1. FIRST ORDER DIFFERENTIAL EQUATIONS

2. In this lecture we discuss various methods of solving first order differential equations.
These include:
Variables separable
Homogeneous equations
Exact equations
Equations that can be made exact by multiplying by an integrating factor

3. Transform to separable
Linear
Non-linear
Integrating Factor
Integrating
Factor
Separable
Homogeneous
Exact
Transform to Exact
Transform to separable

4. Variables Separable The first-order differential equation
(1)
is called separable provided that f(x,y) can be written as the product of a function of x and a function of y.

5. Suppose we can write the above equation as
We then say we have “ separated ” the variables. By taking h(y) to the LHS, the equation becomes

6. Integrating, we get the solution as
where c is an arbitrary constant.

7. Example 1. Consider the DE
Separating the variables, we get
Integrating we get the solution as or
an arbitrary constant.

8. Example 2. Consider the DE
Separating the variables, we get
Integrating we get the solution as or
an arbitrary constant.

9. Homogeneous equations
Definition A function f(x, y) is said to be homogeneous of degree n in x, y if
for all t, x, y
Examples
is homogeneous of degree 2.
is homogeneous of degree 0.

10. clearly homogeneous of degree 0.
A first order DE
is called homogeneous if
are homogeneous functions of x and y of the same degree.
This DE can be written in the form
where is
clearly homogeneous of degree 0.

11. The substitution y = z x converts the given equation into “variables separable” form and hence can be solved. (Note that z is also a (new) variable,) We illustrate by means of examples.

12. Example 3. Solve the DE
That is
Let y = z x. Hence we get

13. or
Separating the variables, we get
Integrating we get

14. We express the LHS integral by partial fractions. We getor
an arbitrary constant.
Noting z = y/x, the solution is:
c an arbitrary constant or
c an arbitrary constant

15. Working Rule to solve a HDE:
1. Put the given equation in the form
2. Check M and N are Homogeneous function of the same degree.

16. 4. Differentiate y = z x to get
5. Put this value of dy/dx into (1) and solve the equation for z by separating the variables.
6. Replace z by y/x and simplify.

17. Example 4. Solve the DE
Let y = z x. Hence we get or
Separating the variables, we get
Integrating we get
cosec z – cot z = c x
where
and c an arbitrary constant.

18. Non-homogeneous equations
We shall now see that some equations can be brought to homogeneous form by appropriate substitution.
Example 5 Solve the DE
That is

19. We shall now put x = u+h, y = v+k
where h, k are constants ( to be chosen).
Hence the given DE becomes
We now choose h, k such that
Hence

20. Hence the DE becomes
which is homogeneous in u and v.
Let v = z u. Hence we get

21. or
Separating the variables, we get
Integrating we get
i.e. or

22. Example 6 Solve the DE
That is
Now the previous method does not work as the lines
are parallel. We now put u = 3x + 2y.

23. The given DE becomes or
Separating the variables, we get

24. Integrating, we get
i.e.

25. EXACT DIFFERENTIAL EQUATIONS
A first order DE
is called an exact DE if there exits a function f(x, y) such that
Here df is the ‘total differential’ of f(x, y) and equals

26. Hence the given DE becomes df = 0
Integrating, we get the solution as
f(x, y) = c, c an arbitrary constant
Thus the solution curves of the given DE are the ‘level curves’ of the function f(x, y) .
Example 8 The DE
is exact as it is d (xy) = 0
Hence the solution is: x y = c

27. Example 7 The DE
is exact as it is d (x2+ y2) = 0
Hence the solution is: x2+ y2 = c
Example 9 The DE
is exact as it is
Hence the solution is:

28. Test for exactness Suppose
is exact. Hence there exists a function f(x, y)
such that
Hence
Assuming all the 2nd order mixed derivatives of f(x, y) are continuous, we get

29. Thus a necessary condition for exactness is

30. We saw a necessary condition for exactness is
We now show that the above condition is also sufficient for M dx + N dy = 0 to be exact. That is, if
then there exists a function f(x, y) such that

31. Integrating
partially w.r.t. x, we get
……. (*)
where g(y) is a function of y alone
We know that for this f(x, y),
Differentiating (*) partially w.r.t. y, we get
= N gives

32. … (**) or
We now show that the R.H.S. of (**) is independent of x and thus g(y) (and so f(x, y)) can be found by integrating (**) w.r.t. y.
= 0
Q.E.D.

33. Note (1) The solution of the exact DE d f = 0 is f(x, y) = c.
Note (2) When the given DE is exact, the solution f(x, y) = c is found as we did in the previous theorem. That is, we integrate M partially w.r.t. x to get
We now differentiate this partially w.r.t. y and equating to N, find g (y) and hence g(y).
The following examples will help you in understanding this.

34. Example 8 Test whether the following DE is exact. If exact, solve it.
Here
Hence exact.
Now

35. Differentiating partially w.r.t. y, we get
Hence
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is or
c an arb const.

36. Example 9 Test whether the following DE is exact. If exact, solve it.
Here
Hence exact.
Now

37. Differentiating partially w.r.t. y, we get
Hence
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is or
c an arb const.

38. In the above problems, we found f(x, y) by integrating M partially w.r.t. x and then
equated
We can reverse the roles of x and y. That is we can find f(x, y) by integrating N partially
w.r.t. y and then equate
The following problem illustrates this.

39. Example 10 Test whether the following DE is exact. If exact, solve it.
Here
Hence exact.
Now

40. Differentiating partially w.r.t. x, we get
gives
Integrating, we get
(Note that we have NOT put the arb constant )
Hence
Thus the solution of the given d.e. is or
c an arb const.

41. Integrating Factors The DE is NOT exact but becomes exact when
as it becomes
multiplied by
i.e.
We say
is an Integrating Factor of the given DE

42. Definition If on multiplying by (x, y), the DE
becomes an exact DE, we say that (x, y) is an Integrating Factor of the above DE
are all integrating factors of
the non-exact DE
We give some methods of finding integrating factors of an non-exact DE

43. Problem Under what conditions will the DE
have an integrating factor that is a function of x alone ?
Solution. Suppose  = h(x) is an I.F.
Multiplying by h(x) the above d.e. becomes
Since (*) is an exact DE, we have

44. i.e. or or or

45. Hence if
is a function of x alone, then
is an integrating factor of the given DE

46. Rule 2: Consider the DE
, a function of y alone, If
then
is an integrating factor of the given DE

47. Problem Under what conditions will the DE
have an integrating factor that is a function of the product z = x y ?
Solution. Suppose  = h(z) is an I.F.
Multiplying by h(z) the above d.e. becomes
Since (*) is an exact DE, we have

48. i.e. or or or

49. Hence if
is a function of z = x y alone, then
is an integrating factor of the given DE

50. Example 11 Find an I.F. for the following DE and hence solve it.
Here
Hence the given DE is not exact.

51. Now
a function of x alone. Hence
is an integrating factor of the given DE
Multiplying by x2, the given DE becomes

52. which is of the form
Note that now
Integrating, we easily see that the solution is
c an arbitrary constant.

53. Example 12 Find an I.F. for the following DE and hence solve it.
Here
Hence the given DE is not exact.

54. Now
a function of y alone. Hence
is an integrating factor of the given DE
Multiplying by sin y, the given DE becomes

55. which is of the form
Note that now
Integrating, we easily see that the solution is
c an arbitrary constant.

56. Example 13 Find an I.F. for the following DE and hence solve it.
Here
Hence the given DE is not exact.

57. Now
a function of z =x y alone. Hence
is an integrating factor of the given DE

58. Multiplying by
the given DE becomes
which is of the form
Integrating, we easily see that the solution is
c an arbitrary constant.

59. Problem Under what conditions will the DE
have an integrating factor that is a function of the sum z = x + y ?
Solution. Suppose  = h(z) is an I.F.
Multiplying by h(z) the above DE becomes
Since (*) is an exact DE, we have

60. i.e. or or or

61. Hence if
is a function of z = x + y alone, then
is an integrating factor of the given DE

62. Linear Equations
A linear first order equation is an equation that can be expressed in the form
where a1(x), a0(x), and b(x) depend only on the independent variable x, not on y.

63. We assume that the function a1(x), a0(x), and b(x) are continuous on an interval and that a1(x)  0on that interval. Then, on dividing by a1(x), we can rewrite equation (1) in the standard form
where P(x), Q(x) are continuous functions on the interval.

64. Let’s express equation (2) in the differential form
If we test this equation for exactness, we find
Consequently, equation(3) is exact only when P(x) = 0. It turns out that an integrating factor , which depends only on x, can easily obtained the general solution of (3).

65. Multiply (3) by a function (x) and try to determine (x) so that the resulting equation
is exact.
We see that (4) is exact if  satisfies the DE
Which is our desired IF

66. and so (7) can be written in the form
In (2), we multiply by (x) defined in (6) to obtain
We know from (5)
and so (7) can be written in the form

67. Integrating (8) w.r.t. x gives
and solving for y yields

68. Working Rule to solve a LDE:
1. Write the equation in the standard form
2. Calculate the IF (x) by the formula
3. Multiply the equation in standard form by (x) and recalling that the LHS is just
obtain

69. 4. Integrate the last equation and solve for y by dividing by (x).

70. Ex 1. Solve
Solution :- Dividing by x cos x, throughout, we get

71. yields
Multiply by
Integrate both side we get

72. Problem (2g p. 62): Find the general solution of the equation
Ans.:

73. The usual notation implies that x is independent variable & y the dependent variable. Sometimes it is helpful to replace x by y and y by x & work on the resulting equation.
* When diff equation is of the form

74. Q. 4 (b) Solve