Errata

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• 1. pag. 30 and following: in the proof of Lemma 2.5.2: replace all the integrals

  $$\int_{{t_0}}^t { \cdots ds} \;\,by\;\,\left| {\int_{{t_0}}^t { \cdots ds} } \right|$$

• 2. Ch.2, Sec.2.5, pg.31, after (2.11) up to the end of the proof, replace the text by:

  Consider the series

  $$\left( {{x_1}\left( t \right) - {x_0}} \right) + \left( {{x_2}\left( t \right) - {x_1}\left( t \right)} \right) + \ldots + \left( {{x_k}\left( t \right) - {x_{k - 1}}\left( t \right)} \right) + \ldots $$

  The uniform upper bound (2.11) and the fact that the numerical series $\Sigma \tfrac{M}{L}\tfrac{{{{\left( {b - a} \right)}^k}{L^k}}}{{k!}}$ converges to $\tfrac{M}{L}{e^{L\left( {b - a} \right)}}$ allows us to use the Weierstrass M-test to infer that $\sum \left( {{x_k}\left( t \right) - {x_{k - 1}}\left( t \right)} \right)$ is uniformly (and absolutely) convergent to a function ψ(t) in [a; b]. Since

  $${x_k}\left( t \right) = {x_0} + \left( {{x_1}\left( t \right) - {x_0}} \right) + \ldots + \left( {{x_{k - 1}}\left( t \right) - {x_{k - 2}}\left( t \right)} \right) + \left( {{x_k}\left( t \right) - {x_{k - 1}}\left( t \right)} \right),$$

  then the uniform convergence of $\sum _{j = 1}^{j = k}\left( {{x_j}\left( t \right) - {x_{j - 1}}\left( t \right)} \right)$ to ψ(t) implies the uniform convergence of x _k (t) to x ₀+ψ(t). This completes the proof.

• 3. pag.32, line 4: replace

  $$ \cdots \leqslant \int_{{t_0}}^t {\left| {f\left( {s,x\left( s \right)} \right) - f\left( {s,y\left( s \right)} \right)} \right|ds} \leqslant L\int_{{t_0}}^t {\left| {x\left( s \right) - y\left( s \right)} \right|ds} $$

  by

  $$ \cdots \leqslant \left| {\int_{{t_0}}^t {\left| {f\left( {s,x\left( s \right)} \right) - f\left( {s,y\left( s \right)} \right)} \right|ds} } \right| \leqslant L\left| {\int_{{t_0}}^t {\left| {x\left( s \right) - y\left( s \right)} \right|ds} } \right|$$

• 4. pag.38, second equation: replace

  $$\frac{1}{{x(\alpha - \beta x}}\;\,by\;\,\frac{1}{{x\left( {\alpha - \beta x} \right)}}$$

• 5. pag.38, Ch.3, Sec. 3.1.1: replace the last 5 lines by .... from which we obtain

  $$\left| {\frac{{x\left( t \right)}}{{\alpha - \beta x\left( t \right)}}} \right| = k{e^{\alpha t}},\quad k = {e^{\alpha c}}.$$

  This is the general solution in implicit form. To solve for x we recall that either (i) α-βx(t) > 0 for all t ≥ 0, or (ii) α-βx(t) < 0 for all t ≥ 0. In the case (i) we find

  $$\frac{{x\left( t \right)}}{{\alpha - \beta x\left( t \right)}} = k{e^{\alpha t}}\;\;which\,yields\;\;x\left( t \right) = \frac{{\alpha k{e^{\alpha t}}}}{{1 + \beta k{e^{\alpha t}}}}.$$

  In the case (ii) we find

  $$ - \frac{{x\left( t \right)}}{{\alpha - \beta x\left( t \right)}} = k{e^{\alpha t}}\;\;which\,yields\;\;x\left( t \right) = \frac{{ - \alpha k{e^{\alpha t}}}}{{1 + \beta k{e^{\alpha t}}}}.$$

  This shows....

• 6. pag.51, line -5: replace $\tfrac{{{e^x}}}{{{{\left( {x + 1} \right)}^2}}}$ by $\tfrac{{{e^x}}}{{x + 1}}$

• 7. pag.62, ex. n.1: replace p, q ≠ 0 by p ≥ 0, q ≥ 1.

• 8. pag.62, ex. n.2: delete t, x ≥ 0.

• 9. pag.63, ex. n.17: after ”its singular points” add ”when p = 2, q = 1 and a, b, d 6 ≠ 0.”

• 10. pag.63, ex. n.27: replace ”for any f(y) ≠ 0, g(y)” by ”for any differentiable f(y) ≠ 0 and any continuous g(y)”

• 11. pag.97, C7: in the system replace $z'' = y - 2z$ by $z' = y - z$.

• 12. pag.108, ex. A1: replace t ³ - 3t by x ₂ = t ³ - 3t.

• 13. pag.108, ex. A11: replace $x'' + \tfrac{x}{t} + q\left( t \right)x = 0$ by $x'' + \tfrac{{x'}}{t} + q\left( t \right)x = 0$.

• 14. pag.109, ex. B3: replace $x'' + 8x + 16 = 0$ by $x'' + 8x' + 16x = 0$

• 15. pag.109, ex. B10: replace k ≠ 0 by k > 0

• 16. pag.109, ex.B11: delete x(0) = 0

• 17. pag.109, ex.B14: replace ”only one solution tends to a constant.” by ”only constant solutions tend to constants”.

• 18. pag.110, ex.C14: replace ”has to change sign in (a, b)” by ”cannot be non-negative in (a, b)”

• 19. pag.111, ex. D4: replace q(t) > 0 by p(t) > 0

• 20. pag.121, ex. n.8: replace -3x′ by -x′

• 21. pag.153, C7: in the third equation replace -2z by -z.

• 22. pag.153, D5: replace -3xx′ = -2 by -3x′ = -2x.

• 23. pag.170, ex. n.7: replace ”all the solutions” by ”all the nontrivial solutions”.

• 24. pag.171, ex. n.9: in the second equation of the system replace -x by -2x.

• 25. pag.172, ex. n.21: replace x(0) = 1 by x(0) = 2.

• 26. pag.185, line 4: replace R - t ₀ < t < t ₀ + R by t ₀ - R _i < t < t ₀ + R _i .

• 27. pag.185, lines 4 and 5: replace a(t ₀) by a ₀(t ₀)

• 28. pag.185, line 8: replace ”is the smallest of the” by ”is at least as large as the smallest of the”

• 29. pag.185, line 5: after ”singular point.” add ”The reader should notice that we have already used the term singular point with a different meaning, dealing with exact equations Mdx + Ndy = 0, see Section 3.2.”

• 30. pag.187, lines 6 and 11, delete 1+ (three times); line 12, replace x(t) = c(1 + sinh t) by x(t) = c sinh t

• 31. pag.254, n.5: in the second equation replace x by 3x

• 32. pag.255, n.13: replace ${x'_2} = {x_1} - {x_2} + {x_3}$ by ${x'_2} = - {x_2} + {x_3}$.

• 33. pag.255, n.19 in the second equation replace -3x′ by -3x.

• 34. pag.255, ex. n.20 replace the system by

  

  and ”a < 3” by -3x.

• 35. pag.256, ex. n.27 and 28: in the first equation of the system, replace -y by y.

• 36. pag.257, replace ex. n.35 by

  35. Show that x″ + x + e _-t x = 0 has oscillatory solutions.

• 37. pag.276, ex. n.8: replace ”x′(b) = 1” by ”x′(b) = 0”.

• 38. pag.276, ex. n.11: add ”where k ≠ 0”.

• 39. pag.289, ex. n.17: after the equation, replace the given answer by ”The singular points are (0; 0) and $\left( {\tfrac{{{b^2}}}{{ad}}, - \tfrac{{{b^3}}}{{a{d^2}}}} \right)$.”

• 40. pag.289, ex. n.33: add ± before the fraction.

• 41. pag.290, ex. n.1 and 2: replace ”a lipschitzian first derivative” by ”a continuous first derivative”.

• 42. pag.290, ex. n.37: replace the given solution by $x\left( t \right) = \tfrac{1}{4}{t^2}$

• 43. pag.291, ex. A11: replace $W\left( 7 \right) = \tfrac{{49}}{6}$ by W(7) = 6

• 44. pag.291, ex. B8: replace by ”One root of the characteristic equation is positive (if β > 0) or null (if β = 0)”

• 45. pag.291, ex. B17: replace a-b < 0 by a ²-b<0

• 46. pag.293, ex. n.15: replace the given solution by $x\left( t \right) = \tfrac{{{e^{ - bt}}}}{{a - b}} - \tfrac{{{e^{ - at}}}}{{a - b}}$.

• 47. pag.293, ex. n.20: replace ”has at least one negative root m = -1” by ”has the negative root m = -1.

• 48. pag.293, ex. n.26: replace $x\left( t \right) = 2 + 2t - 2{e^t} + {t^2}$ by $x\left( t \right) = - 1 + \tfrac{1}{2}{e^t} + \tfrac{1}{2}{e^{ - t}} - \tfrac{1}{2}{t^2}.$.

• 49. pag.294, ex. B5: replace $\tfrac{7}{3}{e^t}$ by 2e ^t and $- \tfrac{2}{3}{e^t}$ by $- \tfrac{1}{2}{e^t}$.

• 50. pag.294, ex. B8: replace the given solution by x = c ₁ e ^2t + c ₂ e ^3t, y = c ₁ e ^2t.

• 51. pag.294, ex. B11: replace the given solution by x = 3c ₁ e ^2t + c ₂ e ^-2t, y = c ₁ e ^2t - c ₂ e ^-2t.

• 52. pag.294, ex. C7: replace $\tfrac{5}{6}$ by $\tfrac{1}{3}$.

• 53. pag.294, ex. D2: replace the given solution by $x = \tfrac{3}{2}{t^2} + + \smallint y\left( t \right)dt$, where $y = {e^{ - \tfrac{1}{2}{t^2}}}\left( {{c_1} + 3\smallint {t^2}{e^{\tfrac{1}{2}{t^2}}}dt} \right)$.

• 54. pag.294, ex. D6: replace the given solution by $x = {c_1}t - {e^{{c_1}}},y = t - \tfrac{1}{2}{c_1}{t^2} + {c_2}$ as well as $x = t\ln t - t + {c'_1},y = t - \tfrac{1}{2}{t^2}\ln t + \tfrac{1}{4}{t^2} + {c'_2}$.

• 55. pag.295, ex. n.3: replace α=β by α=-β.

• 56. pag.296, ex. n.3: replace the given solution by

  $$\eqalign{ & x\left( t \right) = t + \sum\limits_{n \geqslant 1} {\frac{{{t^{3n + 1}}}}{{\left( {3n + 1} \right)3n\left( {3n - 2} \right)\left( {3n - 3} \right) \cdots 4 \cdot 3}}} \cr & \quad \quad \; + \frac{{{t^2}}}{2} + \sum\limits_{n \geqslant 1} {\frac{{{t^{3n + 2}}}}{{\left( {3n + 2} \right)\left( {3n + 1} \right)\left( {3n - 1} \right)\left( {3n - 2} \right) \cdots 5 \cdot 4 \cdot 2}}.} \cr} $$

• 57. pag.297, ex. n.8: replace a ₀ by 0:

• 58. pag.298, ex. n.38: replace the given solution by $x = \left( {4 - {\lambda _1}} \right){A_1}{e^{{\lambda _1}t}} + \left( {4 - {\lambda _2}} \right){A_2}{e^{{\lambda _2}t}},y = {A_1}{e^{{\lambda _1}t}} + {A_2}{e^{{\lambda _2}t}}$, where ${A_i} = \tfrac{{{\lambda _i} - 2}}{{2{\lambda _i} + 2}},{\lambda _{1,2}} = - 1 \pm \sqrt 8$.

• 59. pag.299: starting from n.11, all the numbers of the exercises have to be diminished by 1.

• 60. pag.300, ex. n.11: replace by