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**Differential Equations**

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

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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 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

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**4. Solve (x2 – y)dx + (y2 – x)dy = 0, if it passes through the origin.**

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, Since it passes through the origin, 0 + 0 = 0 + c c = 0 The required solution is

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

P = cot x, Q = 2cosx

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

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**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

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

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**8. Solve (D2 – 13D + 12)y = e –2x The given equation is**

The characteristic 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

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**9. Solve (D2 + 6D + 8)y = e –2x The given equation is**

The characteristic 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

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**10. Solve (D2 – 4D + 13)y = e –3x CF = e2x(Acos3x + B sin3x)**

The given equation is (D2 – 4D + 13)y = e –3x The characteristic equation is p2 – 4p + 13 = 0 The general solution is y = CF + PI

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**11. Solve (D2 – 4)y = sin2x The given equation is (D2 – 4)y = sin2x**

The characteristic 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

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**12. Solve (D2 – 2D – 3)y = sinx cosx**

The given equation is (D2 – 2D – 3)y = sinx cosx The characteristic 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

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**The general solution is**

y = CF + PI

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**13. Solve (D2 + 14D + 49)y = e–7x + 4 The given equation is**

The characteristic 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

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**14. Solve (D2 – 13D + 12)y = e–2x + 5ex The given equation is**

The characteristic 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

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**15. Solve (D2 + 4D + 13)y = cos3x The given equation is**

The characteristic equation is p2 + 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: (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

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**17. Solve (D2 – 4D + 1)y = x2 The given equation is**

The characteristic equation is p2 – 4p + 1 = 0 Particular integral is = {1 + (D2 – 4D)}–1 x2 = {1 – (D2 – 4D) + (D2 – 4D)2 – } x2 = {1 – D2 + 4D + 16D2} x2 = x2 – (2x) + 16(2) = x x The general solution is y = CF + PI

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**18. Solve (D2 + 3D – 4)y = x2 The given equation is**

The characteristic equation is p2 + 3p – 4 = 0 (p + 4)(p – 1) = 0 p = 1, – 4 CF = Aex + Be–4x The particular integral is

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**The general solution is**

y = CF + PI

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**Let x grams of the substance remain after t hours.**

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

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taking log 3log c = 4log60 – log21 = 4(1.7782) – = – = log c = /3 = c = antilog = 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

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**Let T be the temperature of the body at any time t **

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 93.4F. After 2 hours he finds the temperature to be 91.4F. If the room temperature is 72F, estimate the time of death. (Assume that the normal temperature of a human body to be 98.6F). 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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When, t = 120, T = 91.4F 91.4F = 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 = 93.4F 93.4F = 72 + c e0 c = 93.4 – 72 = 21.4F T = e–k t ……(2)

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When T = 98.6F, t = t1 98.6F = 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 x = c e k t …….(1) When t = 0, x = x0 x0 = c e0 c = x0 log x = k t + c1

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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

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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? (loge2 = ) 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

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

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**Let A be the amount of radium disappears at any time t.**

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

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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 = A0 Amount of radium at the end of 100years = A0

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THE END

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Differential Equations

Differential Equations

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