Mathematical Methods - Lecture 7
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1 Mathematical Methods - Lecture 7 Yuliya Tarabalka Inria Sophia-Antipolis Méditerranée, Titane team, Tel.: +33 (0) yuliya.tarabalka@inria.fr Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 1 / 31
2 Outline 1 Ordinary Differential Equations 2 First-order Differential Equations Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 2 / 31
3 Ordinary Differential Equations What is a differential equation? Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 3 / 31
4 Ordinary Differential Equations What is a differential equation? Differential equations are the group of equations that contain derivatives. Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 4 / 31
5 Ordinary Differential Equations What is a differential equation? Differential equations are the group of equations that contain derivatives. An ordinary differential equation (ODE): contains only ordinary derivatives (no partial derivatives) describes the relationship between these derivatives of the dependent variable (ex: y) with respect to the independent variable (ex: x) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 4 / 31
6 Ordinary Differential Equations What is a differential equation? Differential equations are the group of equations that contain derivatives. An ordinary differential equation (ODE): contains only ordinary derivatives (no partial derivatives) describes the relationship between these derivatives of the dependent variable (ex: y) with respect to the independent variable (ex: x) The solution to such an ODE is a function of x: y(x) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 4 / 31
7 Ordinary Differential Equations The simplest type of differential equation The simplest ODEs have the form: d n x dt n = G(t), where G(t) depends only on the independent variable t Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 5 / 31
8 Ordinary Differential Equations The simplest type of differential equation The simplest ODEs have the form: d n x dt n = G(t), where G(t) depends only on the independent variable t Newton s law: F = ma m d 2 x dt 2 = mg, x = height of the object over the ground, m = its mass, g = constant gravitational acceleration Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 5 / 31
9 Ordinary Differential Equations The simplest type of differential equation - Example m d 2 x dt 2 = mg d 2 x dt 2 = g Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 6 / 31
10 Ordinary Differential Equations The simplest type of differential equation - Example m d 2 x dt 2 = mg d 2 x dt 2 = g The first integration yields: dx dt = A gt, with A the constant of integration Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 6 / 31
11 Ordinary Differential Equations The simplest type of differential equation - Example m d 2 x dt 2 = mg d 2 x dt 2 = g The first integration yields: dx dt = A gt, with A the constant of integration The second integration yields the general solution: x = B + At 1 2 gt2, with B the second constant of integration Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 6 / 31
12 Ordinary Differential Equations The simplest type of differential equation - Example x = B + At 1 2 gt2 Two constants of integration A and B can be found from initial conditions: dx x(0) = x 0, ; dt (0) = v 0 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 7 / 31
13 Ordinary Differential Equations The simplest type of differential equation - Example x = B + At 1 2 gt2 Two constants of integration A and B can be found from initial conditions: dx x(0) = x 0, ; dt (0) = v 0 We get: x(0) = x 0 = B Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 7 / 31
14 Ordinary Differential Equations The simplest type of differential equation - Example x = B + At 1 2 gt2 Two constants of integration A and B can be found from initial conditions: dx x(0) = x 0, ; dt (0) = v 0 We get: x(0) = x 0 = B dx dt (0) = v 0 = A Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 7 / 31
15 Ordinary Differential Equations The simplest type of differential equation - Example General solution: x = B + At 1 2 gt2 Particular solution (unique solution that satisfies both the ODE and the initial conditions): x(t) = x 0 + v 0 t 1 2 gt2 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 8 / 31
16 Ordinary Differential Equations The simplest type of differential equation - Example General solution: x = B + At 1 2 gt2 Particular solution (unique solution that satisfies both the ODE and the initial conditions): x(t) = x 0 + v 0 t 1 2 gt2 Example: We drop a ball off from the top of a 50 meter building with v 0 = 0. When will the ball hit the ground? Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 8 / 31
17 Ordinary Differential Equations Order of differential equations The order of an ODE is the order of the highest derivative it contains dy dx d 2 y dx 2 d 3 y dx 3 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 9 / 31
18 First-order differential equations The general first-degree first-order differential equation for the function y = y(x) can be written in either of two standard forms: dy dx = f (x, y), A(x, y)dx + B(x, y)dy = 0, f (x, y), A(x, y), B(x, y) are in general functions of both x and y Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 10 / 31
19 First-order differential equations The general first-degree first-order differential equation for the function y = y(x) can be written in either of two standard forms: dy dx = f (x, y), A(x, y)dx + B(x, y)dy = 0, f (x, y), A(x, y), B(x, y) are in general functions of both x and y How to solve? Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 10 / 31
20 The Euler method First-order Differential Equations The general first-order differential equation for the function y = y(x): dy dx = f (x, y) (1) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 11 / 31
21 The Euler method First-order Differential Equations The general first-order differential equation for the function y = y(x): dy dx = f (x, y) (1) It is not always possible to find an analytical solution of (1) for y = y(x) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 11 / 31
22 The Euler method First-order Differential Equations The general first-order differential equation for the function y = y(x): dy dx = f (x, y) (1) It is not always possible to find an analytical solution of (1) for y = y(x) It is always possible to determine a unique NUMERICAL solution given: an initial vaue y(x 0 ) = y 0 provided f (x, y) is a well-behaved function Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 11 / 31
23 The Euler method = the simplest Runge-Kutta method ODE dy dx = f (x, y) gives the slope f (x 0, y 0 ) of the tangent line to the solution curve y = y(x) at the point (x 0, y 0 ): dy y(x 0 + x) y(x 0 ) (0) = lim = f (x 0, y 0 ) dx x 0 x Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 12 / 31
24 The Euler method = the simplest Runge-Kutta method 1 The first point: (x 0, y 0 ) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 13 / 31
25 The Euler method = the simplest Runge-Kutta method 1 The first point: (x 0, y 0 ) 2 The next point is obtained by choosing a small x, and computing the next y-coordinate (along the tangent line): x = x 0 + x, y(x 0 + x) = y(x 0 ) + xf (x 0, y 0 ) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 13 / 31
26 The Euler method = the simplest Runge-Kutta method 1 The first point: (x 0, y 0 ) 2 The next point is obtained by choosing a small x, and computing the next y-coordinate (along the tangent line): x = x 0 + x, y(x 0 + x) = y(x 0 ) + xf (x 0, y 0 ) 3 (x 0 + x, y 0 + y) becomes the initial condition and we repeat step 2 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 13 / 31
27 The Euler method = the simplest Runge-Kutta method For small enough x, the numerical solution converges to the exact solution! Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 14 / 31
28 Analytical solution First-order Differential Equations The general first-order differential equation for the function y = y(x): dy dx = f (x, y), Some special forms of this equation can be solved analytically: separable equations exact equations inexact equations linear equations... Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 15 / 31
29 Separable equations First-order Differential Equations A first-order ODE is separable if it can be written in the form: or g(y) dy dx = f (x) dy dx = f (x)g(y) where f (x) is independent of y and g(y) is indepedent of x Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 16 / 31
30 Separable equations First-order Differential Equations A first-order ODE is separable if it can be written in the form: g(y) dy dx = f (x) 1 Rearrange (factorise if needed) the equation so that the terms depending on x and y appear on opposite sides: 2 Integrate: g(y)dy = f (x)dx g(y)dy = f (x)dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 17 / 31
31 Separable equations - Example Solve: dy dx = x + xy Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 18 / 31
32 Separable equations - Example Solve: Since x + xy = x(1 + y): dy dx = x + xy dy 1 + y = xdx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 18 / 31
33 Separable equations - Example Solve: Since x + xy = x(1 + y): dy dx = x + xy dy 1 + y = xdx ln(1 + y) = x 2 + c, c = const 2 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 18 / 31
34 Separable equations - Example Solve: Since x + xy = x(1 + y): dy dx = x + xy dy 1 + y = xdx ln(1 + y) = x 2 + c, c = const y = exp ( x c) = A exp (x 2 2 ), A = const Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 18 / 31
35 Separable equations - Example Solve: dy dx 2 cos 2x =, y(0) = y Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 19 / 31
36 Separable equations - Example Solve: dy dx 2 cos 2x =, y(0) = y Solution: y(x) = sin 2x Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 19 / 31
37 Linear equations First-order Differential Equations The first-order linear differential equation (linear in y and its derivative) can be written in the form: dy + p(x)y = g(x) dx with (optional) the initial condition y(x 0 ) = y 0 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 20 / 31
38 Linear equations First-order Differential Equations The first-order linear differential equation (linear in y and its derivative) can be written in the form: dy + p(x)y = g(x) dx with (optional) the initial condition y(x 0 ) = y 0 These equations can be integrated using an integrating factor µ(x) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 20 / 31
39 Linear equations First-order Differential Equations dy + p(x)y = g(x) µ(x) [dy + p(x)y] = µ(x)g(x) dx dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 21 / 31
40 Linear equations dy + p(x)y = g(x) µ(x) [dy + p(x)y] = µ(x)g(x) dx dx We need to determine µ(x) so that: µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 21 / 31
41 Linear equations First-order Differential Equations dy + p(x)y = g(x) µ(x) [dy + p(x)y] = µ(x)g(x) dx dx We need to determine µ(x) so that: µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] d [µ(x)y] = µ(x)g(x) dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 21 / 31
42 Linear equations First-order Differential Equations dy + p(x)y = g(x) µ(x) [dy + p(x)y] = µ(x)g(x) dx dx We need to determine µ(x) so that: Using µ(x 0 ) = µ 0 and y(x 0 ) = y 0 : µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] d [µ(x)y] = µ(x)g(x) dx µ(x)y µ 0 y 0 = x 0 x µ(x)g(x)dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 21 / 31
43 Linear equations First-order Differential Equations dy + p(x)y = g(x) µ(x) [dy + p(x)y] = µ(x)g(x) dx dx We need to determine µ(x) so that: Using µ(x 0 ) = µ 0 and y(x 0 ) = y 0 : µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] d [µ(x)y] = µ(x)g(x) dx µ(x)y µ 0 y 0 = 1 y = µ(x) (µ 0y 0 + x 0 x x 0 x µ(x)g(x)dx µ(x)g(x)dx) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 21 / 31
44 Linear equations First-order Differential Equations Next step: Determine µ(x) from µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 22 / 31
45 Linear equations Next step: Determine µ(x) from µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Using product rule for differentiation: µ dy dµ + pµy = dx dx y + µdy dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 22 / 31
46 Linear equations First-order Differential Equations Next step: Determine µ(x) from µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Using product rule for differentiation: µ dy dµ + pµy = dx dx y + µdy dx dµ dx = pµ Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 22 / 31
47 Linear equations First-order Differential Equations Next step: Determine µ(x) from µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Using product rule for differentiation: This equation is separable: µ dy dµ + pµy = dx dx y + µdy dx dµ dx = pµ µ dµ µ 0 µ = x 0 x p(x)dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 22 / 31
48 Linear equations First-order Differential Equations Next step: Determine µ(x) from µ(x) [ dy dx + p(x)y] = d dx [µ(x)y] Using product rule for differentiation: This equation is separable: x µ dy dµ + pµy = dx dx y + µdy dx dµ dx = pµ µ dµ µ 0 µ = x 0 x p(x)dx ln µ = p(x)dx µ(x) = µ 0 exp ( µ 0 x 0 x 0 x p(x)dx) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 22 / 31
49 Linear equations First-order Differential Equations The first-order linear differential equation: dy + p(x)y = g(x) dx Its solution satisfying the initial condition y(x 0 ) = y 0 is written as: with 1 y = µ(x) (y 0 + x 0 x µ(x)g(x)dx) µ(x) = exp ( x 0 x p(x)dx) Frequent use in applied mathematics! Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 23 / 31
50 Linear equations - Example Solve: dy dx + 2y = e x, with y(0) = 3/4 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
51 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
52 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable With p(x) = 2 and g(x) = e x, we have: µ(x) = exp ( 0 x 2dx) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
53 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable With p(x) = 2 and g(x) = e x, we have: µ(x) = exp ( = e 2x 0 x 2dx) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
54 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable With p(x) = 2 and g(x) = e x, we have: µ(x) = exp ( 0 x 2dx) and y = e 2x ( x = e 2x e 2x e x dx) = e 2x ( x e x dx) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
55 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable With p(x) = 2 and g(x) = e x, we have: µ(x) = exp ( 0 x 2dx) and y = e 2x ( x = e 2x e 2x e x dx) = e 2x ( x e x dx) = e 2x ( (ex 1)) = e 2x (e x 1 4 ) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
56 Linear equations - Example dy Solve: dx + 2y = e x, with y(0) = 3/4 This equation is not separable With p(x) = 2 and g(x) = e x, we have: µ(x) = exp ( 0 x 2dx) and y = e 2x ( x = e 2x e 2x e x dx) = e 2x ( x e x dx) = e 2x ( (ex 1)) = e 2x (e x 1 4 ) e x (1 1 4 e x ) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 24 / 31
57 Example 2 First-order Differential Equations Solve: dy 2xy = x, with y(0) = 0 dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 25 / 31
58 Example 2 First-order Differential Equations Solve: dy 2xy = x, with y(0) = 0 dx Solution: y = 1 2 (ex2 1) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 25 / 31
59 Bernoulli s equation First-order Differential Equations Bernoulli s equation has the form: dy dx + P(x)y = Q(x)y n, where n 0 or 1 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 26 / 31
60 Bernoulli s equation First-order Differential Equations Bernoulli s equation has the form: dy dx + P(x)y = Q(x)y n, where n 0 or 1 The equation can be made linear by substituting v = y 1 n and correspondingly: dy dx = ( y n 1 n ) dv dx Bernoulli s equation becomes linear: dv + (1 n)p(x)v = (1 n)q(x) dx Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 26 / 31
61 Business analytics application: compound interest Equation for growth of an investment with continuous compounding of interest? Yuliya Tarabalka Differential Equations 27 / 31
62 Business analytics application: compound interest Equation for growth of an investment with continuous compounding of interest? S(t) = value of the investment at time t r = annual interest rate compounded after every time interval t k = annual deposit amount suppose that an installment is deposited after every time interval t Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 27 / 31
63 Business analytics application: compound interest Equation for growth of an investment with continuous compounding of interest? S(t) = value of the investment at time t r = annual interest rate compounded after every time interval t k = annual deposit amount suppose that an installment is deposited after every time interval t The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 27 / 31
64 Business analytics application: compound interest Equation for growth of an investment with continuous compounding of interest? S(t) = value of the investment at time t r = annual interest rate compounded after every time interval t k = annual deposit amount suppose that an installment is deposited after every time interval t The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t Example: S(t) = $10000, r = 6% per year, t = 1 month = 1/12 year Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 27 / 31
65 Business analytics application: compound interest Equation for growth of an investment with continuous compounding of interest? S(t) = value of the investment at time t r = annual interest rate compounded after every time interval t k = annual deposit amount suppose that an installment is deposited after every time interval t The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t Example: S(t) = $10000, r = 6% per year, t = 1 month = 1/12 year The interest awarded after 1 month is r ts = (0.06/12) $10000 = $50 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 27 / 31
66 Business analytics application: compound interest The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t S(t + t) S(t) t = rs(t) + k Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 28 / 31
67 Business analytics application: compound interest The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t S(t + t) S(t) t = rs(t) + k The ODE for continuous compounding of interest and continuous deposits is obtained by taking the limit t 0: Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 28 / 31
68 Business analytics application: compound interest The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t S(t + t) S(t) t = rs(t) + k The ODE for continuous compounding of interest and continuous deposits is obtained by taking the limit t 0: ds dt = rs + k Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 28 / 31
69 Business analytics application: compound interest The value of the investment at the time t + t is given by: S(t + t) = S(t) + (r t)s(t) + k t S(t + t) S(t) t = rs(t) + k The ODE for continuous compounding of interest and continuous deposits is obtained by taking the limit t 0: ds dt = rs + k It can be solved with the initial condition S(0) = S 0 S 0 = initial capital Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 28 / 31
70 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
71 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k S 0 S ds rs + k = t dt 0 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
72 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k S 0 S ds rs + k = t dt 0 1 rs + k ln ( r rs 0 + k ) = t Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
73 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k S 0 S ds rs + k = t dt 0 1 rs + k ln ( r rs 0 + k ) = t rs + k = (rs 0 + k)e rt Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
74 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k S 0 S ds rs + k = t dt 0 1 rs + k ln ( r rs 0 + k ) = t rs + k = (rs 0 + k)e rt S = rs 0e rt + ke rt k r Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
75 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k S 0 S ds rs + k = t dt 0 1 rs + k ln ( r rs 0 + k ) = t rs + k = (rs 0 + k)e rt S = rs 0e rt + ke rt k r S = S 0 e rt + k r ert (1 e rt ) Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 29 / 31
76 Business analytics application: compound interest The ODE for the growth of an investment: ds dt = rs + k The solution: S = S 0 e rt + k r ert (1 e rt ) The first term on the right-hand side comes from the initial invested capital The second term comes from deposits (or withdrawals) Compounding results in the exponential growth of an investment Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 30 / 31
77 Business analytics application: compound interest The solution to the ODE for the growth of an investment: S = S 0 e rt + k r ert (1 e rt ) Exercise: A 25-year old plans to set aside a fixed amount every year, invests at a real return of 6%, and retires at age 65. How much must he invest every year to have $8,000,000 at retirement? Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 31 / 31
78 Business analytics application: compound interest The solution to the ODE for the growth of an investment: S = S 0 e rt + k r ert (1 e rt ) Exercise: A 25-year old plans to set aside a fixed amount every year, invests at a real return of 6%, and retires at age 65. How much must he invest every year to have $8,000,000 at retirement? Solution: k = rs(t) e rt 1 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 31 / 31
79 Business analytics application: compound interest The solution to the ODE for the growth of an investment: S = S 0 e rt + k r ert (1 e rt ) Exercise: A 25-year old plans to set aside a fixed amount every year, invests at a real return of 6%, and retires at age 65. How much must he invest every year to have $8,000,000 at retirement? Solution: k = rs(t) e rt 1 k = , 000, 000 e = $47, 889year 1 Yuliya Tarabalka (yuliya.tarabalka@inria.fr) Differential Equations 31 / 31
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