Tab-integrais

Mathematics 1. If ≠ ′, no solution, if = ′ = , unique solution if = ′ < , many solutions. (non-homogeneous) 2. If = , trivial solution, if < ,then (−) linearly independent solutions. (Many solutions) and if < , then many solutions. 3. (+ ℎ) = () + ℎ ′ () + ℎ 2 2! ′′ () + ℎ 3 3! ′′′ () + … … … ∞ 4. If − 2 > 0 and < 0 (,) have maximum, if − 2 > 0 and > 0 (,) have minimum at(,) and if − 2 < 0, then saddle point. If − 2 = 0, ℎ investigation is required to decide. 5. ∫ (∅ +) = ∫ ∫ (− ∅) (Green's) 6. ∫. ℝ=∫. (Stokes) 7. ∫. = ∫ (Gauss) 8. ∫ = ∫ ∫ + 9. If + = 0 be a homogeneous equation in and , then 1 + an integrating factor 10. If the equation of the type 1 () + 2 () = 0. If the equation + = 0 be of this type then 1 − is an integrating factor 11. If − be a function of x only = () say then ∫ () is an integrating factor 12. If − be a function of y only = () say then ∫ () is an integrating factor. 13. ∫ = + ∫ terms of N not containing x dy = c 14.. . = 1 () = 1 () , () ≠ 0, if () = 0, ℎ. . = 1 ′ () , ′ () ≠ 0 15.. = 1 (2) sin(+) = 1 (− 2) , (− 2) ≠ 0, if (− 2) = 0, then. . = 1 ′ (− 2) sin(+), ′ (− 2) ≠ 0 16.. . = 1 () = 1 (+) 17.. = 1 () = [ ()] −1 , 18. (1 +) −1 = 1 − + 2 − ⋯ 19. (1 −) −1 = 1 + + 2 + ⋯ 20. + 1 −1 −1 −1 + ⋯ −1 + = , = , = , 2 2 2 = (− 1) , 3 3 3 = D(D − 1)(D − 2) 21. (∫ ℎ(,) () ()) = ∫ ℎ(,) () () + ℎ[ (), ] − ℎ[ (), ] 22. { ()} = ∫ − ∞ 0 () 23. (1) = 1 24. () = ! +1 25. () = 1 − 26. (sin) = 2 + 2 27. (cos) = 2 + 28. (sinh) = 2 − 2 29. (cosh) = 2 − 2 30. { ()} = ̅ (−) 31. (+) = () then { ()} = ∫ − () 0 1− − 32. { ′ ()} = ̅ () − (0) 33. { ()} = ̅ () − −1 (0) − −2 ′ (0) − ⋯ … …. . −1 (0) 34. {∫ () 0 } = 1 ̅ () 35. { ()} = (−1). [ ̅ (s)] 36. { 1 ()} = ∫ ̅ (s) ∞ 37. () = 0 2 + ∑ ∞ =1 cos + ∑ ∞ =1 sin 38. 0 = 1 ∫ () +2 , = 1 ∫ () cos +2 , = 1 ∫ () sin +2 39. () = 0 2 + ∑ ∞ =1 cos + ∑ ∞ =1 sin 40. 0 = 1 ∫ () +2 , = 1 ∫ () cos +2 , = 1 ∫ () sin +2 41. () = ∑ ∞ =1 sin , where = 2 ∫ () sin 0 42. () = 0 2 + ∑ ∞ =1 cos where, 0 = 2 ∫ () 0 , = 2 ∫ () cos 0 43. = ∑ () = ∫ () ∞ −∞ 44. 2 = ∑ (−) 2 () 2 = ∫ (−) 2 () ∞ −∞ 45. : = : 2 = (Poisson's distribution) 46. () = 1 √2 − 1 2 (−) 2 (Normal distribution) 47. = + , ∑ = + ∑ , ∑ = ∑ + ∑ 2 48. 50. () 0 = 1 ℎ [∆ 0 − 1 2 ∆ 2 0 + 1 3 ∆ 3 0 − 1 4 ∆ 4 0 + ⋯ ] 51. () = 1 ℎ [∇ + 1 2 ∇ 2 + 1 3 ∇ 3 + 1 4 ∇ 4 + ⋯ ] 52. +1 = − () ′ () (Newton-Raphson) 53. ∫ () 0 + ℎ 0 = ℎ 2 [ 0 + + 2(1 + 2 + ⋯. . + −1)] (Trapezoidal) 54. ∫ () 0 + ℎ 0 = ℎ 3 [(0 +) + 4(1 + 3 + ⋯ −1) + 2(2 + 4 + ⋯ −2)] (Simpson's) 55. = − − 12 ℎ 2 ′′ () = (ℎ 2) (Trapezoidal) 56. = − − 180 ℎ 4 () = (ℎ 4) (Simpson's) 57. +1 = + ℎ. (,) where = (,) (Euler's) 2

Mathematics of Computation, 1968

1. Second order; linear 2. Third order; nonlinear because of (dy/dx) 4 3. Fourth order; linear 4. Second order; nonlinear because of cos(r + u) 5. Second order; nonlinear because of (dy/dx) 2 or 1 + (dy/dx) 2 6. Second order; nonlinear because of R 2 7. Third order; linear 8. Second order; nonlinear because of ˙ x 2 9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear in y because of y 2. However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is linear in x. 10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ue u we see that it is linear in v. However, writing it in the form (v + uv − ue u)(du/dv) + u = 0, we see that it is nonlinear in u. 11. From y = e −x/2 we obtain y = − 1 2 e −x/2. Then 2y + y = −e −x/2 + e −x/2 = 0. 12. From y = 6 5 − 6 5 e −20t we obtain dy/dt = 24e −20t , so that dy dt + 20y = 24e −20t + 20 6 5 − 6 5 e −20t = 24. 13. From y = e 3x cos 2x we obtain y = 3e 3x cos 2x − 2e 3x sin 2x and y = 5e 3x cos 2x − 12e 3x sin 2x, so that y − 6y + 13y = 0. 14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and y = tan x + cos x ln(sec x + tan x). Then y + y = tan x. 15. The domain of the function, found by solving x + 2 ≥ 0, is [−2, ∞). From y = 1 + 2(x + 2) −1/2 we have (y − x)y = (y − x)[1 + (2(x + 2) −1/2 ] = y − x + 2(y − x)(x + 2) −1/2 = y − x + 2[x + 4(x + 2) 1/2 − x](x + 2) −1/2 = y − x + 8(x + 2) 1/2 (x + 2) −1/2 = y − x + 8.

American Journal of Physics, 1988

This study accessed the effect of government entrepreneurial policies on Nigeria's economic growth with Ondo State as a case study. Specifically, the study examined the effect of entrepreneurship policies on the productivity of small business owners. Three policy areas considered by the study are credit availability, supply of factor inputs, and training/orientation. Data were collected using questionnaires and responses were analysed with the aid of chi-square statistic. The study revealedthat only twenty-five percent of entrepreneurs have benefitted from government entrepreneurship policies and programmes in Ondo State. The study discovered that meeting the requirements for accessing government entrepreneurship programmes was the most serious difficulty encountered by entrepreneurs in benefiting from government programmes; and that government entrepreneurial policies have been ineffective due mainly to lack of continuity by successive governments. The study found that government policies at present do notimprove economic growth. The study concluded that achievement of the desired economic growth requires review of credit requirements to less stringent terms, harmonisation of government entrepreneurial policies with other fiscal and monetary policies, continuity of government policies by successive government, de-emphasis on political affiliation as condition for accessing government programmes and improved sensitisation of the public on various government entrepreneurial policies and programmes