Chapter 46

Solution of first order

differential equations by

separation of variables

46.1 Family of curves

Integrating both sides of the derivative

dy

dx

=3 with

respect toxgivesy=

∫
3dx, i.e.,y= 3 x+c,wherec
is an arbitrary constant.

y= 3 x+crepresents afamily of curves, each of the
curves in the family depending on the value ofc.
Examples includey= 3 x+8,y= 3 x+3,y= 3 x and
y= 3 x−10 and these are shown in Fig. 46.1.

216

212

28

24

24 23 22 21

4

x

8

12

16

y y 53 x 18

y 53 x 13

y 53 x 210

y 53 x

01 234

Figure 46.1

Each are straight lines of gradient 3. A particular curve

of a family may be determined when a point on the curve

is specified. Thus, ify= 3 x+cpasses through the point

(1, 2) then 2= 3 ( 1 )+c, from which,c=−1. The equa-

tion of the curve passing through (1, 2) is therefore

y= 3 x−1.

Problem 1. Sketch the family of curves given by

the equation

dy

dx

= 4 xand determine the equation of

one of these curves which passes through the point

(2, 3).

Integrating both sides of

dy

dx

= 4 x with respect tox

gives:

∫

dy

dx

dx=

∫

4 xdx, i.e.,y= 2 x^2 +c

Some members of the family of curves having

an equation y= 2 x^2 +c include y= 2 x^2 +15,

y= 2 x^2 +8,y= 2 x^2 and y= 2 x^2 −6, and these are

shown in Fig. 46.2. To determine the equation of the

curve passing through the point (2, 3),x=2andy= 3

are substituted into the equationy= 2 x^2 +c.

Thus 3= 2 ( 2 )^2 +c, from whichc= 3 − 8 =−5.

Hence the equation of the curve passing through the

point (2, 3) isy= 2 x^2 − 5.