1. 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) Integrating
Which is the required solution.

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

5. 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

6. 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

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

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

10. 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

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

12. 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

13. 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

14. 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

15. 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

16. 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

17. The general solution is
y = CF + PI

18. 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

19. 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

20. 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

21. The general solution is
y = CF + PI

22. 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

23. 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

24. 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

25. The general solution is
y = CF + PI

26. 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

27. 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

28. 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 93.4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

29. When, t = 120, T = 91.4F
91.4F = 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 = e–k t ……(2)

30. When T = 98.6F, t = t1
98.6F = 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

31. 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

32. 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

33. 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

34. 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

35. 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

36. 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

37. THE END