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### Week 2

1. 1. KNF1023 Engineering Mathematics II First Order ODEs Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2008/2009
2. 2. Learning Objectives Explain the first order of ODEs Explain the separable equations Apply first order in real life application
3. 3. First Order differential equations ODEs in y=y(x) are 1st order if y’(x) is the highest order derivative of y present in the equations. First order ODEs in y=y(x) may be written in the form dy = G ( x, y ) dx For cases where G(x,y) assumes certain specific forms, methods of solving the ODEs are available.
4. 4. 1st Order ODEs in Separable Form A 1st order ODE is said to be in separable form if it can be written as dy = f ( x) • f ( y ) dx Which is simplified to dy ∫ f ( y) = ∫ f ( x)dx This provides us with a method for solving the ODE if we can work out the integrals on both sides. Notice the integration on the left hand side is with respect y to and that on the right hand side is with respect to x.
5. 5. Example 1 Solve the ODE y , ( x) = ( y + 1) 2 (3x 2 + 2) subject to y(0)=1 The ODE is 1st order and in separable form Rewrite it as y , ( x) ∫ ( y + 1) 2 = (3 x 2 + 2) 1 ∫ ( y + 1) 2 ( ) dy = ∫ 3 x 2 + 2 dx
6. 6. Continue… −1 − ( y + 1) = x 3 + 2 x + C − (1 + 1) −1 = 0 3 + 2(0) + C 1Then C=− 2 −1 3 1 − ( y + 1) = x + 2x − 2For the explicit form, we can write it as 2x3 + 4x + 1 y= − 2x3 − 4x + 1
7. 7. Example 2 , 2x 1Solve y ( x) = = ( 2x) y +1 y +1The ODE is 1st order and separable formRewrite it as dy 2x = dx y + 1 ( y + 1)dy = 2 xdxIntegrate on Both sides ∫ ( y + 1)dy = ∫ 2 xdx
8. 8. Continue… ∫ ydy + ∫ dy = 2∫ xdx y2 2x 2 +y= +C 2 2 2 2 y + 2 y = 2 x + 2C
9. 9. Example 3 x2 + 7 x + 3Solve the initial value problem y = y2with y(0)=3 y 2 y = x2 + 7 x + 3 dy 2 y = x2 + 7x + 3 dx 2 2 y dy = ( x + 7 x + 3)dx ∫ ( ) y 2 dy = ∫ x 2 + 7 x + 3 dx
10. 10. Continue… y 3 x3 x2 = + 7 + 3x + C 3 3 2 3 21 2 3 y = x + x + 9 x + 3C 2 21 2 3 y = x + x + 9 x + 3C 3 2From the initial condition y(0)=3. 3 = y (0) = 3 0 + 0 + 0 + 3C = 3 3C 27 C= =9 3
11. 11. Continue… Thus C=9 and substitute it into equation we get 21 2 3 y = x + x + 9 x + 3(9) 3 2 21 2 3 y = x + x + 9 x + 27 3 2
12. 12. 1ST ORDER ODEs REDUCIBLE TOSEPARATE FORM
13. 13. 1ST ORDER ODEs REDUCIBLE TOSEPARATE FORM
14. 14. 1ST ORDER ODEs REDUCIBLE TOSEPARATE FORM
15. 15. Example 4 3  y  xSolve the ODE y =  +  subject to  x  yy(1)=1Rewrite the equation as 3 dy  y   x  = +  dx  x   y   Use y y = x.u ( x) or u= x
16. 16. Continue… Substitute it in the equation 3 dy  y   x  = +  dx  x   y    3 dy 1 =u+  dx u Differential the y = x.u ( x ) , thus we get dy = x.u , ( x) + u dx
17. 17. Continue 3 1 x ⋅ u ( x) + u = u +   u 3 du 1 x =u +  −u dx u 3 du  1  x =  dx  u This is a separable ODE and can be writtenas u 3 du = dx xIntegrate both sides, we get ∫ u 3 du = ∫ dx x
18. 18. Continue… u4 + C1 = ln( x) + C 2 4 u4 = ln( x) + (C 2 − C1 ) 4 u4 = ln( x) + C 4Replacing u back by y/x we obtain 4 1 y   = ln( x) + C 4 x or y 4 = 4 x 4 ln( x) + 4Cx 4
19. 19. Continue… as the general solution of the ODE (in implicit form) Now y(1)=1 gives 14 = 4(1) 4 ln(1) + 4C (1) 4 or 1 C= 4Thus, the required particular solution is y 4 = 4 x 4 ln( x) + ( x) 4
20. 20. Example 5 Solve the first order differential equation. dy 2 y x = y + x cos   dx x
21. 21. Application in Growth and Decay First order ODEs in separable form can be found in population dynamics which is concerned with the growth and decay of a population. The population may be a lump of decaying radioactive substance, a colony of bacteria thriving on an agar culture or a group of people living in a community.
22. 22. Theory The rate at which the “Something” grows or decay is directly proportional to the current population (until such time as resources become scarce or overcrowding becomes a limiting factor). If we let y(t) represent the number of “something” at time t, then the rate of change of the “something” with respect to time is y’(t). Thus, since y’(t) is , proportional to y(t), we have y (t ) = ky(t )
23. 23. Theory For k > 0, equation is called an exponential growth law and for k < 0 it is an exponential decay law.
24. 24. Example 1: Exponential Growth of aBacterial ColonyA freshly inoculated bacterial culturecontains 100 cells. When the culture ischecked 100 minutes later, it isdetermined that there are 450 cellspresent. Assuming exponential growth,determine the number of cells presentat any time t (measured in minutes).
25. 25. Solution:Exponential growth means that , y (t ) = ky (t ) dy = ky (t ) dt dy = kdt y (t ) dy ∫ y (t ) = ∫ kdt ln y (t ) = kt + C y (t ) = e kt +C
26. 26. Continue… y (t ) = e kt .e CThen the equation become y (t ) = Ae ktWhere A and k are constants to bedetermined. Notice that if we set thestarting time be t=0, we have y(0) = 100
27. 27. Continue…Setting t = 0, we now have100 = y(0) = Ae0=AAnd hence, y(t) = 100 ektWe can use the second observation todetermine the value of the growth constant,k. We have 450 = y(100) = 100e100k
28. 28. Continue…Dividing both sides by 100, we hav 4.5 = e100k ln 4.5 = 100k ln 4.5 k= ≈ 0.01504 100We now have a formula representing the numberof cells present at any time t: kt  ln 4.5  y (t ) = 100e = 100 exp  t  100 
29. 29. Example 2: Radioactive Decay If you have 50 grams of 14 C today, how much will be left in 100 years? (half-life for 14 C is approximately 5730 years)
30. 30. Solution:Let y(t) be the number of grams of 14 C presentat time t. Then, we have y , (t ) = ky(t ) dy = ky (t ) dt dy = kdt y (t ) dy ∫ y (t ) = ∫ kdt
31. 31. Continue... ln y (t ) = kt + C kt + C y (t ) = e kt C y (t ) = e .eAssume C e =A kt y (t ) = Ae
32. 32. Continue...Our initial condition is y(0) = 50 and so k ( 0) y (0) = 50 = AeThus, A = 50Substitute A=50 and we get kt y (t ) = 50eIn order to determine the value of the decayconstant k, we use the half-life: 5730 k 25 = y (5730) = 50e
33. 33. Continue... 25 = e 5730k 50 1 ln = 5730k 2 The value of K can be calculated using the scientific calculator 1 ln k= 2 ≈ −1.20968 ×10 − 4 5730To see how much will be left in 100 years, we substitute thevalue that we calculated and into the equation − 1 . 20968 × 10 − 4 × 100 y (100 ) = 50 e ≈ 49.3988 grams
34. 34. Prepared By Annie ak JosephPrepared ByAnnie ak Joseph Session 2008/2009