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Ingénierie

ordinary differential equations (for half semester course)

sujeet singhSuivre

- 1. MA 108 - Ordinary Diﬀerential Equations Neela Nataraj Department of Mathematics, Indian Institute of Technology Bombay, Powai, Mumbai 76 neela@math.iitb.ac.in February 29, 2016 Neela Nataraj Lecture 1: D3
- 2. Outline of the lecture Basic Concepts Classiﬁcation based on 1 Type 2 Order 3 Linearity Solution(s) 1 Explicit 2 Implicit 3 Formal First Order ODEs, IVP Separable ODEs Neela Nataraj Lecture 1: D3
- 3. Diﬀerential equations Deﬁnition An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a diﬀerential equation (DE). 1 Dennis G. Zill, Diﬀerential Equations Neela Nataraj Lecture 1: D3
- 4. Diﬀerential equations Deﬁnition An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a diﬀerential equation (DE). DE’s occur naturally in physics, engineering, biology, economics and so on. 1 Dennis G. Zill, Diﬀerential Equations Neela Nataraj Lecture 1: D3
- 5. Diﬀerential equations Deﬁnition An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a diﬀerential equation (DE). DE’s occur naturally in physics, engineering, biology, economics and so on. A few examples of physical systems which lead to DE 1 : Radio active decay Population dynamics Newton’s law of cooling Spread of a contagious disease Chemical reactions Falling bodies Suspended cables 1 Dennis G. Zill, Diﬀerential Equations Neela Nataraj Lecture 1: D3
- 6. Classiﬁcation based on TYPE: ODE/PDE Deﬁnition Let y(x) denote a function in the variable x. An ordinary diﬀerential equation (ODE) is an equation containing one or more derivatives of an unknown function y. Neela Nataraj Lecture 1: D3
- 7. Classiﬁcation based on TYPE: ODE/PDE Deﬁnition Let y(x) denote a function in the variable x. An ordinary diﬀerential equation (ODE) is an equation containing one or more derivatives of an unknown function y. In general, a diﬀerential equation involving one or more dependent variables with respect to a single independent variable is called an ODE. Neela Nataraj Lecture 1: D3
- 8. Classiﬁcation based on TYPE: ODE/PDE Deﬁnition Let y(x) denote a function in the variable x. An ordinary diﬀerential equation (ODE) is an equation containing one or more derivatives of an unknown function y. In general, a diﬀerential equation involving one or more dependent variables with respect to a single independent variable is called an ODE. Deﬁnition A diﬀerential equation involving partial derivatives of one or more dependent variables with respect to more than one independent variable is called a partial diﬀerential equation (PDE). Neela Nataraj Lecture 1: D3
- 9. Examples dy dx + 5y = ex , d2y dx2 − dy dx + 6y = 0, ODEs with one dependent variable Neela Nataraj Lecture 1: D3
- 10. Examples dy dx + 5y = ex , d2y dx2 − dy dx + 6y = 0, ODEs with one dependent variable dx dt + dy dt = 2x + y. an ODE with two dependent variables Neela Nataraj Lecture 1: D3
- 11. Examples dy dx + 5y = ex , d2y dx2 − dy dx + 6y = 0, ODEs with one dependent variable dx dt + dy dt = 2x + y. an ODE with two dependent variables ∂2u ∂x2 + ∂2u ∂y2 = 0. PDE with two independent variables- Laplace equation Neela Nataraj Lecture 1: D3
- 12. Examples dy dx + 5y = ex , d2y dx2 − dy dx + 6y = 0, ODEs with one dependent variable dx dt + dy dt = 2x + y. an ODE with two dependent variables ∂2u ∂x2 + ∂2u ∂y2 = 0. PDE with two independent variables- Laplace equation ∂2u ∂t2 − ( ∂2u ∂x2 + ∂2u ∂y2 ) = 0. PDE with three independent variables- Wave equation Neela Nataraj Lecture 1: D3
- 13. ODE- general form Along with derivatives of y (the dependent variable) with respect to x (the independent variable), note that, an an ODE may contain y itself (the 0th derivative), and known functions of x (including constants). Neela Nataraj Lecture 1: D3
- 14. ODE- general form Along with derivatives of y (the dependent variable) with respect to x (the independent variable), note that, an an ODE may contain y itself (the 0th derivative), and known functions of x (including constants). In other words, an ODE is a relation between the derivatives y, y or dy dx , . . . , y(n) or dny dxn Neela Nataraj Lecture 1: D3
- 15. ODE- general form Along with derivatives of y (the dependent variable) with respect to x (the independent variable), note that, an an ODE may contain y itself (the 0th derivative), and known functions of x (including constants). In other words, an ODE is a relation between the derivatives y, y or dy dx , . . . , y(n) or dny dxn and functions of x: F(x, y, y , . . . , y(n) ) = 0. Neela Nataraj Lecture 1: D3
- 16. ODE- general form Along with derivatives of y (the dependent variable) with respect to x (the independent variable), note that, an an ODE may contain y itself (the 0th derivative), and known functions of x (including constants). In other words, an ODE is a relation between the derivatives y, y or dy dx , . . . , y(n) or dny dxn and functions of x: F(x, y, y , . . . , y(n) ) = 0. Neela Nataraj Lecture 1: D3
- 17. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 Neela Nataraj Lecture 1: D3
- 18. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, Neela Nataraj Lecture 1: D3
- 19. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) Neela Nataraj Lecture 1: D3
- 20. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t Neela Nataraj Lecture 1: D3
- 21. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, Neela Nataraj Lecture 1: D3
- 22. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) Neela Nataraj Lecture 1: D3
- 23. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v Neela Nataraj Lecture 1: D3
- 24. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, Neela Nataraj Lecture 1: D3
- 25. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) Neela Nataraj Lecture 1: D3
- 26. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) 4 ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 Neela Nataraj Lecture 1: D3
- 27. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) 4 ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 (PDE, Neela Nataraj Lecture 1: D3
- 28. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) 4 ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 (PDE, 2nd order) Neela Nataraj Lecture 1: D3
- 29. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) 4 ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 (PDE, 2nd order) 5 dx dt = f (x, y), dy dt = g(x, y), x = x(t), y = y(t). Neela Nataraj Lecture 1: D3
- 30. Classiﬁcation based on ORDER Further classiﬁcation according to the appearance of the highest derivative appearing in the equation is done now. Deﬁnition The order of a diﬀerential equation is the order of the highest derivative in the equation. Examples : 1 d2y dx2 + xy dy dx 2 = 0 (ODE, 2nd order) 2 d4x dt4 + 5 d2x dt2 + 3x = sin t (ODE, 4th order) 3 ∂v ∂t + ∂v ∂s = v (PDE, 1st order) 4 ∂2u ∂x2 + ∂2u ∂y2 + ∂2u ∂z2 = 0 (PDE, 2nd order) 5 dx dt = f (x, y), dy dt = g(x, y), x = x(t), y = y(t). (System of ODEs, 1st order) Neela Nataraj Lecture 1: D3
- 31. Linear equations Linear equations Neela Nataraj Lecture 1: D3
- 32. Linear equations Linear equations - F(x, y, y , . . . , y(n) ) = 0 is linear if F is a linear function of the variables y, y , . . . , y(n) . Neela Nataraj Lecture 1: D3
- 33. Linear equations Linear equations - F(x, y, y , . . . , y(n) ) = 0 is linear if F is a linear function of the variables y, y , . . . , y(n) . Thus, a linear ODE of order n is of the form Neela Nataraj Lecture 1: D3
- 34. Linear equations Linear equations - F(x, y, y , . . . , y(n) ) = 0 is linear if F is a linear function of the variables y, y , . . . , y(n) . Thus, a linear ODE of order n is of the form a0(x)y(n) + a1(x)y(n−1) + . . . + an(x)y = b(x) Neela Nataraj Lecture 1: D3
- 35. Linear equations Linear equations - F(x, y, y , . . . , y(n) ) = 0 is linear if F is a linear function of the variables y, y , . . . , y(n) . Thus, a linear ODE of order n is of the form a0(x)y(n) + a1(x)y(n−1) + . . . + an(x)y = b(x) where a0, a1, . . . , an, b are functions of x and a0(x) = 0. Check list : If the dependent variable is y, derivatives occur upto ﬁrst degree only, no products of y and/or its derivatives are there. Neela Nataraj Lecture 1: D3
- 36. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Neela Nataraj Lecture 1: D3
- 37. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Let y(t) be the amount present at time t. Neela Nataraj Lecture 1: D3
- 38. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Let y(t) be the amount present at time t. Then dy dt = −k · y Neela Nataraj Lecture 1: D3
- 39. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Let y(t) be the amount present at time t. Then dy dt = −k · y where k is a physical constant whose value is found by experiments Neela Nataraj Lecture 1: D3
- 40. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Let y(t) be the amount present at time t. Then dy dt = −k · y where k is a physical constant whose value is found by experiments (−k is called the decay constant). Neela Nataraj Lecture 1: D3
- 41. Example : Radioactive decay A radioactive substance decomposes at a rate proportional to the amount present. Let y(t) be the amount present at time t. Then dy dt = −k · y where k is a physical constant whose value is found by experiments (−k is called the decay constant). Linear ODE of ﬁrst order. Neela Nataraj Lecture 1: D3
- 42. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Neela Nataraj Lecture 1: D3
- 43. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. Neela Nataraj Lecture 1: D3
- 44. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. Neela Nataraj Lecture 1: D3
- 45. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. d2θ dt2 + g L sin θ = 0. Neela Nataraj Lecture 1: D3
- 46. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. d2θ dt2 + g L sin θ = 0. Neela Nataraj Lecture 1: D3
- 47. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. d2θ dt2 + g L sin θ = 0. ODE of Neela Nataraj Lecture 1: D3
- 48. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. d2θ dt2 + g L sin θ = 0. ODE of second order. Neela Nataraj Lecture 1: D3
- 49. Examples - The motion of an oscillating pendulum Consider an oscillating pendulum of length L. Let θ be the angle it makes with the vertical direction. d2θ dt2 + g L sin θ = 0. ODE of second order. not linear - Non-linear DE. Neela Nataraj Lecture 1: D3
- 50. Example : A falling object A body of mass m falls under the force of gravity. Neela Nataraj Lecture 1: D3
- 51. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is Neela Nataraj Lecture 1: D3
- 52. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 Neela Nataraj Lecture 1: D3
- 53. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Neela Nataraj Lecture 1: D3
- 54. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . Neela Nataraj Lecture 1: D3
- 55. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Neela Nataraj Lecture 1: D3
- 56. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? Neela Nataraj Lecture 1: D3
- 57. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Neela Nataraj Lecture 1: D3
- 58. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 Neela Nataraj Lecture 1: D3
- 59. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 - 2nd order, linear Neela Nataraj Lecture 1: D3
- 60. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 - 2nd order, linear 2 y(4) + x2 y(3) + x3 y = xex Neela Nataraj Lecture 1: D3
- 61. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 - 2nd order, linear 2 y(4) + x2 y(3) + x3 y = xex - 4th order, linear Neela Nataraj Lecture 1: D3
- 62. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 - 2nd order, linear 2 y(4) + x2 y(3) + x3 y = xex - 4th order, linear 3 y + 5(y )3 + 6y = 0 Neela Nataraj Lecture 1: D3
- 63. Example : A falling object A body of mass m falls under the force of gravity. The drag force due to air resistance is c · v2 where v is the velocity and c is a constant Then, m dv dt = mg − c · v2 . An ODE of ﬁrst order. Linear or non-linear? (NL) Examples : 1 y + 5y + 6y = 0 - 2nd order, linear 2 y(4) + x2 y(3) + x3 y = xex - 4th order, linear 3 y + 5(y )3 + 6y = 0 - 2nd order, non-linear. Neela Nataraj Lecture 1: D3
- 64. Can we solve it? Given an equation, you would like to Neela Nataraj Lecture 1: D3
- 65. Can we solve it? Given an equation, you would like to solve it. Neela Nataraj Lecture 1: D3
- 66. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Neela Nataraj Lecture 1: D3
- 67. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: Neela Nataraj Lecture 1: D3
- 68. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? Neela Nataraj Lecture 1: D3
- 69. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? Neela Nataraj Lecture 1: D3
- 70. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? Neela Nataraj Lecture 1: D3
- 71. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? Neela Nataraj Lecture 1: D3
- 72. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? Neela Nataraj Lecture 1: D3
- 73. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? 4 How much can we proceed in a systematic manner? Neela Nataraj Lecture 1: D3
- 74. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? 4 How much can we proceed in a systematic manner? order Neela Nataraj Lecture 1: D3
- 75. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? 4 How much can we proceed in a systematic manner? order - ﬁrst, second, ..., nth , ... Neela Nataraj Lecture 1: D3
- 76. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? 4 How much can we proceed in a systematic manner? order - ﬁrst, second, ..., nth , ... linear Neela Nataraj Lecture 1: D3
- 77. Can we solve it? Given an equation, you would like to solve it. At least, try to solve it. Questions: 1 What is a solution? 2 Does an equation always have a solution? If so, how many? 3 Can the solutions be expressed in a nice form? If not, how to get a feel for it? 4 How much can we proceed in a systematic manner? order - ﬁrst, second, ..., nth , ... linear or non-linear? Neela Nataraj Lecture 1: D3
- 78. What is a solution? Consider F(x, y, y , . . . , y(n) ) = 0. Neela Nataraj Lecture 1: D3
- 79. What is a solution? Consider F(x, y, y , . . . , y(n) ) = 0. We assume that it is always possible to solve a diﬀerential equation for the highest derivative, obtaining y(n) = f (x, y, y , · · · , y(n−1) ) and study equations of this form. Neela Nataraj Lecture 1: D3
- 80. What is a solution? Consider F(x, y, y , . . . , y(n) ) = 0. We assume that it is always possible to solve a diﬀerential equation for the highest derivative, obtaining y(n) = f (x, y, y , · · · , y(n−1) ) and study equations of this form. This is to avoid the ambiguity which may arise because a single equation F(x, y, y , . . . , y(n) ) = 0 may correspond to several equations of the form y(n) = f (x, y, y , · · · , y(n−1) ). Neela Nataraj Lecture 1: D3
- 81. What is a solution? Consider F(x, y, y , . . . , y(n) ) = 0. We assume that it is always possible to solve a diﬀerential equation for the highest derivative, obtaining y(n) = f (x, y, y , · · · , y(n−1) ) and study equations of this form. This is to avoid the ambiguity which may arise because a single equation F(x, y, y , . . . , y(n) ) = 0 may correspond to several equations of the form y(n) = f (x, y, y , · · · , y(n−1) ). For example, the equation y 2 + xy + 4y = 0 leads to the two equations y = −x + x2 − 16y 2 or y = −x − x2 − 16y 2 . Neela Nataraj Lecture 1: D3
- 82. What is a solution? Consider F(x, y, y , . . . , y(n) ) = 0. We assume that it is always possible to solve a diﬀerential equation for the highest derivative, obtaining y(n) = f (x, y, y , · · · , y(n−1) ) and study equations of this form. This is to avoid the ambiguity which may arise because a single equation F(x, y, y , . . . , y(n) ) = 0 may correspond to several equations of the form y(n) = f (x, y, y , · · · , y(n−1) ). For example, the equation y 2 + xy + 4y = 0 leads to the two equations y = −x + x2 − 16y 2 or y = −x − x2 − 16y 2 . Deﬁnition A explicit solution of the ODE y(n) = f (x, y, y , · · · , y(n−1) ) on the interval α < x < β is a function φ(x) such that φ , φ , · · · , φ(n) exist and satisfy φ(n) (x) = f (x, φ, φ , · · · , φ(n−1) ), for every x in α < x < β. Neela Nataraj Lecture 1: D3
- 83. Implicit solution & Formal solution Deﬁnition A relation g(x, y) = 0 is called an implicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) if this relation deﬁnes at least one function φ(x) on an interval α < x < β, such that, this function is an explicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) in this interval. Neela Nataraj Lecture 1: D3
- 84. Implicit solution & Formal solution Deﬁnition A relation g(x, y) = 0 is called an implicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) if this relation deﬁnes at least one function φ(x) on an interval α < x < β, such that, this function is an explicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) in this interval. Examples : 1 x2 + y2 − 25 = 0 is an implicit solution of x + yy = 0 in −5 < x < 5, because it deﬁnes two functions φ1(x) = 25 − x2, φ2(x) = − 25 − x2 which are solutions of the DE in the given interval. Neela Nataraj Lecture 1: D3
- 85. Implicit solution & Formal solution Deﬁnition A relation g(x, y) = 0 is called an implicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) if this relation deﬁnes at least one function φ(x) on an interval α < x < β, such that, this function is an explicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) in this interval. Examples : 1 x2 + y2 − 25 = 0 is an implicit solution of x + yy = 0 in −5 < x < 5, because it deﬁnes two functions φ1(x) = 25 − x2, φ2(x) = − 25 − x2 which are solutions of the DE in the given interval. Verify! 2 Consider x2 + y2 + 25 = 0 =⇒ x + yy = 0 =⇒ y = − x y . Neela Nataraj Lecture 1: D3
- 86. Implicit solution & Formal solution Deﬁnition A relation g(x, y) = 0 is called an implicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) if this relation deﬁnes at least one function φ(x) on an interval α < x < β, such that, this function is an explicit solution of y(n) = f (x, y, y , · · · , y(n−1) ) in this interval. Examples : 1 x2 + y2 − 25 = 0 is an implicit solution of x + yy = 0 in −5 < x < 5, because it deﬁnes two functions φ1(x) = 25 − x2, φ2(x) = − 25 − x2 which are solutions of the DE in the given interval. Verify! 2 Consider x2 + y2 + 25 = 0 =⇒ x + yy = 0 =⇒ y = − x y . We say x2 + y2 + 25 = 0 formally satisﬁes x + yy = 0. But it is NOT an implicit solution of DE as this relation doesn’t yield φ which is an explicit solution of the DE on any real interval I. Neela Nataraj Lecture 1: D3
- 87. First order ODE & Initial Value Problem for ﬁrst order ODE We now consider ﬁrst order ODE of the form F(x, y, y ) = 0 or y = f (x, y) . Neela Nataraj Lecture 1: D3
- 88. First order ODE & Initial Value Problem for ﬁrst order ODE We now consider ﬁrst order ODE of the form F(x, y, y ) = 0 or y = f (x, y) . Consider a linear ﬁrst order ODE of the form y + a(x)y = b(x) . Neela Nataraj Lecture 1: D3
- 89. First order ODE & Initial Value Problem for ﬁrst order ODE We now consider ﬁrst order ODE of the form F(x, y, y ) = 0 or y = f (x, y) . Consider a linear ﬁrst order ODE of the form y + a(x)y = b(x) . If b(x) = 0, then we say that the equation is homogeneous. Neela Nataraj Lecture 1: D3
- 90. First order ODE & Initial Value Problem for ﬁrst order ODE We now consider ﬁrst order ODE of the form F(x, y, y ) = 0 or y = f (x, y) . Consider a linear ﬁrst order ODE of the form y + a(x)y = b(x) . If b(x) = 0, then we say that the equation is homogeneous. MA106 nostalgia : Do the solutions of a homogeneous diﬀerential equation form a vector space under usual addition and scalar multiplication? Neela Nataraj Lecture 1: D3
- 91. First order ODE & Initial Value Problem for ﬁrst order ODE We now consider ﬁrst order ODE of the form F(x, y, y ) = 0 or y = f (x, y) . Consider a linear ﬁrst order ODE of the form y + a(x)y = b(x) . If b(x) = 0, then we say that the equation is homogeneous. MA106 nostalgia : Do the solutions of a homogeneous diﬀerential equation form a vector space under usual addition and scalar multiplication? Deﬁnition Initial value problem (IVP) : A DE along with an initial condition is an IVP. y = f (x, y), y(x0) = y0. Neela Nataraj Lecture 1: D3
- 92. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. Neela Nataraj Lecture 1: D3
- 93. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Neela Nataraj Lecture 1: D3
- 94. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Initial amount given is 1 gm at time t = 0. Neela Nataraj Lecture 1: D3
- 95. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Initial amount given is 1 gm at time t = 0. i.e., y(0) = 1. Neela Nataraj Lecture 1: D3
- 96. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Initial amount given is 1 gm at time t = 0. i.e., y(0) = 1. By inspection, y = ce−kt , for an arbitrary constant c, is a solution of the above ODE. Neela Nataraj Lecture 1: D3
- 97. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Initial amount given is 1 gm at time t = 0. i.e., y(0) = 1. By inspection, y = ce−kt , for an arbitrary constant c, is a solution of the above ODE. The initial condition determines c = Neela Nataraj Lecture 1: D3
- 98. Examples Given an amount of a radioactive substance, say 1 gm, ﬁnd the amount present at any later time. The relevant ODE is dy dt = −k · y. Initial amount given is 1 gm at time t = 0. i.e., y(0) = 1. By inspection, y = ce−kt , for an arbitrary constant c, is a solution of the above ODE. The initial condition determines c = 1. Hence y = e−kt is a particular solution to the above ODE with the given initial condition. Neela Nataraj Lecture 1: D3
- 99. Examples Find the curve through the point (1, 1) in the xy-plane having at each of its points, the slope − y x . Neela Nataraj Lecture 1: D3
- 100. Examples Find the curve through the point (1, 1) in the xy-plane having at each of its points, the slope − y x . The relevant ODE is y = − y x . Neela Nataraj Lecture 1: D3
- 101. Examples Find the curve through the point (1, 1) in the xy-plane having at each of its points, the slope − y x . The relevant ODE is y = − y x . By inspection, y = c x is its general solution for an arbitrary constant c; that is, a family of hyperbolas. Neela Nataraj Lecture 1: D3
- 102. Examples Find the curve through the point (1, 1) in the xy-plane having at each of its points, the slope − y x . The relevant ODE is y = − y x . By inspection, y = c x is its general solution for an arbitrary constant c; that is, a family of hyperbolas. The initial condition given is y(1) = 1, which implies c = 1. Neela Nataraj Lecture 1: D3
- 103. Examples Find the curve through the point (1, 1) in the xy-plane having at each of its points, the slope − y x . The relevant ODE is y = − y x . By inspection, y = c x is its general solution for an arbitrary constant c; that is, a family of hyperbolas. The initial condition given is y(1) = 1, which implies c = 1. Hence the particular solution for the above problem is y = 1 x . Neela Nataraj Lecture 1: D3
- 104. Examples A ﬁrst order IVP can have 1 NO solution : Neela Nataraj Lecture 1: D3
- 105. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. Neela Nataraj Lecture 1: D3
- 106. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : Neela Nataraj Lecture 1: D3
- 107. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. Neela Nataraj Lecture 1: D3
- 108. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. What is the solution? 3 Inﬁnitely many solutions: Neela Nataraj Lecture 1: D3
- 109. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. What is the solution? 3 Inﬁnitely many solutions: xy = y − 1, y(0) = 1 Neela Nataraj Lecture 1: D3
- 110. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. What is the solution? 3 Inﬁnitely many solutions: xy = y − 1, y(0) = 1 The solutions are y = 1 + cx. Neela Nataraj Lecture 1: D3
- 111. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. What is the solution? 3 Inﬁnitely many solutions: xy = y − 1, y(0) = 1 The solutions are y = 1 + cx. Motivation to study conditions under which the solution would exist and the conditions under which it will be unique! Neela Nataraj Lecture 1: D3
- 112. Examples A ﬁrst order IVP can have 1 NO solution : |y | + |y| = 0, y(0) = 3. 2 Precisely one solution : y = x, y(0) = 1. What is the solution? 3 Inﬁnitely many solutions: xy = y − 1, y(0) = 1 The solutions are y = 1 + cx. Motivation to study conditions under which the solution would exist and the conditions under which it will be unique! We ﬁrst start with a few methods for ﬁnding out the solution of ﬁrst order ODEs, discuss the geometric meaning of solutions and then proceed to study existence-uniqueness results. Neela Nataraj Lecture 1: D3
- 113. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Neela Nataraj Lecture 1: D3
- 114. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Neela Nataraj Lecture 1: D3
- 115. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Substituting in the DE, we obtain H1(x) + H2(y)y = 0. Neela Nataraj Lecture 1: D3
- 116. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Substituting in the DE, we obtain H1(x) + H2(y)y = 0. Using chain rule, d dx H2(y) = H2(y) dy dx . Neela Nataraj Lecture 1: D3
- 117. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Substituting in the DE, we obtain H1(x) + H2(y)y = 0. Using chain rule, d dx H2(y) = H2(y) dy dx . Hence, d dx (H1(x) + H2(y)) = 0. Neela Nataraj Lecture 1: D3
- 118. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Substituting in the DE, we obtain H1(x) + H2(y)y = 0. Using chain rule, d dx H2(y) = H2(y) dy dx . Hence, d dx (H1(x) + H2(y)) = 0. Integrating, H1(x) + H2(y) = c, where c is an arbirtary constant. Note: Neela Nataraj Lecture 1: D3
- 119. Separable ODE’s An ODE of the form M(x) + N(y)y = 0 is called a separable ODE. Let H1(x) and H2(y) be any functions such that H1(x) = M(x) and H2(y) = N(y). Substituting in the DE, we obtain H1(x) + H2(y)y = 0. Using chain rule, d dx H2(y) = H2(y) dy dx . Hence, d dx (H1(x) + H2(y)) = 0. Integrating, H1(x) + H2(y) = c, where c is an arbirtary constant. Note: This method many times gives us an implicit solution and not necessarily an explicit one! Neela Nataraj Lecture 1: D3
- 120. Separable ODE - Example 1 Solve the DE : y = −2xy. Neela Nataraj Lecture 1: D3
- 121. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : Neela Nataraj Lecture 1: D3
- 122. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Neela Nataraj Lecture 1: D3
- 123. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Integrating both sides, we obtain : Neela Nataraj Lecture 1: D3
- 124. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Integrating both sides, we obtain : ln |y| = −x2 + c1. Neela Nataraj Lecture 1: D3
- 125. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Integrating both sides, we obtain : ln |y| = −x2 + c1. Thus, the solutions are Neela Nataraj Lecture 1: D3
- 126. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Integrating both sides, we obtain : ln |y| = −x2 + c1. Thus, the solutions are y = ce−x2 . Neela Nataraj Lecture 1: D3
- 127. Separable ODE - Example 1 Solve the DE : y = −2xy. Separating the variables, we get : dy y = −2xdx. Integrating both sides, we obtain : ln |y| = −x2 + c1. Thus, the solutions are y = ce−x2 . How do they look? Neela Nataraj Lecture 1: D3
- 128. If we are given an initial condition y(x0) = y0, Neela Nataraj Lecture 1: D3
- 129. If we are given an initial condition y(x0) = y0, then we get: c = y0ex2 0 Neela Nataraj Lecture 1: D3
- 130. If we are given an initial condition y(x0) = y0, then we get: c = y0ex2 0 and y = y0ex2 0 −x2 . Neela Nataraj Lecture 1: D3
- 131. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Neela Nataraj Lecture 1: D3
- 132. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Neela Nataraj Lecture 1: D3
- 133. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Neela Nataraj Lecture 1: D3
- 134. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, Neela Nataraj Lecture 1: D3
- 135. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, ln |y| + y2 = sin x + c. Neela Nataraj Lecture 1: D3
- 136. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, ln |y| + y2 = sin x + c. As y(0) = 1, we get c = 1. Neela Nataraj Lecture 1: D3
- 137. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, ln |y| + y2 = sin x + c. As y(0) = 1, we get c = 1. Hence a particular solution to the IVP is ln |y| + y2 = sin x + 1. Neela Nataraj Lecture 1: D3
- 138. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, ln |y| + y2 = sin x + c. As y(0) = 1, we get c = 1. Hence a particular solution to the IVP is ln |y| + y2 = sin x + 1. Note: y ≡ 0 is a solution to the DE Neela Nataraj Lecture 1: D3
- 139. Separable ODE - Example 2 Find the solution to the initial value problem: dy dx = y cos x 1 + 2y2 ; y(0) = 1. Assume y = 0. Then, 1 + 2y2 y dy = cos x dx. Integrating, ln |y| + y2 = sin x + c. As y(0) = 1, we get c = 1. Hence a particular solution to the IVP is ln |y| + y2 = sin x + 1. Note: y ≡ 0 is a solution to the DE but it is not a solution to the given IVP. Neela Nataraj Lecture 1: D3
- 140. Separable ODE - Example 3 Escape velocity. Neela Nataraj Lecture 1: D3
- 141. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Neela Nataraj Lecture 1: D3
- 142. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. Neela Nataraj Lecture 1: D3
- 143. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Neela Nataraj Lecture 1: D3
- 144. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x Neela Nataraj Lecture 1: D3
- 145. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) Neela Nataraj Lecture 1: D3
- 146. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) Neela Nataraj Lecture 1: D3
- 147. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , Neela Nataraj Lecture 1: D3
- 148. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , where R is the radius of the earth. Neela Nataraj Lecture 1: D3
- 149. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , where R is the radius of the earth. Neglect force due to air resistance and other celestial bodies. Neela Nataraj Lecture 1: D3
- 150. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , where R is the radius of the earth. Neglect force due to air resistance and other celestial bodies. Therefore, the equation of motion is Neela Nataraj Lecture 1: D3
- 151. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , where R is the radius of the earth. Neglect force due to air resistance and other celestial bodies. Therefore, the equation of motion is m dv dt = − mgR2 (R + x)2 ; Neela Nataraj Lecture 1: D3
- 152. Separable ODE - Example 3 Escape velocity. A projectile of mass m moves in a direction perpendicular to the surface of the earth. Suppose v0 is its initial velocity. We want to calculate the height the projectile reaches. Its weight at height x (from the surface of the earth) is given by, w(x) = − mgR2 (R + x)2 , where R is the radius of the earth. Neglect force due to air resistance and other celestial bodies. Therefore, the equation of motion is m dv dt = − mgR2 (R + x)2 ; v(0) = v0. Neela Nataraj Lecture 1: D3
- 153. Separable ODE’s By chain rule, dv dt Neela Nataraj Lecture 1: D3
- 154. Separable ODE’s By chain rule, dv dt = dv dx · dx dt Neela Nataraj Lecture 1: D3
- 155. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Neela Nataraj Lecture 1: D3
- 156. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx Neela Nataraj Lecture 1: D3
- 157. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . Neela Nataraj Lecture 1: D3
- 158. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is Neela Nataraj Lecture 1: D3
- 159. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Neela Nataraj Lecture 1: D3
- 160. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? Neela Nataraj Lecture 1: D3
- 161. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Neela Nataraj Lecture 1: D3
- 162. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Separating the variables and integrating, we get: Neela Nataraj Lecture 1: D3
- 163. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Separating the variables and integrating, we get: v2 2 = gR2 R + x + c. Neela Nataraj Lecture 1: D3
- 164. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Separating the variables and integrating, we get: v2 2 = gR2 R + x + c. For x = 0, we get v2 0 2 = gR + c, Neela Nataraj Lecture 1: D3
- 165. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Separating the variables and integrating, we get: v2 2 = gR2 R + x + c. For x = 0, we get v2 0 2 = gR + c, hence, c = v2 0 2 − gR, Neela Nataraj Lecture 1: D3
- 166. Separable ODE’s By chain rule, dv dt = dv dx · dx dt = v · dv dx . Thus, v · dv dx = − gR2 (R + x)2 . This ODE is separable. Linear or non-linear? (NL) Separating the variables and integrating, we get: v2 2 = gR2 R + x + c. For x = 0, we get v2 0 2 = gR + c, hence, c = v2 0 2 − gR, and, v = ± v2 0 − 2gR + 2gR2 R + x . Neela Nataraj Lecture 1: D3
- 167. Separable ODE’s Suppose the body reaches the maximum height H. Neela Nataraj Lecture 1: D3
- 168. Separable ODE’s Suppose the body reaches the maximum height H. Then v = Neela Nataraj Lecture 1: D3
- 169. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. Neela Nataraj Lecture 1: D3
- 170. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Neela Nataraj Lecture 1: D3
- 171. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Thus, v2 0 = 2gR − 2gR2 R + H = 2gR H R + H . Neela Nataraj Lecture 1: D3
- 172. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Thus, v2 0 = 2gR − 2gR2 R + H = 2gR H R + H . The escape velocity is found by taking limit as Neela Nataraj Lecture 1: D3
- 173. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Thus, v2 0 = 2gR − 2gR2 R + H = 2gR H R + H . The escape velocity is found by taking limit as H → ∞. Neela Nataraj Lecture 1: D3
- 174. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Thus, v2 0 = 2gR − 2gR2 R + H = 2gR H R + H . The escape velocity is found by taking limit as H → ∞. Thus, ve = 2gR Neela Nataraj Lecture 1: D3
- 175. Separable ODE’s Suppose the body reaches the maximum height H. Then v = 0 at this height. v2 0 − 2gR + 2gR2 (R + H) = 0. Thus, v2 0 = 2gR − 2gR2 R + H = 2gR H R + H . The escape velocity is found by taking limit as H → ∞. Thus, ve = 2gR ∼ 11 km/sec. Neela Nataraj Lecture 1: D3