MAT238/ MAT438 : Foundation of Applied Mathematics 294 TUTORIAL 4.2.2 : 2nd order ODE (non-homogeneous equation) Solve the differential equation 1. ′′ − 8 ′ + 16 = 2 2 − 1 Ans : = 1 4 + 2 4 + 1 8 2 + 1 8 − 1 64 2. 2 2 − = 2 2 − − 3 Ans : = 1 + 2 −− 2 2 + − 1 3. 2 2 − 5 + 4 = 2 Ans : = 1 + 2 4 + 5 17 + 3 17 4. ′′ + 2 ′ − 3 = 4 Ans : = 1 + 2 −3− 4 5 + 2 5 5. ′′ + 14 ′ + 49 = 8 3 , (0) = 2 ′(0) = 1 Ans : = 48 25 −7 + 71 5 −7 + 2 25 3 6. 2 2 − 10 + 25 = 25 5 Ans : = 1 5 + 2 5 + 25 2 2 5 7. ′′ + 2 ′ = 24 + −2 Ans : = 1 + 2 −2 + 6 2 − 6 − 1 2 −2
Chapter 4 : Ordinary Differential Equation (ODE) 295 4.3 Formation of Differential Equations by eliminating arbitrary constant To understand the formation of differential equations in a better way, there are a few suitable differential equations examples that are given below along with important steps. The formulas of differential equations are important as they help in solving the problems easily. Arbitrary constant is a constant whose value could be assumed to be anything, and it doesn’t depend on the other variables in an equation or expression. To achieve the differential equation from this equation let's look at the following examples. Example 1 : Find the differential equation from the function = + 4 , where A and B are arbitrary constants. Solution : = + 4 ′ = + 4 4 ′′ = + 16 4 To eliminate A, From ① : = − 4 ①a into ② : ′ = − 4 + 4 4 ′ − = 3 4 ( 5) : 5 ′ − 5 = 15 4 ①a into ③ : ′′ = − 4 + 16 4 ′′ − = 15 4 ③a−②a : ( ′′ − ) − (5 ′ − 5) = 15 4− 15 4 ′′ − 5 ′ + 4 = 0 # ① ② ③ ①a ②a ③a Rules to follow in eliminating arbitrary constant 1. The number of differentiating the given equation is the same as the number of arbitrary constants. 2. The order of the differential equation is the same as the number of arbitrary constants. 3. the desired equation is free from arbitrary constants. The number of arbitrary constants are two! So.. differentiate two times! The differential equation must free from arbitrary constants
MAT238/ MAT438 : Foundation of Applied Mathematics 296 Example 2 : Find the differential equation from the function = 3 + 2 , where A and B are arbitrary constants Solution : = 3 + 2 ′ = 3 3 + 2 2 ′′ = 9 3 + 4 2 To eliminate B, ①4 : 4 = 4 3 + 4 2 ②2 : 2′ = 6 3 + 4 2 ③−①a : ′′ − 4 = 5 3 ③−②a : ′′ − 2′ = 3 3 To eliminate A, ④3 : 3 ′′ − 12 = 15 3 ⑤5 : 5 ′′ − 10′ = 15 3 ④a−⑤a : (3 ′′ − 12) − (5 ′′ − 10′) = 0 3 ′′ − 5 ′′ + 10′ − 12 = 0 −2 ′′ + 10′ − 12 = 0 ÷ (−2) ∶ ′′ − 5 ′ + 6 = 0 # ① ② ③ ①a ②a ④ ⑤ ④a ⑤a The number of arbitrary constants are two! So.. differentiate two times! The differential equation must free from arbitrary constants
Chapter 4 : Ordinary Differential Equation (ODE) 297 Example 3 : Find the differential equation from the function = + −3 , where A and B are arbitrary constants. Solution : Example 4 : Find the differential equation from the function = 3 2 + −2 , where A is an arbitrary constant. Solution : = + −3 ′ = −3 −3 ′′ = 9 −3 To eliminate B, ②3 : 3′ = −9 −3 ③+②a : ′′ + 3 ′ = 9 −3 + (−9 −3 ) ′′ + 3′ = 0 ① ② ③ ②a = 3 2 + −2 ′ = 6 − 2 −2 To eliminate A, ①2 : 2 = 6 2 + 2 −2 ②+①a : ′ + 2 = (6 − 2 −2 ) + (6 2 + 2 −2 ) ′ + 2 = 6 + 6 2 ① ② ①a The number of arbitrary constants are two! So.. differentiate two times! The differential equation must free from arbitrary constants The differential equation must free from arbitrary constants
MAT238/ MAT438 : Foundation of Applied Mathematics 298 Example 5 : Find the differential equation from the function = 2 + 2 , where A and B are arbitrary constants. Solution : = 2 + 2 ′ = 2 2 + ′ + ′ ′ = 2 2 + 2 + 2 2 ′′ = 4 2 + 2 2 + ′ + ′ ′′ = 4 2 + 2 2 + 2 2 + 42 ′′ = 4 2 + 4 2 + 42 ②4 : 4 ′ = 8 2 + 4 2 + 8 2 ③−②a : ′′ − 4 ′ = −4 2− 42 ①4: 4 = 4 2 + 4 2 ④+①a : ′′ − 4 ′ + 4 = (−4 2− 42 ) + (4 2 + 4 2 ) ′′ − 4 ′ + 4 = 0 # ① ② ③ ②a ④ ①a = = 2 ′ = ′ = 2 2 = 2 = 2 ′ = 2 ′ = 2 2 The number of arbitrary constants are two! So.. differentiate two times! The differential equation must free from arbitrary constants
Chapter 4 : Ordinary Differential Equation (ODE) 299 Example 6 : Find the differential equation from the function = (4) + (4) + , where A and B are arbitrary constants. Solution : Example 7 : Find the differential equation from the function = ( + ) − , where A is an arbitrary constant. Solution : = (4) + (4) + ′ = 4 (4) − 4(4) ′′ = −16 (4) − 16(4) ①16 : 16 = 16 (4) + 16(4) + 16 ③+①a : ′′ = −16 (4) − 16(4) +(16 = 16 (4) + 16(4) + 16) ′′ + 16 = 16 ′′ + 16 − 16 = 0 # ① ② ③ ①a = ( + ) − ′ = ′ + ′ = − + ( + )(− − ) ′ = −− ( + ) − ②+① : ′ + = −− ( + ) − + ( + ) − ′ + = − # ① ② = + = − ′ = 1 ′ = − − The number of arbitrary constants are two! So.. differentiate two times! The number of arbitrary constant is one! So.. differentiate one time only! The differential equation must free from arbitrary constants The differential equation must free from arbitrary constants
MAT238/ MAT438 : Foundation of Applied Mathematics 300 TUTORIAL 4.3 : Formation of Differential Equations Find the differential equation of the following 1. = + 5 , where and re arbitrary constants. Ans : ′′ − 6′ + 5 = 0 2. = 3 + 3 , where and re arbitrary constants. Ans : ′′ − 6′ + 9 = 0 3. = 2 ( + ), where and re arbitrary constants. Ans : ′′ − 4′ + 5 = 0 4. = 5 + 5 , where and re arbitrary constants. Ans : ′′ + 25 = 0
Chapter 4 : Ordinary Differential Equation (ODE) 301 1. Solve x y e dx dy dx d y 2 12 2 2 − + = 2. Solve y x dx d y sin2 2 2 + = 3. Solve the following initial value problem of the second order differential equation using the method of undetermined coefficients. 2 4 , (0) 3 '(0) 3 3 2 2 − + y = e y = − and y = − dx dy dx d y x 4. Find the solution that satisfies the following non-homogeneous second order differential equation. 2 y'' − 2y' + 4y = 2x 5. Determine the solution for the second order differential equation given below y'' + y' − 6y = 4x; y(0) = 6, y'(0) = 15 6. Use the method of undetermined coefficients to solve x dx dy dx d y 2sin2 2 2 + = . 7. Solve y e x dx dy dx d y x + − = + 3 2 2 2 5 8. Use the method of undetermined coefficients to solve the following second order differential equation. y'' + 2y' + y = 10sinx . 9. Use the method of undetermined coefficients to solve the following second order differential equation. y'' + y' − 2y = cos2x .
MAT238/ MAT438 : Foundation of Applied Mathematics 302 10. Solve the following second order differential equation. 6 5 1 2 2 + − y = x + dx dy dx d y . 11. Find the general solution for the following second order differential equation. x y e dx dy dx d y − − − 2 = 2 2 . 1. x x x y C e C xe x e 2 = 1 + 2 + 6 2. y = C1cos x + C2 sinx − 3 1 sin2x 3. x x y e e 3 = − 4 + 4. y e (C cos x C sin x) x x x 2 1 2 1 3 3 2 = 1 + 2 + + 5. 9 1 3 2 45 31 5 34 2 3 = − − − − y e e x x x 6. y C C e sin x x 2 3 2 = 1 + 2 − − . 7. 4 1 2 1 2 1 3 2 2 = 1 + + − − − y C e C e e x x x x 8. y C e C xe x x x = 1 + 2 − 5cos − − 9. y C e C e x x x x cos2 20 3 sin2 20 2 1 = 1 + 2 + − − 10. 36 11 6 3 5 2 2 = 1 + − − − y C e C e x x x 11. x x x y C e C e xe − − = + − 3 2 1 1 2 END OF CHAPTER 4 END OF SYLLABUS MAT238/MAT438 (Foundation of Applied Mathematics) ALL THE BEST FOR YOUR COMING FINAL ☺ ☺
MAT238 APPENDIX 1 (1) TABLE OF INTEGRALS 1. ln | ax b | C; n 1 a 1 C; n 1 a(n 1) (ax b) (ax b) dx n 1 n 2. dx ln | x | C x 1 3. cos(ax) C a 1 sin(ax)dx 4. sin(ax) C a 1 cos(ax) dx 5. tan(ax) C a 1 sec (ax)dx 2 6. ln | sec(ax) tan(ax)| C a 1 sec(ax)dx 7. sec(ax) C a 1 sec(ax)tan(ax)dx 8. cosh(ax) C a 1 sinh(ax) dx 9. sinh(ax) C a 1 cosh(ax)dx 10. tanh(ax) C a 1 sech (ax)dx 2 11. coth(ax)C a 1 csch (ax)dx 2 12. sech(ax)C a 1 sech(ax)tanh(ax)dx 13. csch(ax) C a 1 csch(ax)coth(ax) dx 14. C a x dx sin a x 1 1 2 2
MAT238 APPENDIX 1 (2) 15. C a x tan a 1 dx a x 1 1 2 2 16. C a x sec a 1 dx x x a 1 1 2 2 17. C ln | x a x | C a x dx sinh a x 1 1 2 2 2 2 18. C ln | x x a | C, if x a a x dx cosh x a 1 1 2 2 2 2 19. C, if | x | a a x coth a 1 C, if | x | a a x tanh a 1 C x a x a ln 2a 1 dx a x 1 1 1 2 2 20. C, if 0 x a x a a x ln a 1 C a x sech a 1 dx x a x 1 2 2 1 2 2 21. C, if x 0 x a a x ln a 1 C a x csch a 1 dx x a x 1 2 2 1 2 2 TRIGONOMETRIC IDENTITIES 1. sin x cos x 1 2 2 2. sin2x 2sinxcosx 3. cos2x cos x sin x 2 2 HYPERBOLIC FUNCTIONS 1. 2 e e sinh x x x 2. 2 e e cosh x x x 3. cosh x sinh x 1 2 2