MAT238/ MAT438 : Foundation of Applied Mathematics 194 Reminder : c x ln x dx = + 1 (don’t ever do this!!) But, dx ln x c x = + 1 (this is the correct formula) ✓ Because ( ) x ln x dx d 1 = ✓ the correct method to integrate ln x dx is using integration by parts (refer example 1, in page 84 above) Example 2 ( ) NR • NR dx Evaluate x ln x dx 2 . Solution : ∫ 2 = ∫ = − ∫ = 3 3 ∙ − ∫ 3 3 ∙ 1 = 1 3 3 − 1 3 ∫ 2 = 1 3 3 − 1 3 ( 3 3 ) + = 1 3 3 − 1 9 3 + # = = 2 = 1 = 3 3
Chapter 3 : Integration Techniques 195 Example 3 ( ) NR • NR dx Evaluate − x tan x dx 3 1 . Solution : ∫ 3 −1 = ∫ −1 ∙ 3 = ∫ = − ∫ = 4 4 −1 − ∫ 4 4 • 1 1 + 2 = 1 4 4 −1 − 1 4 ∫ 4 1 + 2 ∗∗ = 1 4 4 −1 − 1 4 ( 3 3 − + −1 ) + = 1 4 4 −1 − 1 12 3 + 1 4 − 1 4 −1 + # = −1 = 3 = 1 1 + 2 = 4 4 2 + 0 + 1 4 + 0 3 + 0 2 + 0 + 0 2 − 1 −( 4 + 0 3 + 2 ) 1 quotient remainder divisor 4 1 + 2 = − 2 + 0 + 0 −(− 2 − 0 − 1) 4 1 + 2 = + 4 1 + 2 = ( 2 − 1) + 1 1 + 2 ∗∗ ∫ 4 1 + 2 = ∫( 2 − 1) + ∫ 1 1 + 2 = 3 3 − + −1 +
MAT238/ MAT438 : Foundation of Applied Mathematics 196 Example 4 ( ) NR dx Evaluate − sin 2x dx 1 . Solution : ∫ −1 (2) = ∫ = − ∫ = −1 (2) − ∫ ∙ 2 √1 − 4 2 = −1 (2) − ∫ 2 √1 − 4 2 ∗∗ = −1 (2) − (− 1 2 √1 − 4 2) + = −1 (2) + 1 2 √1 − 4 2 + # = −1 (2) = = 1 √1− (2) 2 ∙ 2 = 2 √1− 4 2 = ∗∗ ∫ 2 √1 − 4 2 = ∫ 2 √ ∙ −8 = − 1 4 ∫ 1 √ = − 1 4 ∫ − 1 2 = − 1 4 ∙ 1 2 ( 1 2 ) + = − 1 4 ∙ 2 1 √ + = − 1 2 √ + = − 1 2 √1− 4 2 + = 1− 4 2 = −8 = −8
Chapter 3 : Integration Techniques 197 3.5 A Shortcut : Tabular Integration Tabular Integration works for integrals of the form ( ) NR • R dx or ( ) R • R dx Example 5 ( ) NR • R dx Evaluate x e dx 2 3x . Solution : Method I (using Tabular Integration) Sign u (differentiate) dv (integrate) + 2 x x e 3 − 2x 3 3x e + 2 9 3x e − 0 27 3x e x e dx 2 3x = • − • + • − • dx e e e x e x x x x x 27 0 27 2 9 2 3 3 3 3 3 2 = − e + e − dx x e x x x x 0 27 2 9 2 3 3 3 3 2 = x e xe e C x x x − + + 2 3 3 3 27 2 9 2 3 1 # c a e e dx ax ax = +
MAT238/ MAT438 : Foundation of Applied Mathematics 198 Method II : using u dv = uv − v du x e dx 2 3x = u dv = uv − v du = • − • dx e x e x x x 3 2 3 3 3 2 = • − x e dx e x x x 3 3 2 3 2 3 = x e − u dv x 3 2 3 1 2 3 = − − x e uv v du x 3 2 3 1 2 3 = − • − dx e e x e x x x x 3 3 3 2 3 1 3 3 2 3 = x e − x e + e dx 2 3x 3x 3x 9 2 9 2 3 1 = C e x e x e x x x − + • + 9 3 2 9 2 3 1 3 2 3 3 = x e x e e C x x x − + + 2 3 3 3 27 2 9 2 3 1 # Try this !! How about x e dx x 2 3 ?? u x dv e dx 3x = = 3 3x e du =dx v = u x dv e dx 2 3x = = 3 2 3x e du = x dx v =
Chapter 3 : Integration Techniques 199 Example 6 ( ) NR • R dx Evaluate ( ) x sin 4x dx 2 . Solution : Method I (using Tabular Integration) ( ) x sin 4x dx 2 = − x • x + x • x + • x − • cos4x dx 64 1 cos4 0 64 1 sin4 2 16 1 cos4 2 4 2 1 = − x x + x x + cos4x − 0dx 32 1 sin4 8 1 cos4 4 1 2 = − x x + x x + cos4x + C 32 1 sin4 8 1 cos4 4 1 2 # Sign u (differentiate) dv (integrate) + − 2x + 2 − 0
MAT238/ MAT438 : Foundation of Applied Mathematics 200 Method II : using u dv = uv − v du ( ) x sin 4x dx 2 = u dv = uv − v du = ( ) ( ) − x x − x • − cos 4x dx 4 1 cos 4 2 4 1 2 = ( ) ( ) − x x + x cos 4x dx 2 1 cos 4 4 1 2 = ( ) − x x + u dv 2 1 cos 4 4 1 2 = ( ) − + − x x uv v du 2 1 cos 4 4 1 2 = ( ) ( ) ( ) − + • − x x x x sin 4x dx 4 1 sin 4 4 1 2 1 cos 4 4 1 2 = ( ) ( ) ( ) − x x + x x − sin 4x dx 8 1 sin 4 8 1 cos 4 4 1 2 = x ( x) x ( x) cos(4x) 4 1 8 1 sin 4 8 1 cos 4 4 1 2 − + − • − = x ( x) x ( x) cos(4x) 32 1 sin 4 8 1 cos 4 4 1 2 − + + # u = x dv = cos(4x)dx du dx v sin(4x) 4 1 = = u x dv sin(4x)dx 2 = = du x dx v cos(4x) 4 1 = 2 = −
Chapter 3 : Integration Techniques 201 Example 7 ( ) NR • R dx Evaluate ( ) x cos 2x dx 2 . Solution : Method I (using Tabular Integration) Sign u (differentiate) dv (integrate) + 2 x cos 2x − 2x sin2x 2 1 + 2 cos2x 4 1 − − 0 sin2x 8 1 − ( ) x cos 2x dx 2 = x • x + x • x − • x − • sin2x dx 8 1 sin2 0 8 1 cos2 2 4 1 sin2 2 2 2 1 = x x + x x − sin2x − 0dx 4 1 cos2 2 1 sin2 2 1 2 = x x + x x − sin2x + C 4 1 cos2 2 1 sin2 2 1 2 # c a cos ax sinax dx + − = c a sinax cos ax dx = +
MAT238/ MAT438 : Foundation of Applied Mathematics 202 Method II : using u dv = uv − v du ( ) x cos 2x dx 2 = u dv = uv − v du = ( ) ( ) x x − x • sin 2x dx 2 1 sin 2 2 2 1 2 = ( ) ( ) x sin 2x − x sin 2x dx 2 1 2 = ( ) x sin 2x − u dv 2 1 2 = ( ) − − x sin 2x uv v du 2 1 2 ( ) x cos 2x dx 2 = ( ) ( ) ( ) − • − − − x x x x cos 2x dx 2 1 cos 2 2 1 sin 2 2 1 2 = ( ) ( ) ( ) x x + x x − cos 2x dx 2 1 cos 2 2 1 sin 2 2 1 2 = x ( x) + x ( x) − • sin(2x) + C 2 1 2 1 cos 2 2 1 sin 2 2 1 2 = x ( x) + x ( x) − sin(2x) + C 4 1 cos 2 2 1 sin 2 2 1 2 # u = x dv = sin(2x)dx du dx v cos(2x) 2 1 = = − u x dv cos(2x)dx 2 = = du x dx v sin(2x) 2 1 = 2 =
Chapter 3 : Integration Techniques 203 Example 8 ( ) R • R dx Evaluate e x dx x cos2 2 . Solution : Method I (using Tabular Integration) e x dx e x e x e x dx x x x x = + • − • 4cos2 4 1 2sin2 4 1 cos2 2 1 cos2 2 2 2 2 e x dx e x e x e x dx x x x x = + sin2 − cos2 2 1 cos2 2 1 cos2 2 2 2 2 e x dx e x dx e x e x x x x x sin2 2 1 cos2 2 1 cos2 cos2 2 2 2 2 + = + e x dx e ( x x) x x cos2 sin2 2 1 2 cos2 2 2 = + e x dx e ( x x) C x x = + + cos2 sin2 4 1 cos2 2 2 # c a e e dx ax ax = + Sign u (differentiate) dv (integrate) + cos 2x x e 2 − − 2sin2x x e 2 2 1 + − 4cos2x x e 2 4 1
MAT238/ MAT438 : Foundation of Applied Mathematics 204 Example 9 ( ) R • R dx Evaluate e x dx x sin2 4 . Solution : Method I (using Tabular Integration) e x dx e x e x e x dx x x x x = − • − • 4sin2 16 1 2cos2 16 1 sin2 4 1 sin2 4 4 4 4 e x dx e x e x e x dx x x x x = − − sin2 4 1 cos2 8 1 sin2 4 1 sin2 4 4 4 4 e x dx e x dx e x e x x x x x cos2 8 1 sin2 4 1 sin2 sin2 4 1 4 4 4 4 + = − e x dx e ( x x) x x 2sin2 cos2 8 1 sin2 4 5 4 4 = − e x dx e ( x x) C x x = • − + 2sin2 cos2 8 1 5 4 sin2 4 4 e x dx e ( x x) C x x = − + 2sin2 cos2 10 1 sin2 4 4 # c a e e dx ax ax = + Sign u (differentiate) dv (integrate) + sin2x x e 4 − 2cos 2x x e 4 4 1 + − 4sin2x x e 4 16 1
Chapter 3 : Integration Techniques 205 Example 10/ MAT238/ MAR 2005/ Q1c (10 marks) Use integration by parts to evaluate the integral ( ) cos ln x dx . Ans : x x C x sin(ln ) + cos(ln ) + 2 Solution : Using u dv = uv − v du ( ) cos ln x dx = u dv ( ) cos ln x dx = uv − v du ( ) cos ln x dx = ( ) ( ) − • − dx x x x x x sin ln cos ln ( ) cos ln x dx = ( ) ( ) x cos ln x + sin ln x dx ( ) cos ln x dx = ( ) x cos ln x + u dv ( ) cos ln x dx = ( ) x cos ln x + uv − v du ( ) cos ln x dx = ( ) ( ) ( ) + − • dx x x x x x x x cos ln cos ln sin ln ( ) cos ln x dx = ( ) ( ) ( ) x cos ln x + x sin ln x − cos ln x dx ( ) ( ) cos ln x dx + cos ln x dx = x cos(ln x) + x sin(ln x) ( ) 2 cos ln x dx = xcos(ln x) + sin(ln x) Thus, ( ) cos ln x dx = ( x) ( x) C x cos ln + sin ln + 2 # u = sin(ln x) dv =dx ( ) dx v x x x du = = cos ln u = cos(ln x) dv =dx ( ) dx v x x x du = − = sin ln
MAT238/ MAT438 : Foundation of Applied Mathematics 206 TUTORIAL 3.3 : Integration by Parts Use integration by parts to evaluate 1. ∫ 2 2 4 0 ∶ 8 − 1 4 = 0.1427 2. ∫ 2 ℎ 0 ∶ 63.03 3. ∫ 4 2 1 ∶ 3.196 4. ∫ 3 2 − 1 0 ∶ 0.4818 5. ∫ (3) 2 ∶ 1 2 (3) − 1 2 + 6. 1 2 ∫ ℎ −1 2 ∶ 1 2 ℎ −1 2 − 1 4 √4 2 − 1 + 7. ∫(2 − 1)( + 5) 6 ∶ 1 7 (2 − 1)( + 5) 7 − 1 28 ( + 5) 8 + 8. ∫ −1 2 ∶ 1 2 2 −1 2 − 1 4 ȁ1 + 4 ȁ + 9. ∫ 2 ℎ(4) ∶ − 1 6 2 (ℎ(4) + 2 ℎ(4)) + 10. ∫ 2 (3) ∶ − 1 3 (3) + 1 9 ȁ(3)ȁ + 11. ∫ ( − 1) 2 ∶ 1 − − + ȁ − 1ȁ + 12. ∫ ( 1 ) ∶ − 1 2 2 + 1 4 2 + 13. ∫ ℎ −1 ∶ ℎ −1 + 1 2 ȁ1 − 2 ȁ + 14. ∫ ( ) ∶ 2 (( ) − ( )) + 15. ∫ 2 4 ∶ 1 4 2 4− 1 8 4 + 1 32 4 + 16. ∫ 2 ∶ 1 2 2 + 1 4 2 + 17. ∫ 2 (3 + 1) ∶ 2 3 (3 + 1) + 2 9 (3 + 1) +
Chapter 3 : Integration Techniques 207 18. ∫ − ∶ − −− − + 19. ∫(4 2 + 1) (2) ∶ 2 2 (2) + 2 (2) − 1 2 (2) + 20. ∫( + 1) +2 ∶ +2 + 21. ∫ 2 ∶ 1 3 3 − 1 9 3 + 22. ∫ ( 2 + 2) 2 0 ∶ = 3.1416 23. ∫ −12 ∶ −1 (2) + 1 2 √1 − 4 2 + 24. ∫ 2() 2 . ∶ 2 ( ) 2 − 2 + 2 2 + 25. ∫ ( + 7) . ∶ 2 2 − 7 + 49 ȁ + 7ȁ + 26. ∫ − (2) 0 . ∶ 0.1914 27. ∫( + 2) . ∶ ( + 2) + + 28. ∫( − 1) . ∶ (1 − ) + + 29. ∫ √ 4 1 . 2014/3 ∶ 4.2824
MAT238/ MAT438 : Foundation of Applied Mathematics 208 END OF CHAPTER 3 SUMMARY Useful formula, Differentiation Integration ( ) ax ax e ae dx d = c a e e dx ax ax = + ( ax) a ax c dx d sin = cos + c a cos ax sinax dx + − = ( ax) a ax c dx d cos = − sin + c a sinax cos ax dx = + Reminder : to solve integrals, don’t ever do these : i) use product rule ii) use quotient rule iii) generalized power rule or chain rule iv) bring the term containing x (or any variable) outside integration sign
Chapter 4 : Ordinary Differential Equation (ODE) 209 CHAPTER 4 ORDINARY DIFFERENTIAL EQUATION (ODE) List of Topics First Order ODE 3 methods to solve the first order ODE i) Separable equation ii) Homogeneous equation iii) Linear equation Applications of First Order ODE i) Population Growth and Bacterial Colony ii) Radioactive Decay iii) Newton’s Law of Cooling iv) Mixing Problem Second Order ODE i) Homogeneous equation ii) Non-homogeneous equation 4.0 Introduction 4.0.1 Definition of Degree i) y x y dx dy + = ………………. degree ____ ii) xy x y dx dy 2 2 + = …………… degree ____ iii) xy xy y dx dy 2 + = …………… degree ____ v) 3 2 2 y x y xy dx dy + = …………… degree ____ v) 2 2 3 xy x y y dx dy + = …………… degree ____
MAT238/ MAT438 : Foundation of Applied Mathematics 210 4.0.2 Definition of Order (highest derivative) i) 1 2 = x + dx dy ………………….. order ____ ii) y sin x dx dy dx d y + + = 2 2 ………. order ____ iii) x y e dx dy dx d y + + = 4 4 ………. order ____ iv) y ln x dx dy dx d y + = + 3 2 2 …… order ____ 4.1 First Order ODE 4.1.1 Three methods to solve the first order ODE • Separable Equation • Homogeneous Equation • Linear Equation 4.1.1.1 Separable equation Procedure to solve separable equation Step I : separating variables x and y LHS : f(y) dy RHS : f(x) dx Step II : Integrating both sides (using an appropriate technique) LHS : f(y ) dy RHS : ( ) f x dx (if the initial condition is given) Step III : Applying the given initial condition to find the value of C.
Chapter 4 : Ordinary Differential Equation (ODE) 211 Example 1 Solve 0 2 2 − = + + e dx e dy x y x y Ans : e e C y x = − + − Solution : 0 2 2 − = + + e dx e dy x y x y I) Separating variables x and y, LHS : f(y )dy and LHS : f(x)dx e dx e dy x+y 2x+2y = Or e dy e dx x+ y x+y = 2 2 e e dy e e dx x y x y • = • 2 2 dx e e dy e e x x y y 2 2 = e dy e dx 2y −y x−2x = e dy e dx y −x = II) Integrating both sides, LHS : ( ) f y dy and LHS : ( ) f x dx − e dy = e dx y x C e e x y + − = − 1 e e C y x = − + − #
MAT238/ MAT438 : Foundation of Applied Mathematics 212 Example 2 Solve (1 ) (1 ) 0 2 3 2 2 y + x y' + x + y = ans : y − y = − + x + C −1 3 ln 1 3 1 tan Solution : (1 ) (1 ) 0 2 3 2 2 y + x y' + x + y = (1 ) (1 ) 0 2 3 2 2 + + x + y = dx dy y x I) Separating variables x and y, LHS : f(y )dy and LHS : f(x)dx ( ) ( ) 2 3 2 2 1 x 1 y dx dy y + x = − + y ( x )dy x ( y )dx 2 3 2 2 1+ = − 1+ dx x x dy y y 3 2 2 2 1 1+ = − + II) Integrating both sides, LHS : ( ) f y dy and LHS : ( ) f x dx + = − + dx x x dy y y 3 2 2 2 1 1 --------------- Thus, the solution is, y y 1 tan− − = − + x + C 3 ln1 3 1 # RHS : Using u-substitution = = = = = (into ) LHS : By doing long division = = = Hence, = = (into ) dx x C x = + + − 1 2 tan 1 1 because ( ) 2 1 1 1 tan x x dx d + = −
Chapter 4 : Ordinary Differential Equation (ODE) 213 Example 3 Solve 0 2 xy + y' e ln y = x ans : ( y ) e C x = + − 2 2 1 ln 2 1 2 Solution : 0 2 xy + y' e ln y = x ln 0 2 + e y = dx dy xy x I) Separating variables x and y, LHS : f(y )dy and LHS : f(x)dx e y xy dx dy x ln = − 2 dx e x dy y y x 2 ln = − dy xe dx y y x ln 2 − = − II) Integrating both sides, LHS : ( ) f y dy and LHS : ( ) f x dx − dy = − xe dx y y x ln 2 ------------- Thus, the solution is, y 2 ln 2 1 = e C x + − 2 2 1 # RHS : Using u-substitution = = = = = (into ) LHS : Using u-substitution = = = = = (into )
MAT238/ MAT438 : Foundation of Applied Mathematics 214 Example 4 – with initial condition Solve 1 0 1 2 2 + = − dx dy y tan x , when 2 2 x = , y = ans : ln sin ln 3 1 1 ln = − + − + x y y Solution : 1 0 1 2 2 + = − dx dy y tan x I) Separating variables x and y, LHS : f(y )dy and LHS : f(x)dx 1 1 2tan 2 = − − dx dy y x dx x dy y tan 1 1 2 2 = − − dy x dx y cot 1 2 2 = − − II) Integrating both sides, LHS : ( ) f y dy and LHS : ( ) f x dx = − − dy x dx y cot 1 2 2 ------------- LHS : Using partial fraction ------- (i) = ------- (ii) When → When → LHS : Using appendix (no. 20) = = = (into ) = = = = = (into )
Chapter 4 : Ordinary Differential Equation (ODE) 215 Therefore, the general solution is, ln y − 1 − ln y + 1= − ln sin x + C ------------------ III) Applying the given initial condition into (to find C) By substituting 2 2 x = , y = into , ln 2 − 1 − ln 2 + 1= + C − 2 ln sin ln1 − ln3= − ln(1)+ C 0 − ln3 = 0 + C C = − ln3 (into ) RHS : Change to basic trigonometry and use u-substitution, = = = = = = (into ) LHS : Substitute and into (i) = = (into ) Thus, the particular solution is, ln y −1 − ln y + 1 = −ln sinx − ln3 #
MAT238/ MAT438 : Foundation of Applied Mathematics 216 Example 5– with initial condition Solve ( 1) 0 2 ln y cos x − dy − cos x dx = , when 1 2 x = , y = ans : y ln y − y = cosec x − 2 Solution : ( 1) 0 2 ln y cos x − dy − cos x dx = I) Separating variables x and y, LHS : f(y )dy and LHS : f(x)dx ln y (cos x 1) dy cos x dx 2 − = ( ) dx x x y dy cos 1 cos ln 2 − = ( ) dx x x y dy 2 1 cos cos ln − − = dx x x y dy 2 sin cos ln − = ( ) dx x x y dy 2 sin cos ln = − II) Integrating both sides, LHS : ( ) f y dy and LHS : ( ) f x dx ( ) = − dx x x y dy 2 sin cos ln ------------- RHS : = = = = = = (into ) LHS : (integration by parts) = = = = = (into ) (integration using u- substitution)
Chapter 4 : Ordinary Differential Equation (ODE) 217 Therefore, the general solution is, C x y y − y = + sin 1 ln ------------------ III) Applying the given initial condition into (to find C) By substituting , 1 2 x = y = into , ( ) ( ) − = + C 2 sin 1 1 ln 1 1 − = + C 1 1 1 − 1 = 1 + C C = − 2 (into ) Thus, the particular solution is, 2 sin 1 ln − = − x y y y # or y ln y − y = cosecx − 2
MAT238/ MAT438 : Foundation of Applied Mathematics 218 TUTORIAL 4.1.1 : Separable Equation 1. ℎ ℎ ℎ (0) = 3 . √4 2 + 16 = ()() . ∶ 2 = ℎ −1 ( 2 ⁄ ) + 4 2. ℎ (4 − 2 2 ) = 2 . ∶ | + | = 4 − 2 3 3 + 3. ℎ 2 + 3 = 2 . ∶ − | + | = 1 2 | 2 + 3| + 4. ℎ 4 + 2 = ( + 1)( + 2) 2 . ∶ 8 ( + 2) − 4 ( + 1) = ( + 2) 3 3 + 5. ℎ = 25 − 4 2 3 + 1 . ∶ 1 10 ℎ −1 ( 2 5 ) = 1 3 |3 + 1| + 6. ℎ ( 2 − 4) ( 2 + 4) − ( + 1) = 1 . ∶ 1 2 −1 ( 2 ) = ( − 2) + 7. ℎ = (1 + 2 )( − 2), (0) = 1 . ∶ (1 + 2 ) = 2 − 4 + (2) 8. ℎ = (4 + 2 ) ∶ 1 2 |4 + 2 | = − | + | +
Chapter 4 : Ordinary Differential Equation (ODE) 219 4.1.1.2 Homogeneous equation Procedure to solve Homogeneous equation Step I : write dx dy as a subject, and name as equation Step II : Substitute = + and = into Step III : factorize and simplify x Step IV : separating variables x and v LHS : f(v) dv RHS : f(x) dx Step V : Integrating both sides (using an appropriate technique) ∶ ∫ () ∶ ∫ () Step VI : Replace u by x y (if the initial condition is given) Step VII : Applying the given initial condition to find the value of C.
MAT238/ MAT438 : Foundation of Applied Mathematics 220 Example 1 Solve (x + 2y )dy = (y − 2x)dx ans : ( ) ( x ) x C y x y + + = − + − ln 1 tan 2ln 1 2 Solution : ( + 2) = ( − 2) = − 2 + 2 + = () − 2 + 2() + = − 2 + 2 + = ( − 2) (1 + 2) + = − 2 1 + 2 = − 2 1 + 2 − = − 2 − (1 + 2) 1 + 2 = − 2 − − 2 2 1 + 2 = −2 − 2 2 1 + 2 = −2(1 + 2 ) 1 + 2 1 + 2 1 + 2 = −2 Integrating both sides ∫ 1 + 2 1 + 2 = ∫ −2 ∫ 1 + 2 1 + 2 = −2 ∫ 1 ∫ ( 1 1 + 2 + 2 1 + 2 ) = −2 + −1 () + |1 + 2 | = −2 + −1 ( ) + |1 + ( ) 2 | = −2 + #
Chapter 4 : Ordinary Differential Equation (ODE) 221 Example 2 Solve 2 2 2 2 x 2xy y y x dx dy + + − = ans : ( ) ( x ) x C y x y + + = − + − ln 1 tan ln 2 1 1 2 Solution : = 2 − 2 2 + 2 + 2 + = () 2 − 2 2 + 2() + () 2 + = 2 2 − 2 2 + 2 2 + 2 2 + = 2 ( 2 − 1) 2(1 + 2 + 2) + = 2 − 1 2 + 2 + 1 + = ( + 1)( − 1) ( + 1)( + 1) + = − 1 + 1 = − 1 + 1 − = − 1 − ( + 1) + 1 = − 1 − 2 − + 1 = −1 − 2 + 1 = −(1 + 2 ) + 1 + 1 1 + 2 = −1 Integrating both sides ∫ + 1 1 + 2 = ∫ − 1 ∫ ( 1 + 2 + 1 1 + 2 ) = − ∫ 1 ∫ 1 + 2 + ∫ 1 1 + 2 = − + 1 2 ∫ 2 1 + 2 + ∫ 1 1 + 2 = − + 1 2 |1 + 2 | + −1 () = − + 1 2 |1 + ( ) 2 | + −1 ( ) = − + #
MAT238/ MAT438 : Foundation of Applied Mathematics 222 Example 3 Solve dx x y x dy y x cos = + 2 ans : ( x ) x C y tan = ln + Solution : = ( + 2 ( )) = + 2 ( ) + = + 2 ( ) + = + 2 () + = ( + 2 ()) + = + 2 () = + 2 () − = 2 () 1 2() = 1 2 () = 1 Integrating both sides ∫ 2 () = ∫ 1 () = + ( ) = + #
Chapter 4 : Ordinary Differential Equation (ODE) 223 Example 4 – with initial condition Solve (x xy )y' y x y 3 2 3 2 + 6 = 8 + 2 , when x = 1, y = 2. ans : ln 2( ) ( ) ln ln18 3 + = x + x y x y Solution : ( 3 + 6 2 )′ = 8 3 + 2 2 ( 3 + 6 2 ) = 8 3 + 2 2 = 8 3 + 2 2 3 + 6 2 + = 8() 3 + 2 2 () 3 + 6() 2 + = 8 3 3 + 2 3 3 + 6 2 3 + = 3 (8 3 + 2) 3(1 + 6 2) + = 8 3 + 2 1 + 6 2 = 8 3 + 2 1 + 6 2 − = 8 3 + 2 − (1 + 6 2 ) 1 + 6 2 = 8 3 + 2 − − 6 3 1 + 6 2 = + 2 3 1 + 6 2 1 + 6 2 + 2 3 = 1 Integrating both sides ∫ 1 + 6 2 + 2 3 = ∫ 1 | + 2 3 | = + |( ) + 2 ( ) 3 | = + When = 1, = 2 |( 2 1 ) + 2 ( 2 1 ) 3 | = (1) + |2 + 16| = 0 + = (18) The particular solution is, |( ) + 2 ( ) 3 | = + (18) #
MAT238/ MAT438 : Foundation of Applied Mathematics 224 Example 5 – with initial condition Solve (3x − y )dx + (x + y )dy = 0 , when x = 3 , y = 0. ans : ( ) x x y x y ln3 ln 3 tan 3 1 ln 3 2 1 1 2 = − + + − Solution : (3 − ) + ( + ) = 0 ( + ) = −(3 − ) ( +) = ( − 3) = −3 + + = − 3 + + = ( − 3) ( + 1) + = − 3 + 1 = −3 +1 − = −3 − ( +1) + 1 = −3 − 2 − +1 = − 2 −3 +1 = −( 2 + 3) +1 + 1 2 + 3 = − 1 When = √3, = 0 1 2 |(0) 2 + 3| + 1 √3 −1 (0) = −√3 + 1 2 (3) + 1 √3 (0) = −√3 + √3+ 0 = −√3 + = 2√3 = (√3) 2 = (3) The particular solution is, 1 2 |( ) 2 + 3|+ 1 √3 −1 ( √3 ) = − + (3) # Integrating both sides ∫ + 1 2 + 3 = ∫ − 1 ∫ ( 2 + 3 + 1 2 + 3 ) = − ∫ 1 ∫ 2 + 3 + ∫ 1 2 + 3 = − + 1 2 ∫ 2 2 + 3 + ∫ 1 2 + (√3) 2 = − + 1 2 | 2 +3| + 1 √3 −1 ( √3 ) = − + 1 2 |( ) 2 +3| + 1 √3 −1 ( ( ) √3 ) = − + 1 2 |( ) 2 +3| + 1 √3 −1 ( √3 ) = − + #
Chapter 4 : Ordinary Differential Equation (ODE) 225 TUTORIAL 4.1.2 : Homogeneous Equation 1. ℎ ℎ = 4 − 2 2( − 2) . ∶ |2 − 2( ⁄ ) + 2( ⁄ ) 2 | = −2 + or |( ⁄ ) 2 − ( ⁄ ) + 1| = −2 + 2. ℎ ℎ = 2 2 + 2 . ∶ 1 2 |( ⁄ ) 2 + 1| = + 3. ℎ ℎ = 3 + 6 4 . ∶ 2 |2( ⁄ ) + 3| = + 4. ℎ ℎ 3 = 2 + 3 2 . ∶ − 3 4 |3 − 2( ⁄ ) 2 | = || + 5. ℎ ℎ = 2 2 + 2 . ∶ 1 2 |( ⁄ ) 2 + 1| = + 6. ℎ ℎ = 3 + 6 4 . ∶ 2 |2( ⁄ ) + 3| = + 7. ℎ ℎ 4 = (2 + 8) . ∶ |1 + 2( ⁄ )| = || + 8. ℎ ℎ 4 = ( 2 − 2 ) . ∶ − 2 3 |3( ⁄ ) 2 + 1| = || + − 2 3 |−3( ⁄ ) 2 − 1| = || + 9. ℎ ℎ = 2 + 2 . ∶ 1 2 ( ⁄ ) 2 = + 10. ℎ ℎ = + 4 2 . ∶ |1 + 2( ⁄ )| = || + 11. ℎ ℎ ( 2 − 2) = 2 − 2 . ∶ − (( ⁄ ) 2 − ⁄ ) = 3 +
MAT238/ MAT438 : Foundation of Applied Mathematics 226 4.1.1.3 Linear equation Procedure to solve Linear equation Step I : write into the form of P(x)y Q(x) dx dy + = Step II : Identify P(x) and Q(x) Step III : Find P(x)dx Step IV : Find the Integrating Factor (IF) Where IF = ( ) P x dx e Step V : the solution is ( ) y • IF = Q x • IF dx (if the initial condition is given) Step VI : Applying the given initial condition to find the value of C. Example 1 Solve sin x y cot x dx dy − 2 = . ans : y 2sin x C sin x 2 = + Solution : sin x y cot x dx dy − 2 = y x x dx dy − cot = sin2 ( x)y x dx dy + −cot = sin2 P(x)y Q(x) dx dy + = The solution is given by, ( ) y • IF = Q x • IF dx • = • dx x x x y sin 1 sin2 sin 1 = • dx x x x x y sin 1 2sin cos sin = x dx x y 2 cos sin x C x y = 2sin + sin y 2sin x C sinx 2 = + # P(x) = −cot x , Q(x) = sin2x P(x)dx x dx = −cot = − dx x x sin cos = − ln(sinx) = ( ) 1 ln sin − x IF = P(x )dx e ( ) 1 ln sin − = x e = ( ) 1 sin − x = sinx 1 compare Linear equation : • first term → term containing dx dy • Second term→ term containing y • RHS → term not containing y • Coefficient of dx dy must +ve1
Chapter 4 : Ordinary Differential Equation (ODE) 227 Example 2 Solve + (1+ ) − = 0 x x y e dx dy x . ans : x Ce x e y x −x = + 2 Solution : + (1+ ) − = 0 x x y e dx dy x ( ) x x y e dx dy x + 1+ = ( ) x e x x y dx dy x x x = + + 1 x e y x x dx dy x = + + 1 P(x)y Q(x) dx dy + = The solution is given by, ( ) y • IF = Q x • IF dx • = • xe dx x e y xe x x x y • xe = e • e dx x x x y • xe = e dx x 2x C e y xe x x • = + 2 2 x x x xe C xe e y = + 2 2 x Ce x e y x −x = + 2 # ( ) x x P x + = 1 , ( ) x e Q x x = ( ) dx x x P x dx + = 1 = + dx x x x 1 = + dx x 1 1 = ln x + x @ x + ln x IF = P(x )dx e x x e +ln = = x x e e ln • = e x x • = x xe compare Linear equation : • first term → term containing dx dy • Second term→ term containing y • RHS → term not containing y • Coefficient of dx dy must +ve1
MAT238/ MAT438 : Foundation of Applied Mathematics 228 Example 3 Solve cos x x sin x x y dx dy + − = . ans : x C y = sin x + Solution : cos x x sin x x y dx dy + − = x x y x dx x dy sin cos 1 = + + P(x)y Q(x) dx dy + = The solution is given by, ( ) y • IF = Q x • IF dx • = + x dx x x y x x sin cos ( ) xy = x cosx + sinx dx xy = x cos x dx + sin x dx xy = x sinx + cos x − cos x + C xy = x sinx + C x C x x x y = + sin x C y = sinx + # ( ) x P x 1 = , ( ) x x Q x x sin = cos + ( ) dx x P x dx = 1 = ln x IF = P(x )dx e x e ln = = x compare using Tabular Integration Sign u (differentiate) dv (integrate) + x − 1 + 0 = = Linear equation : • first term → term containing dx dy • Second term→ term containing y • RHS → term not containing y • Coefficient of dx dy must +ve1
Chapter 4 : Ordinary Differential Equation (ODE) 229 Example 4 − with initial condition Solve (2y − xy − 3)dx + x dy = 0 , when y(1) = 2 ans : 2 1 2 3 3 8 x e x x y x− = − − + Solution : (2y − xy − 3)dx + x dy = 0 Dividing by dx (to make dx dy symbol) ( ) dx dx x dy dx 2y xy 3 dx 0 + = − − (2 − − 3) + = 0 dx dy y xy x + 2y − xy = 3 dx dy x + (2 − x)y = 3 dx dy x Dividing by x (to make coefficient dx dy + ve1) ( ) x x x y dx dy x x 2 3 = − + x y x x dx dy 2 3 = − + P(x)y Q(x) dx dy + = The solution is given by, ( ) y • IF = Q x • IF dx − − • = • x e dx x y x e 2 x 3 2 x − = xe dx e x y x x 3 2 xe e C e x y x x x = − − + − − 3 3 2 compare ( ) x x P x − = 2 , ( ) x Q x 3 = ( ) dx x x P x dx − = 2 = − dx x x x 2 = − dx x 1 2 = dx − dx x 1 2 = dx − dx x 1 1 2 = 2ln x − x = x − x 2 ln IF = P(x )dx e x x e − = 2 ln = x x e e − • 2 ln = x x e 2 − using Tabular Integration Sign u (differentiate) dv (integrate) + 3x − 3 + 0 = = Linear equation : • first term → term containing dx dy • Second term→ term containing y • RHS → term not containing y • Coefficient of dx dy must +ve1
MAT238/ MAT438 : Foundation of Applied Mathematics 230 C e e x e x y x x x = − − + 3 3 2 2 2 2 3 3 x e C x e x e e e x y x x x x x = − • − • + • 2 2 3 3 x Ce x x y x = − − + ------------------ III) Applying the given initial condition into (to find C) By substituting y(1) = 2 (@ x = 1, y = 2) into , 1 1 3 1 3 2 1 Ce = − − + 2 = − 3 − 3 + Ce 8 = CeC e = 8 1 8 − C = e (into ) Thus, the particular solution is, 2 1 2 3 3 8 x e e x x y x • = − − + − 2 1 2 3 3 8 x e x x y x − =− − + #
Chapter 4 : Ordinary Differential Equation (ODE) 231 Example 5 − with initial condition Solve sec 2 0 ; ( ) 0 4 3 2 − x x − y = y = dx dy x ans : 2 2 y = x tan x − x Solution : sec 2 0 3 2 − x x − y = dx dy x y x x dx dy x 3 2 − 2 = sec x x x x y dx dy x x 3 2 2 sec − = y x x dx x dy 2 2 sec 2 = + − P(x)y Q(x) dx dy + = The solution is given by, ( ) y • IF = Q x • IF dx • = • dx x x x x y 2 2 2 2 1 sec 1 = x dx x y 2 2 secx C x y = tan + 2 ---------------- III) Applying the given initial condition into (to find C) By substituting ( ) 0 4 y = (@ x = 4 , y = 0) into , 0 = tan( 4 )+ C 0 = 1+ C C = − 1 (into ) Thus, the particular solution is, tan 1 2 = x − x y 2 2 y = x tan x − x # compare ( ) x P x 2 = − , Q(x) x x 2 2 = sec ( ) dx x P x dx = − 2 = − dx x 1 2 = − 2ln x = 2 ln − x IF = P(x )dx e 2 ln − = x e = −2 x = 2 1 x Linear equation : • first term → term containing dx dy • Second term→ term containing y • RHS → term not containing y • Coefficient of dx dy must +ve1
MAT238/ MAT438 : Foundation of Applied Mathematics 232 TUTORIAL 4.1.3 : Linear Equation 1. ℎ + 2 = ( 2 ) . ∶ = 1 2 −2 ( 2 ) + −2 2. ℎ + 2 = ∶ = 2 5 2 + √ 3. ℎ + 4 = 8 3 . ∶ = 4 + −4 4. ℎ − 3 = 2 (2) 2 . ∶ = 1 3 3 ( 2 ) 3 + 3 5. ℎ − 2 = 3 2 3 . ∶ = 2 3 + 2 6. ℎ − = 2 2 . ∶ = 1 2 2 2− 1 4 2 + 7. ℎ + ( 4 ) = 2 + 1 . ∶ = 2 3 + 5 + 4 8. ℎ + 2 = . ∶ = 1 2 + 2 9. ℎ + = 2 . ∶ = − 2 2 +
Chapter 4 : Ordinary Differential Equation (ODE) 233 Solve the first order differential equation given below (question 1 – 21) : 1. dx (x )(y ) dy x x 2 1 2 2 4 = + + + 2. 2 2 x y dx dy xy = − 3. x x y dx dy x ln + = 4. ( 1) ( 1) 0 2 2 x y + dx + y x − dy = 5. , 0 sin 3 2 + = x x x y dx dy x 6. ( ) ( ) 1 1 2 2 2 + + − + = x e x x y dx dy x x 7. 2 2 cos 1 2y dx dy y x = + 8. ' 3 , (0) 1 2 y − y = e y = x 9. (x y )dx xy dy 3 3 2 + = 6 10. dx dy x y y 2 tan sec = sin 11. 2 2 x xy y dx dy − = 12. 4 , (1) 0 3 + y = x − x y = dx dy x 13. 1 3 , (1) 3 3 − = + x y = x y dx dy 14. 2 3 , ( 1) 2 4 − y = x y − = dx dy x 15. 2 2 2 y x xy dx dy − = 16. ( ) 2 3 3 2 2 3 y xy x y x dx dy xy + x = + + − 17. ( )( ) 4 0 2 2 8 2 − + = + + y dx dy x y xy 18. ( + 1) − y = 1, y(0)= 2 dx dy x 19. , (1) 2 2 2 2 = y − xy + x y = dx dy x 20. − = x y x x x y y dx dy sin sin 21. x xe x y dx dy 2 + = 22. Give the order and the degree of the equation 2 0 4 3 3 + = − y dx dy dx d y x 23. Verify that the function x y x x e − = − cos + 10 2 1 sin 2 1 is a solution of the differential equation y x dx dy + = sin . 24. Explain why the differential equation (4 3 ) 0 2 2 x − xy dy − x dx = is homogeneous. 25. Given y = 2 sin x + Acos x , where A is an arbitrary constant. Form the corresponding differential equation. 26. Without solving, identify the following differential equation as separable, homogeneous or linear : i) x y e dx dy − + = ii) y x x dx dy x − = + 3 2 iii) 2 2 x y dx dy x = +
MAT238/ MAT438 : Foundation of Applied Mathematics 234 Answers : 1. ( ) x x C y = + − + + + 8ln 2 4ln 1 3 2 3 2. ( x ) x C y − ln 1− 2 = ln + 4 1 2 3. ( ) x C x x y = + 2 ln 2 4. x x 2 1 y y ln y 1 2 1 2 2 − + + = − − − ln|x| + C 5. 3 3 x C x cosx y + − = 6. 2 2 x x Ce x 1 e y + + = − 7. ln 1 2y tan x C 4 1 2 + = + 8. y = 3e2x − 2ex 9. ln 1 5( ) ln x C 5 2 3 x y − − = + 10. sin y C 3 1 ln cosx 3 − = + 11. ln x C y x = + 12. 4 3 35 x 2 x 5 1 x 2 1 y = − + 13. y = x ln x + x4 + 2x 14. 4 2 x 2 1 x 2 3 y = + 15. ln 3( ) ( ) ln x C 3 1 3 x y x y − − = + 16. ln x C y x y x ln x y = + − + − 17. ln(y 2) x 2ln x C y 2 4 ln(y 2) + − = + + + + − 18. y = 3x + 2 19. x = (y − 1)(1 − lnx) 20. x C x y = − + − cos ln 21. C x e x e e x y x x x = − + + 2 2 4 2 2 2 2 22. order : 3, degree : 1 25. x dx dy y = 2sinx + 2cot x cosx − cot
Chapter 4 : Ordinary Differential Equation (ODE) 235 4.1.2 Applications of First Order ODE Applications of Separable Equation ODE Population Growth and Bacterial Colony Radioactive Decay Newton’s Law of Cooling Mixing Problem Note : Let P0 is an original P (when t = 0) • Doubling time : when P = 2P0 , t = ? • Triple : when P = 3P0 , t = ? • Quadruple : when P = 4P0 , t = ? • Half-life : when P = 2 1 P0 , t = ? = By using separable equation, 1 = Integrating both sides, න 1 = න = + = + = • = (ℎ = ) = ( −) By using separable equation, 1 − = Integrating both sides, න 1 − = න ( − ) = + − = + − = • − = (ℎ = ) Mixing Problem using Separable Equation (rate in = rate out) = 1 − ൬ 0 +(1 − 2 ) ൰ 2 ,(0) = 0 Mixing Problem using Linear Equation (rate in = rate out and rate in ≠ rate out) + ൬ 2 0 +(1 − 2 ) ൰ = 1 Bacterial colony/
MAT238/ MAT438 : Foundation of Applied Mathematics 236 4.1.2.1 Population Growth and Bacterial Colony Example 1 (Population Growth) The town of Tembering has a population of 15,000 in 1995 and 20,000 in 2000. Assuming that the rate increase of the population is proportionate to the current population, estimate i) the population in 2020. ii) the doubling time for the town’s population. Solution : = By using separable equation, 1 = Year 1995 → t = 0, P = 15,000 (to find C) Year 2000 → t = 5 years P = 20,000 (to find k) i) Year 2020 → t = 25 years, P = ? ii) t = ? P = 2P0 = 2(15,000) = 30,000 (doubling time) to find k, substitute t = 5 years, P = 20,000 into ② 20,000 = 15,000 5 5 = 20,000 15,000 5 = 4 3 5 = ( 4 3 ) = ( 4 3 ) 5 = 0.0575 = 15,000 0.0575 − − − − ③ a) in year 2020, = 25 = 15,000 0.0575 (25) = 63,152 # the population in 2020 is estimate to 63,152 b) Doubling time, when = 30,000 30,000 = 15,000 0.0575 0.0575 = 2 0.0575 = (2) = (2) 0.0575 = 12.05 # it took 12.05 years for the population to become double from the original population Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = to find C, substitute t = 0, P = 15,000 into ① 15,000 = 0 = 15,000 ( ①) = 15000 − − − − ②
Chapter 4 : Ordinary Differential Equation (ODE) 237 Example 2 (Population Growth) Suppose that the population of Town PJ has grown to 50,000 after 3 years and after 6 years the population has tripled the original number. If the population grows at the rate proportional to the number of population, i) obtain the expression for the population at any time t, ii) determine . Solution : = By using separable equation, 1 = t = 0, P = P0 t = 3 years, P = 50,000 t = 6 years, P = 3P0 (tripled the original number) i) = (Substitute the values of C and k into = ) ii) t = ?, P = 500,000 substitute t = 6 years, P = 3P0 into ② 30 = 0 6 6 = 3 6 = (3) = (3) 6 = 0.1831 ( ③) 0 = 50,000 −3(0.1831) 0 = 28,868 = 0 = 28,868 i) the expression for the population at any time t, = 28,868 0.1831 ii) the time taken for the population to reach to 500,000 500,000 = 28,868 0.1831 0.1831 = 500,000 28,868 0.1831 = ( 500,000 28,868 ) = ( 500,000 28,868 ) 0.1831 = 15.58 # Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 − − − − ② substitute t = 3 years, P = 50,000 into ② 50,000 = 0 3 0 = 50,000 3 0 = 50,000 −3− − − −③
MAT238/ MAT438 : Foundation of Applied Mathematics 238 Example 3 (Population Growth) The population of Green Valley grows at a rate proportional to the population at any time. The town was opened in the year 1997. In the year 2000, its population was 28,000 and in the year 2002, the population was 32,500. i) Write an expression for the population at any time. ii) What was its initial population? Solution : Year 1997 → t = 0, P = P0 (initial population) Year 2000 → t = 3 years P = 28,000 Year 2002 → t = 5 years, P = 32,500 i) = (Substitute the values of C and k into = ) ii) t = 0, P = ? (initial population) = By using separable equation, 1 = Solve ③ and ④ simultaneously (subs ③ into ④) 28,000 3= 32,500 5 5 3= 32,500 28,000 2 = 65 56 2 = ( 65 56) = ( 65 56) 2 = 0.0745 ( ③) 0 = 28,000 3= 28,000 −3 0 = 28,000 −3(0.0745) 0 = 22,392 = 0 = 22392 i) the expression for the population at any time t, = 22,392 0.0745 ii) the initial population 0 = 22,392 Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 −− − − ② substitute t = 3 years, P = 28,000 into ② 28,000 = 0 3 0 = 28,000 3− − −−③ substitute t = 5 years, P = 32,500 into ② 32,500 = 0 5 0 = 32,500 5− − − −④
Chapter 4 : Ordinary Differential Equation (ODE) 239 Example 4 (Bacterial Colony) The initial number of a bacteria colony is N0. After two hours, the numbers increases to 2 5 N0. Assuming that the number of bacteria increases at a rate proportional to the number of the bacteria at time t, how long does it take for the number of bacteria to become quadruple? Solution : = By using separable equation, 1 = t = 0, P = N0 (initial number) t = 2 hours, = 5 2 0 t = ? hours, P = 4N0 (quadruple) substitute t = 2 hours, = 5 2 0 into ② 5 2 0 = 0 2 2 = 2.5 2 = (2.5) = (2.5) 2 = 0.4581 ( ②) = 0 0.4581 when, = 40 (into ③) 40 = 0 0.4581 0.4581 = 4 0.4581 = (4) = (4) 0.4581 = 3.026 ℎ # Thus, it took 3.026 hours for the bacteria to become quadruple. Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute t = 0, P = N0 into ① 0 = 0 = 0 ( ①) = 0 − − − − ②
MAT238/ MAT438 : Foundation of Applied Mathematics 240 Example 5 (Bacterial Colony) A population of protozoa develops with a constant growth rate of 0.7944 members per day. On day zero, the population consists of two members. Find the population size after 6 days. Solution : = By using separable equation, 1 = = 0.7944 t = 0, P = 2 (to find C) t = 6 days, P = ? to find c, substitute t = 0 days, P = 2 into ② 2 = 0 = 2 ( ②) = 2 0.7944 − − − − ③ After 6 days, the population is = 2 0.7944 (6) = 235 # Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute = 0.7944 into ① = 0.7944 − − − − ②
Chapter 4 : Ordinary Differential Equation (ODE) 241 Example 6 (Bacterial Colony) A colony of bacteria is grown under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 hours there are 10,000 bacteria. At the end of 5 hours there are 40,000 bacteria. How many bacteria were present initially? Solution : t = 0, P = P0 (initial population) t = 3 hours P = 10,000 t = 5 hours, P = 40,000 t = 0, P = ? (initial population) = By using separable equation, 1 = Solve ③ and ④ simultaneously (subs ③ into ④) 10,000 3= 40,000 5 5 3= 40,000 10,000 2 = 4 2 = (4) = (4) 2 = 0.6931 ( ③) 0 = 10,000 3= 10,000 −3 0 = 10,000 −3(0.6931) 0 = 1,250 the initial population 0 = 1,250 Integrating both sides, න 1 = න = + = + = • = −− − − ① ℎ = substitute t = 0, P = P0 into ① 0 = 0 = 0 ( ①) = 0 −− − − ② substitute t = 3 hours, P = 10,000 into ② 10,000 = 0 3 0 = 10,000 3− − −−③ substitute t = 5 hours, P = 40,000 into ② 40,000 = 0 5 0 = 40,000 5− − − −④
MAT238/ MAT438 : Foundation of Applied Mathematics 242 TUTORIAL 4.1.4 : Population growth/bacterial colony 1. Azalea City had a population of 20,000 in 2002 and a population of 50,000 in 2010. Assume that its population will continue to grow exponentially at a constant rate, estimate the number of population at Azalea City in 2012. Ans : = 62,849 2. A population of Ameba is known to growth at rate proportional to the amount present. Given that the Ameba has a doubling time of 5 hours, and after 8 hours there are 15,000 of Ameba. What was its initial population? Ans : Po = 4,948 3. Cyber City had a population of 25 000 in 2001 and a population of 35 000 in 2011. Assume that its population grows exponentially at a constant rate, estimate the number of populations in the year 2020. Ans : = 47,336 4. A biologist observes that a certain bacterial colony triples every 4 hours. After 12 hours, the population reached 150,000. Assuming that the population, P will continue to grow exponentially at a constant rate = , what is the initial population of the colony? Ans : = 5, 556 5. Suppose a bacterial culture initially has 400 cells. After 1 hour, the population has increased to 800. Assuming that the growth rate of bacterial culture is proportional to the present population, find the number of cells after 10 hours. Ans : = (2), = 409,600 6. The number of populations in Kampung Bali is increasing from 150 to 200 in 7 years. By using exponential growth model, find the number of populations at any time t. Ans:. = 150 0.041097
Chapter 4 : Ordinary Differential Equation (ODE) 243 4.1.2.2 Radioactive Decay Example 1 Initially 100 mg of a radioactive substance was present. After 6 hours the mass had decreased by 10%. If the rate of decay is proportional to the amount of substance present at time t, find the amount of the remaining radioactive after 24 hours. Solution : = By using separable equation, 1 = t = 0, P = 100 mg (to find C) t = 6 hours, P = 90% of 100 mg = 0.9(100) = 90 (decreased by 10%) t = 24 hours, P = ? 6 = (0.9) = (0.9) 6 = −0.0176 ② The amount of substance at any time t, = 100 −0.0176 the amount of the remaining radioactive after 24 hours, = 100 −0.0176 (24) = 65.55 # Integrating both sides, න 1 = න = + = + = • = − − − − ① ℎ = substitute t = 0, P = 100 mg into ① 100 = 0 = 100 ( ①) = 100 − − − − ② substitute t = 6 hours, P = 90 mg into ② 90 = 100 6 6 = 90 100 6 = 0.9