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Differential Equations

Differential Equations. Differential Equations. PREPARED BY: R.RAJENDRAN. M.A., M . Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21. 1.Solve;. = (1 + x) + y(1 + x) = (1 + x)(1 + y). Which is the required solution. Integrating. 2.Solve;. Put x + y = z.

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Differential Equations

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  1. Differential Equations Differential Equations PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

  2. 1.Solve; = (1 + x) + y(1 + x) = (1 + x)(1 + y) Which is the required solution. Integrating

  3. 2.Solve; Put x + y = z The given eqn becomes Integrating

  4. 3. Solve 3ex tan y dx + (1 + ex) sec 2y dy = 0 Given equation is 3ex tan y dx + (1 + ex) sec2y dy = 0 3ex tan y dx = – (1 + ex) sec2y dy Integrating (1+ex)3(tan y) = c Which is the required solution 3 log(1+ex) = – log(tan y) + log c log(1+ex)3 + log(tan y) = log c log(1+ex)3(tan y) = log c

  5. 4. Solve (x2 – y)dx + (y2 – x)dy = 0, if it passes through the origin. Since it passes through the origin, 0 + 0 = 0 + c c = 0 The required solution is The given equation is (x2 – y)dx + (y2 – x)dy = 0 x2 dx – ydx + y2 dy – xdy = 0 x2 dx + y2 dy = ydx + xdy Integrating,

  6. 5. Solve: The solution is It is a linear diff eqn with P = cot x, Q = 2cosx

  7. 6. Solve: The solution is It is a linear diff eqn with

  8. 7. Solve: (1 + y2)dx = (tan–1 y – x )dy The given equation (1+y2)dx = (tan–1y – x)dy can be written as The solution is It is a linear diff eqn in x with

  9. Put u = tan – 1 y Eqn (1) becomes It is the required soln.

  10. 8. Solve (D2 – 13D + 12)y = e –2x The given equation is (D2 – 13D + 12)y = e –2x Thecharacteristic equation is p2 – 13p + 12 = 0 (p – 12)(p – 1) = 0 p = 1, 12 The complementary function is CF = A ex + B e 12x Particular integral is The general solution is y = CF + PI

  11. 9. Solve (D2 + 6D + 8)y = e –2x The given equation is (D2 + 6D + 8)y = e –2x Thecharacteristic equation is p2 + 6p + 8 = 0 (p + 2)(p + 4) = 0 p = – 2, – 4 The complementary function is CF = A e–2x + B e– 4x Particular integral is The general solution is y = CF + PI

  12. 10. Solve (D2 – 4D + 13)y = e –3x CF = e2x(Acos3x + B sin3x) The given equation is (D2 – 4D + 13)y = e –3x Thecharacteristic equation is p2 – 4p + 13 = 0 The general solution is y = CF + PI

  13. 11. Solve (D2 – 4)y = sin2x The given equation is (D2 – 4)y = sin2x Thecharacteristic equation is p2 – 4 = 0 (p – 2)(p + 2) = 0 p = –2, 2 The complementary function is CF = A e –2x + B e 2x Particular integral is The general solution is y = CF + PI

  14. 12. Solve (D2 – 2D – 3)y = sinx cosx The given equation is (D2 – 2D – 3)y = sinx cosx Thecharacteristic equation is p2 – 2p – 3 = 0 (p – 3)(p + 1) = 0 p = –1, 3 The complementary function is CF = A e –x + B e 3x Particular integral is

  15. The general solution is y = CF + PI

  16. 13. Solve (D2 + 14D + 49)y = e–7x + 4 The given equation is (D2 + 14D + 49)y = e–7x + 4 Thecharacteristic equation is p2 + 14p + 49 = 0 (p + 7)(p + 7) = 0 p = – 7, – 7 CF = (Ax + B)e–7x Particular integral is The general solution is y = CF + PI

  17. 14. Solve (D2 – 13D + 12)y = e–2x + 5ex The given equation is (D2 – 13D + 12)y = e–2x + 5ex Thecharacteristic equation is p2 – 13p + 12 = 0 (p – 12)(p – 1) = 0 p = 12, 1 CF = A e12x + B ex Particular integral is The general solution is y = CF + PI

  18. 15. Solve (D2 + 4D + 13)y = cos3x CF = e–2x(Acos3x + Bsin3x) Particular integral is The given equation is (D2 + 4D + 13)y = cos 3x Thecharacteristic equation is p2 + 4p + 13 = 0

  19. The general solution is y = CF + PI

  20. 16. Solve: (D2 – 1)y = cos 2x – 2sin 2x The given equation is (D2 – 1)y = cos 2x – 2sin 2x The characteristic equation is p2 – 1 = 0 (p – 1)(p + 1) = 0 p = 1, – 1 CF = A ex + B e–x The particular integral is The general solution is y = CF + PI

  21. 17. Solve (D2 – 4D + 1)y = x2 Particular integral is The given equation is (D2 – 4D + 1)y = x2 Thecharacteristic equation is p2 – 4p + 1 = 0 = {1 + (D2 – 4D)}–1 x2 = {1 – (D2 – 4D) + (D2 – 4D)2 – } x2 = {1 – D2 + 4D + 16D2} x2 = x2 – 2 + 4 (2x) + 16(2) = x2 + 30 + 8x The general solution is y = CF + PI

  22. 18. Solve (D2 + 3D – 4)y = x2 The given equation is (D2 + 3D – 4)y = x2 Thecharacteristic equation is p2 + 3p – 4 = 0 (p + 4)(p – 1) = 0 p = 1, – 4 CF = Aex + Be–4x The particular integral is

  23. The general solution is y = CF + PI

  24. 19. In a chemical reaction the rate of conversion of a substance at time t is proportional to the quantity of the substance still untransformed at that instant. At the end of one hour, 60grams remain and at the end of 4 hours 21 grams. How many grams of the first substance was there initially? Let x grams of the substance remain after t hours. Where k is constant of proportionality Integrating x = c e– k t …….(1) When t = 1, x = 60 60 = c e–k ……….(2) log x = – k t + c1

  25. taking log 3log c = 4log60 – log21 = 4(1.7782) – 1.3222 = 7.1128 – 1.3222 = 5.7906 log c = 5.7906/3 = 1.9302 c = antilog 1.9302 = 85.15  85 Hence initially there was 85gms of the substance. x = c e– k t …….(1) When t = 4, x = 21 21 = c e–4k ……….(3) When t = 0, x = c From (2) e – k = 60/c Sub in (3) we get

  26. 20. For a postmortem report, a doctor requires to know approximately the time of death of the deceased. He records the first temperature at 10.00am to be 93.4F. After 2 hours he finds the temperature to be 91.4F. If the room temperature is 72F, estimate the time of death. (Assume that the normal temperature of a human body to be 98.6F). Given Let T be the temperature of the body at any time t The temperature difference is T – 72 By Newton’s law of cooling, Where k is constant of proportionality

  27. When, t = 120, T = 91.4F 91.4F = 72 + 21.4 e– 120k 21.4e– 120k =91.4 – 72 21.4e– 120k = 19.4 Integrating log (T – 72) = – k t + c1 T = 72 + c e– k t …….(1) Initially, t = 0, T = 93.4F 93.4F = 72 + c e0 c =93.4 – 72 = 21.4F T = 72 + 21.4 e–k t ……(2)

  28. When T = 98.6F, t = t1 98.6F = 72 + 21.4 e–k t1 21.4 e–k t1 = 98.6 – 72 = 26.6 4hours 26min before the first recorded temperature The approximate time of death = 10 – 4:26 = 5:34am

  29. 21. The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If the population of a colony of yeast bacteria triples in 1hour. Show that the number of bacteria at the end of five hours will be 35 times of the population at initial time. Let x be the number of bacteria present in the yeast culture at any time t Where k is constant of proportionality Integrating x = c e k t …….(1) When t = 0, x = x0 x0 = c e0 c = x0 log x = k t + c1

  30. Sub in eqn (1) we get x = x0 e k t………(2) Given x = 3x0, when t = 1 3x0 = x0 ek e k = 3……..(3) When t = 5, let x = x1 x1 = x0 e 5k = x0 (ek)5 = x0 35 =35 times of the population at initial time

  31. 22. The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four percent per annum. In how many years will the amount be twice the original principal? (loge2 = 0.6931) Let x be the principal log x = 0.04 t + c1 x = c e 0.04 t …….(1) When t = 0, x = 1000 1000 = c e0 c = 1000 x = 1000e 0.04t……..(2) Integrating

  32. x = 1000 e 0.04 t …….(2) When x = 2000, t = ? 2000 = 1000 e0.04t e0.04t = 2 0.04t = loge2 0.04t = 0.6931 In 17 years the amount will be twice the original principal

  33. 23. Radium disappears at a rate proportional to the amount present. If 5% of the original amount disappears in 50years, how much will remain at the end of 100years. (Take A0 as the initial amount.) Let A be the amount of radium disappears at any time t. Where k is constant of proportionality Integrating Initially, t = 0 and A = A0 A0 = c e0  c = A0 log A = –k t + c1

  34. When t = 50years, A = 95% of A0 = 0.95A0 0.95A0 = A0 e –50k e – 50k = 0.95 When t = 100 years, A = A1 A1 = A0 e – 100k = A0 (e – 50k)2 = A0 (0.95)2 = 0.9025A0 Amount of radium at the end of 100years = 0.9025A0

  35. THE END

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