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Differential Equations

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

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

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

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4. Solve (x 2 – y)dx + (y 2 – x)dy = 0, if it passes through the origin. The given equation is (x 2 – y)dx + (y 2 – x)dy = 0 x 2 dx – ydx + y 2 dy – xdy = 0 x 2 dx + y 2 dy = ydx + xdy Integrating, Since it passes through the origin, 0 + 0 = 0 + c c = 0 The required solution is

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

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

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7. Solve: (1 + y 2 )dx = (tan –1 y – x )dy The given equation (1+y 2 )dx = (tan –1 y – x)dy can be written as It is a linear diff eqn in x with The solution is

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

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

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9. Solve (D 2 + 6D + 8)y = e –2x The given equation is (D 2 + 6D + 8)y = e –2x The characteristic equation is p 2 + 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

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10. Solve (D 2 – 4D + 13)y = e –3x The given equation is (D 2 – 4D + 13)y = e –3x The characteristic equation is p 2 – 4p + 13 = 0 The general solution is y = CF + PI CF = e 2x (Acos3x + B sin3x)

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11. Solve (D 2 – 4)y = sin2x The given equation is (D 2 – 4)y = sin2x The characteristic equation is p 2 – 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

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12. Solve (D 2 – 2D – 3)y = sinx cosx The given equation is (D 2 – 2D – 3)y = sinx cosx The characteristic equation is p 2 – 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

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The general solution is y = CF + PI

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13. Solve (D 2 + 14D + 49)y = e –7x + 4 The given equation is (D 2 + 14D + 49)y = e –7x + 4 The characteristic equation is p 2 + 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

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

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

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The general solution is y = CF + PI

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

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

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18. Solve (D 2 + 3D – 4)y = x 2 The given equation is (D 2 + 3D – 4)y = x 2 The characteristic equation is p 2 + 3p – 4 = 0 (p + 4)(p – 1) = 0 p = 1, – 4 CF = Ae x + Be –4x The particular integral is

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The general solution is y = CF + PI

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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 log x = – k t + c 1 x = c e – k t …….(1) When t = 1, x = 60 60 = c e –k ……….(2)

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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 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.

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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

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Integrating log (T – 72) = – k t + c 1 T = 72 + c e – k t …….(1) Initially, t = 0, T = 93.4 F 93.4 F = 72 + c e 0 c = 93.4 – 72 = 21.4 F T = 72 + 21.4 e –k t ……(2) 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

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When T = 98.6 F, t = t 1 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

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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 log x = k t + c 1 x = c e k t …….(1) When t = 0, x = x 0 x 0 = c e 0 c = x 0

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Sub in eqn (1) we get x = x 0 e k t ………(2) Given x = 3x 0, when t = 1 3x 0 = x 0 e k e k = 3……..(3) When t = 5, let x = x 1 x 1 = x 0 e 5k = x 0 (e k ) 5 = x 0 3 5 = 3 5 times of the population at initial time

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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? (log e 2 = 0.6931) Let x be the principal Integrating log x = 0.04 t + c 1 x = c e 0.04 t …….(1) When t = 0, x = 1000 1000 = c e 0 c = 1000 x = 1000e 0.04t ……..(2)

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

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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 log A = –k t + c 1 Initially, t = 0 and A = A 0 A 0 = c e 0 c = A 0

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When t = 50years, A = 95% of A 0 = 0.95A 0 0.95A 0 = A 0 e –50k e – 50k = 0.95 When t = 100 years, A = A 1 A 1 = A 0 e – 100k = A 0 (e – 50k ) 2 = A 0 (0.95) 2 = 0.9025A 0 Amount of radium at the end of 100years = 0.9025A 0

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