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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 + 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 D + 49)y = e –7x + 4 The given equation is (D D + 49)y = e –7x + 4 The characteristic equation is p p + 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 – (2x) + 16(2) = x x 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 = = c e –k ……….(2)

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x = c e – k t …….(1) When t = 4, x = = 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) – = – = log c = /3 = c = antilog = 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 = e –k t ……(2) When, t = 120, T = 91.4 F 91.4 F = e – 120k 21.4e – 120k = 91.4 – e – 120k = 19.4

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When T = 98.6 F, t = t F = e –k t e –k t1 = 98.6 – 72 = hours 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 = 3 5 times of the population at initial time

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22. The sum of Rs 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 = ) Let x be the principal Integrating log x = 0.04 t + c 1 x = c e 0.04 t …….(1) When t = 0, x = = 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 = t = log e t = 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 A 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 = A 0 Amount of radium at the end of 100years = A 0

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