 2 ◦ . The integral equation has some other (more complicated solutions) of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. 14. ∞ 1 y(t)y(xt) dt = Ax –λ , λ >0, 1≤ x < ∞. This is a special case of equation 6.2.3 with f(t)=1,a = 1, and b = ∞. 1 ◦ . Solutions: y 1 (x)=Bx –λ , y 2 (x)=–Bx –λ , λ > 1 2 ; y 3 (x)=B (2λ – 3)x – 2λ +2 x –λ , y 4 (x)=–B (2λ – 3)x – 2λ +2 x –λ , λ > 3 2 ; where B = √ A(2λ – 1). 2 ◦ . For sufficiently large λ, the integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. See 6.2.2 for some other solutions. Page 373 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 15. ∞ 0 e –λt y(t)y(xt) dt = A, λ >0, 0≤ x < ∞. This is a special case of equation 6.2.2 with f(t)=e –λt , a = 0, and b = ∞. 1 ◦ . Solutions: y 1 (x)= √ Aλ, y 3 (x)= 1 2 Aλ (λx – 2), y 2 (x)=– √ Aλ, y 4 (x)=– 1 2 Aλ (λx – 2). 2 ◦ . The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. See 6.2.2 for some other solutions. 16. 1 0 y(t)y(x + λt) dt = A,0≤ x < ∞. This is a special case of equation 6.2.7 with f(t) ≡ 1, a = 0, and b =1. Solutions: y 1 (x)= √ A, y 3 (x)= 3A/λ (1 – 2x), y 2 (x)=– √ A, y 4 (x)=– 3A/λ (1 – 2x). 17. ∞ 0 y(t)y(x + λt) dt = Ae –βx , A, λ, β >0, 0≤ x < ∞. This is a special case of equation 6.2.9 with f(t) ≡ 1, a = 0, and b = ∞. Solutions: y 1 (x)= Aβ(λ +1)e –βx , y 3 (x)=B β(λ +1)x – 1 e –βx , y 2 (x)=– Aβ(λ +1)e –βx , y 4 (x)=–B β(λ +1)x – 1 e –βx , where B = Aβ(λ +1)/λ. 18. 1 0 y(t)y(x – t) dt = A, –∞ < x < ∞. This is a special case of equation 6.2.10 with f(t) ≡ 1, a = 0, and b =1. 1 ◦ . Solutions with A >0: y 1 (x)= √ A, y 3 (x)= √ 5A(6x 2 – 6x + 1), y 2 (x)=– √ A, y 4 (x)=– √ 5A(6x 2 – 6x + 1). 2 ◦ . Solutions with A <0: y 1 (x)= √ –3A (1 – 2x), y 2 (x)=– √ –3A (1 – 2x). The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. 19. ∞ 0 e –λt y x t y(t) dt = Ax b , λ >0. Solutions: y(x)=± √ Aλ x b . Page 374 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 6.1-3. Equations of the Form y(x)+ b a K(x, t)y 2 (t) dt = F (x) 20. y(x)+A b a x λ y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)=– 2λ +1 A(b 2λ+1 – a 2λ+1 ) x λ . 21. y(x)+A b a x λ t µ y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)=– 2λ + µ +1 A(b 2λ+µ+1 – a 2λ+µ+1 ) x λ . 22. y(x)+A b a e –λx y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)= 2λ A(e –2λb – e –2λa ) e –λx . 23. y(x)+A b a e –λx–µt y 2 (t) dt =0. Solutions: y 1 (x)=0, y 2 (x)= 2λ + µ A[e –(2λ+µ)b – e –(2λ+µ)a ] e –λx . 24. y(x)+A b a x λ e –µt y 2 (t) dt =0. This is a special case of equation 6.2.20 with f(x)=Ax λ and g(t)=e –µt . 25. y(x)+A b a e –µx t λ y 2 (t) dt =0. This is a special case of equation 6.2.20 with f(x)=Ae –µx and g(t)=t λ . 26. y(x)+A 1 0 y 2 (t) dt = Bx µ , µ > –1. This is a special case of equation 6.2.22 with g(t)=A, f(x)=Bx µ , a = 0, and b =1. A solution: y(x)=Bx µ + λ, where λ is determined by the quadratic equation λ 2 + 1 A 1+ 2AB µ +1 λ + B 2 2µ +1 =0. 27. y(x)+A b a t β y 2 (t) dt = Bx µ . This is a special case of equation 6.2.22 with g(t)=At β and f(x)=Bx µ . Page 375 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 28. y(x)+A b a e βt y 2 (t) dt = Be µx . This is a special case of equation 6.2.22 with g(t)=Ae βt and f(x)=Be µx . 29. y(x)+A b a x β y 2 (t) dt = Bx µ . This is a special case of equation 6.2.23 with g(x)=Ax β and f(x)=Bx µ . 30. y(x)+A b a e βx y 2 (t) dt = Be µx . This is a special case of equation 6.2.23 with g(x)=Ae βx and f(x)=Be µx . 6.1-4. Equations of the Form y(x)+ b a K(x, t)y(x)y(t) dt = F (x) 31. y(x)+A b a t β y(x)y(t) dt = Bx µ . This is a special case of equation 6.2.25 with g(t)=At β and f(x)=Bx µ . 32. y(x)+A b a e βt y(x)y(t) dt = Be µx . This is a special case of equation 6.2.25 with g(t)=Ae βt and f(x)=Be µx . 33. y(x)+A b a x β y(x)y(t) dt = Bx µ . This is a special case of equation 6.2.26 with g(x)=Ax β and f(x)=Bx µ . 34. y(x)+A b a e βx y(x)y(t) dt = Be µx . This is a special case of equation 6.2.26 with g(x)=Ae βx and f(x)=Be µx . 6.1-5. Equations of the Form y(x)+ b a G(···) dt = F (x) 35. y(x)+A 1 0 y(t)y(xt) dt =0. This is a special case of equation 6.2.30 with f(t)=A, a = 0, and b =1. 1 ◦ . Solutions: y 1 (x)=– 1 A (2C +1)x C , y 2 (x)= (I 1 – I 0 )x + I 1 – I 2 I 0 I 2 – I 2 1 x C , I m = A 2C + m +1 , m =0,1,2, where C is an arbitrary nonnegative constant. There are more complicated solutions of the form y(x)=x C n k=0 B k x k , where C is an arbitrary constant and the coefficients B k can be found from the corresponding system of algebraic equations. Page 376 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 2 ◦ . A solution: y 3 (x)= (I 1 – I 0 )x β + I 1 – I 2 I 0 I 2 – I 2 1 x C , I m = A 2C + mβ +1 , m =0,1,2, where C and β are arbitrary constants. There are more complicated solutions of the form y(x)=x C n k=0 D k x kβ , where C and β are arbitrary constants and the coefficients D k can be found from the corresponding system of algebraic equations. 3 ◦ . A solution: y 4 (x)= x C (J 1 ln x – J 2 ) J 0 J 2 – J 2 1 , J m = 1 0 t 2C (ln t) m dt, m =0,1,2, where C is an arbitrary constant. There are more complicated solutions of the form y(x)=x C n k=0 E k (ln x) k , where C is an arbitrary constant and the coefficients E k can be found from the corresponding system of algebraic equations. 36. y(x)+A ∞ 1 y(t)y(xt) dt =0. This is a special case of equation 6.2.30 with f(t)=A, a = 1, and b = ∞. 37. y(x)+λ ∞ 1 y(t)y(xt) dt = Ax β . This is a special case of equation 6.2.31 with f(t)=λ, a = 0, and b =1. 38. y(x)+A 1 0 y(t)y(x + λt) dt =0. This is a special case of equation 6.2.35 with f(t) ≡ A, a = 0, and b =1. 1 ◦ . A solution: y(x)= C(λ +1) A[1 – e C(λ+1) ] e Cx , where C is an arbitrary constant. 2 ◦ . There are more complicated solutions of the form y(x)=e Cx n m=0 B m x m , where C is an arbitrary constant and the coefficients B m can be found from the corresponding system of algebraic equations. 39. y(x)+A ∞ 0 y(t)y(x + λt) dt =0, λ >0, 0≤ x < ∞. This is a special case of equation 6.2.35 with f(t) ≡ A, a = 0, and b = ∞. A solution: y(x)=– C(λ +1) A e –Cx , where C is an arbitrary positive constant. Page 377 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 40. y(x)+A ∞ 0 e –λt y x t y(t) dt =0, λ >0. A solution: y(x)=– λ A x C , where C is an arbitrary constant. 41. y(x)+A ∞ 0 e –λt y x t y(t) dt = Bx b , λ >0. Solutions: y 1 (x)=β 1 x b , y 2 (x)=β 2 x b , where β 1 and β 2 are the roots of the quadratic equation Aβ 2 + λβ – Bλ =0. 6.2. Equations With Quadratic Nonlinearity That Contain Arbitrary Functions 6.2-1. Equations of the Form b a G(···) dt = F (x) 1. b a g(t)y(x)y(t) dt = f(x). Solutions: y(x)=±λf(x), λ = b a f(t)g(t) dt –1/2 . 2. b a f(t)y(t)y(xt) dt = A. 1 ◦ . Solutions* y 1 (x)= A/I 0 , y 3 (x)=q(I 1 x – I 2 ), y 2 (x)=– A/I 0 , y 4 (x)=–q(I 1 x – I 2 ), where I m = b a t m f(t) dt, q = A I 0 I 2 2 – I 2 1 I 2 1/2 , m =0,1,2. The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. 2 ◦ . Solutions: y 5 (x)=q(I 1 x C – I 2 ), y 6 (x)=–q(I 1 x C – I 2 ), q = A I 0 I 2 2 – I 2 1 I 2 1/2 , I m = b a t mC f(t) dt, m =0,1,2, where C is an arbitrary constant. The equation has more complicated solutions of the form y(x)= n k=0 B k x kC , where C is an arbitrary constant and the coefficients B k can be found from the corresponding system of algebraic equations. * The arguments of the equations containing y(xt) in the integrand can vary, for example, within the following intervals: (a) 0 ≤ t ≤ 1, 0 ≤ x ≤ 1 for a = 0 and b = 1; (b) 1 ≤ t < ∞,1≤ x < ∞ for a = 1 and b = ∞; (c) 0 ≤ t < ∞,0≤ x < ∞ for a = 0 and b = ∞; or (d) a ≤ t ≤ b,0≤ x < ∞ for arbitrary a and b such that 0 ≤ a < b ≤ ∞. Case (d) is a special case of (c) if f(t) is nonzero only on the interval a ≤ t ≤ b. Page 378 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3 ◦ . Solutions: y 7 (x)=p(J 0 ln x – J 1 ), y 8 (x)=–p(J 0 ln x – J 1 ), p = A J 2 0 J 2 – J 0 J 2 1 1/2 , J m = b a (ln t) m f(t) dt. The equation has more complicated solutions of the form y(x)= n k=0 E k (ln x) k , where the constants E k can be found from the corresponding system of algebraic equations. 3. b a f(t)y(t)y(xt) dt = Ax β . 1 ◦ . Solutions: y 1 (x)= A/I 0 x β , y 3 (x)=q(I 1 x – I 2 ) x β , y 2 (x)=– A/I 0 x β , y 4 (x)=–q(I 1 x – I 2 ) x β , where I m = b a t 2β+m f(t) dt, q = A I 2 (I 0 I 2 – I 2 1 ) , m =0,1,2. 2 ◦ . The substitution y(x)=x β w(x) leads to an equation of the form 6.2.2: b a g(t)w(t)w(xt) dt = A, g(x)=f(x)x 2β . Therefore, the integral equation in question has more complicated solutions. 4. b a f(t)y(t)y(xt) dt = A ln x + B. This equation has solutions of the form y(x)=p ln x +q. The constants p and q are determined from the following system of two second-order algebraic equations: I 1 p 2 + I 0 pq = A, I 2 p 2 +2I 1 pq + I 0 q 2 = B, where I m = b a f(t)(ln t) m dt, m =0,1,2. 5. b a f(t)y(t)y(xt) dt = Ax λ ln x + Bx λ . The substitution y(x)=x λ w(x) leads to an equation of the form 6.2.4: b a g(t)w(t)w(xt) dt = A ln x + B, g(t)=f (t)t 2λ . 6. ∞ 0 f(t)y(t)y x t dt = Ax λ . Solutions: y 1 (x)= A I x λ , y 2 (x)=– A I x λ , I = ∞ 0 f(t) dt. Page 379 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 7. b a f(t)y(t)y(x + λt) dt = A, λ >0. 1 ◦ . Solutions* y 1 (x)= A/I 0 , y 3 (x)=q(I 0 x – I 1 ), y 2 (x)=– A/I 0 , y 4 (x)=–q(I 0 x – I 1 ), where I m = b a t m f(t) dt, q = A λ(I 2 0 I 2 – I 0 I 2 1 ) , m =0,1,2. 2 ◦ . The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. 8. b a f(t)y(t)y(x + λt) dt = Ax + B, λ >0. A solution: y(x)=βx + µ, where the constants β and µ are determined from the following system of two second-order algebraic equations: I 0 βµ + I 1 β 2 = A, I 0 µ 2 +(λ +1)I 1 βµ + λI 2 β 2 = B, I m = b a t m f(t) dt. (1) Multiplying the first equation by B and the second by –A and adding the resulting equations, we obtain the quadratic equation AI 0 z 2 + (λ +1)AI 1 – BI 0 z + λAI 2 – BI 1 =0, z = µ/β. (2) In general, to each root of equation (2) two solutions of system (1) correspond. Therefore, the original integral equation can have at most four solutions of this form. If the discriminant of equation (2) is negative, then the integral equation has no such solutions. The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 β k x k , where the constants β k can be found from the corresponding system of algebraic equations. 9. b a f(t)y(t)y(x + λt) dt = Ae –βx , λ >0. 1 ◦ . Solutions: y 1 (x)= A/I 0 e –βx , y 3 (x)=q(I 0 x – I 1 )e –βx , y 2 (x)=– A/I 0 e –βx , y 4 (x)=–q(I 0 x – I 1 )e –βx , where I m = b a t m e –β(λ+1)t f(t) dt, q = A λ(I 2 0 I 2 – I 0 I 2 1 ) , m =0,1,2. 2 ◦ . The equation has more complicated solutions of the form y(x)=e –βx n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. * The arguments of the equations containing y(x+λt) in the integrand can vary within the following intervals: (a) 0 ≤t < ∞, 0 ≤ x < ∞ for a = 0 and b = ∞ or (b) a ≤ t ≤ b,0≤ x < ∞ for arbitrary a and b such that 0 ≤ a < b < ∞. Case (b) is a special case of (a) if f(t) is nonzero only on the interval a ≤ t ≤ b. Page 380 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 3 ◦ . The substitution y(x)=e –βx w(x) leads to an equation of the form 6.2.7: b a e –β(λ+1)t f(t)w(t)w(x + λt) dt = A. 10. b a f(t)y(t)y(x – t) dt = A. 1 ◦ . Solutions* y 1 (x)= A/I 0 , y 3 (x)=q(I 0 x – I 1 ), y 2 (x)=– A/I 0 , y 4 (x)=–q(I 0 x – I 1 ), where I m = b a t m f(t) dt, q = A I 0 I 2 1 – I 2 0 I 2 , m =0,1,2. 2 ◦ . The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 λ k x k , where the constants λ k can be found from the corresponding system of algebraic equations. For n = 3, such a solution is presented in 6.1.18. 11. b a f(t)y(t)y(x – t) dt = Ax + B. A solution: y(x)=λx + µ, where the constants λ and µ are determined from the following system of two second-order algebraic equations: I 0 λµ + I 1 λ 2 = A, I 0 µ 2 – I 2 λ 2 = B, I m = b a t m f(t) dt, m = 0, 1, 2. (1) Multiplying the first equation by B and the second by –A and adding the results, we obtain the quadratic equation AI 0 z 2 – BI 0 z – AI 2 – BI 1 =0, z = µ/λ. (2) In general, to each root of equation (2) two solutions of system (1) correspond. Therefore, the original integral equation can have at most four solutions of this form. If the discriminant of equation (2) is negative, then the integral equation has no such solutions. The integral equation has some other (more complicated) solutions of the polynomial form y(x)= n k=0 λ k x k , where the constants λ k can be found from the corresponding system of algebraic equations. 12. b a f(t)y(t)y(x – t) dt = n k=0 A k x k . This equation has solutions of the form y(x)= n k=0 λ k x k , (1) where the constants λ k are determined from the system of algebraic equations obtained by substituting solution (1) into the original integral equation and matching the coefficients of like powers of x. * The arguments of the equations containing y(x–t) in the integrand can vary within the following intervals: (a) –∞ < t< ∞, –∞ < x < ∞ for a = –∞ and b = ∞ or (b) a ≤ t ≤ b, –∞ ≤ x < ∞, for arbitrary a and b such that –∞ < a < b < ∞. Case (b) is a special case of (a) if f(t) is nonzero only on the interval a ≤ t ≤ b. Page 381 © 1998 by CRC Press LLC © 1998 by CRC Press LLC 13. b a f(t)y(x – t)y(t) dt = Ae λx . Solutions: y 1 (x)= A/I 0 e λx , y 3 (x)=q(I 0 x – I 1 )e λx , y 2 (x)=– A/I 0 e λx , y 4 (x)=–q(I 0 x – I 1 )e λx , where I m = b a t m f(t) dt, q = A I 0 I 2 1 – I 2 0 I 2 , m =0,1,2. The integral equation has more complicated solutions of the form y(x)=e λx n k=0 B k x k , where the constants B k can be found from the corresponding system of algebraic equations. 14. b a f(t)y(t)y(x – t) dt = A sinh λx. A solution: y(x)=p sinh λx + q cosh λx. (1) Here p and q are roots of the algebraic system I 0 pq + I cs (p 2 – q 2 )=A, I cc q 2 – I ss p 2 = 0, (2) where the notation I 0 = b a f(t) dt, I cs = b a f(t) cosh(λt) sinh(λt) dt, I cc = b a f(t) cosh 2 (λt) dt, I ss = b a f(t) sinh 2 (λt) dt is used. Different solutions of system (2) generate different solutions (1) of the integral equation. It follows from the second equation of (2) that q = ± I ss /I cc p. Using this expression to eliminate q from the first equation of (2), we obtain the following four solutions: y 1,2 (x)=p sinh λx ± k cosh λx , y 3,4 (x)=–p sinh λx ± k cosh λx , k = I ss I cc , p = A (1 – k 2 )I cs ± kI 0 . 15. b a f(t)y(t)y(x – t) dt = A cosh λx. A solution: y(x)=p sinh λx + q cosh λx. (1) Here p and q are roots of the algebraic system I 0 pq + I cs (p 2 – q 2 )=0, I cc q 2 – I ss p 2 = A, (2) where we use the notation introduced in 6.2.14. Different solutions of system (2) generate different solutions (1) of the integral equation. Page 382 © 1998 by CRC Press LLC © 1998 by CRC Press LLC [...]... equation 6. 8.43 with g(x, y) = A coth(βy) and f (t, y) = coth(γy) b 44 y(x) + A y(xt) coth[βy(t)] dt = 0 a This is a special case of equation 6. 8.45 with f (t, y) = A coth(βy) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 397 6. 6 Equations With Logarithmic Nonlinearity 6. 6-1 Integrands With Nonlinearity of the Form ln[βy(t)] b 1 y(x) + A ln[βy(t)] dt = g(x) a This is a special case of equation 6. 8.27... equation 6. 8.43 with g(x, y) = A exp(βy) and f (t, y) = exp(γy) b 11 y(x) + A y(xt) exp[βy(t)] dt = 0 a This is a special case of equation 6. 8.45 with f (t, y) = A exp(βy) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 393 6. 5 Equations With Hyperbolic Nonlinearity 6. 5-1 Integrands With Nonlinearity of the Form cosh[βy(t)] b 1 y(x) + A cosh[βy(t)] dt = g(x) a This is a special case of equation 6. 8.27... equation 6. 8.31 with f (t, y) = A tanh(βy) b 25 y(x) + A sinh(λx + µt) tanh[βy(t)] dt = h(x) a This is a special case of equation 6. 8.32 with f (t, y) = A tanh(βy) b 26 y(x) + A cos(λx + µt) tanh[βy(t)] dt = h(x) a This is a special case of equation 6. 8.33 with f (t, y) = A tanh(βy) b 27 y(x) + A sin(λx + µt) tanh[βy(t)] dt = h(x) a This is a special case of equation 6. 8.34 with f (t, y) = A tanh(βy) 6. 5-4... equation 6. 8.33 with f (t, y) = A coth(βy) b 36 y(x) + A sin(λx + µt) coth[βy(t)] dt = h(x) a This is a special case of equation 6. 8.34 with f (t, y) = A coth(βy) 6. 5-5 Other Integrands b 37 y(x) + A cosh[βy(x)] cosh[γy(t)] dt = h(x) a This is a special case of equation 6. 8.43 with g(x, y) = A cosh(βy) and f (t, y) = cosh(γy) b 38 y(x) + A y(xt) cosh[βy(t)] dt = 0 a This is a special case of equation 6. 8.45... This is a special case of equation 6. 8.44 b 21 y(xt)y β (t) dt = 0 y(x) + A a This is a special case of equation 6. 8.45 with f (t, y) = Ay β © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 392 6. 4 Equations With Exponential Nonlinearity 6. 4-1 Integrands With Nonlinearity of the Form exp[βy(t)] b 1 y(x) + A exp[βy(t)] dt = g(x) a This is a special case of equation 6. 8.27 with f (t, y) = A exp(βy)... of equation 6. 8.31 with f (t, y) = A ln(βy) b 7 y(x) + A sinh(λx + µt) ln[βy(t)] dt = h(x) a This is a special case of equation 6. 8.32 with f (t, y) = A ln(βy) b 8 y(x) + A cos(λx + µt) ln[βy(t)] dt = h(x) a This is a special case of equation 6. 8.33 with f (t, y) = A ln(βy) b 9 y(x) + A sin(λx + µt) ln[βy(t)] dt = h(x) a This is a special case of equation 6. 8.34 with f (t, y) = A ln(βy) 6. 6-2 Other Integrands... equation 6. 8.43 with g(x, y) = A ln(βy) and f (t, y) = ln(γy) b 11 y(x) + A y(xt) ln[βy(t)] dt = 0 a This is a special case of equation 6. 8.45 with f (t, y) = A ln(βy) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 398 6. 7 Equations With Trigonometric Nonlinearity 6. 7-1 Integrands With Nonlinearity of the Form cos[βy(t)] b 1 y(x) + A cos[βy(t)] dt = g(x) a This is a special case of equation 6. 8.27... equation 6. 8.31 with f (t, y) = A tan(βy) b 25 y(x) + A sinh(λx + µt) tan[βy(t)] dt = h(x) a This is a special case of equation 6. 8.32 with f (t, y) = A tan(βy) b 26 y(x) + A cos(λx + µt) tan[βy(t)] dt = h(x) a This is a special case of equation 6. 8.33 with f (t, y) = A tan(βy) b 27 y(x) + A sin(λx + µt) tan[βy(t)] dt = h(x) a This is a special case of equation 6. 8.34 with f (t, y) = A tan(βy) 6. 7-4 Integrands... equation 6. 8.33 with f (t, y) = A cot(βy) b 36 y(x) + A sin(λx + µt) cot[βy(t)] dt = h(x) a This is a special case of equation 6. 8.34 with f (t, y) = A cot(βy) 6. 7-5 Other Integrands b 37 y(x) + A cos[βy(x)] cos[γy(t)] dt = h(x) a This is a special case of equation 6. 8.43 with g(x, y) = A cos(βy) and f (t, y) = cos(γy) b 38 y(x) + A y(xt) cos[βy(t)] dt = 0 a This is a special case of equation 6. 8.45... cot[βy(x)] cot[γy(t)] dt = h(x) a This is a special case of equation 6. 8.43 with g(x, y) = A cot(βy) and f (t, y) = cot(γy) b 44 y(x) + A y(xt) cot[βy(t)] dt = 0 a This is a special case of equation 6. 8.45 with f (t, y) = A cot(βy) © 1998 by CRC Press LLC © 1998 by CRC Press LLC Page 402 6. 8 Equations With Nonlinearity of General Form 6. 8-1 Equations of the Form b a G(· · ·) dt = F (x) b y(x)f t, y(t) dt . case of equation 6. 2. 26 with g(x)=Ax β and f(x)=Bx µ . 34. y(x)+A b a e βx y(x)y(t) dt = Be µx . This is a special case of equation 6. 2. 26 with g(x)=Ae βx and f(x)=Be µx . 6. 1-5. Equations of. special case of equation 6. 2.10 with f(t) ≡ 1, a = 0, and b =1. 1 ◦ . Solutions with A >0: y 1 (x)= √ A, y 3 (x)= √ 5A(6x 2 – 6x + 1), y 2 (x)=– √ A, y 4 (x)=– √ 5A(6x 2 – 6x + 1). 2 ◦ . Solutions. notation introduced in 6. 2.41. Different solutions of system (2) generate different solutions (1) of the integral equation. 6. 3. Equations With Power-Law Nonlinearity 6. 3-1. Equations of the Form b a G(···)