Higher Engineering Mathematics

Differential equations

I

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, where
cis an arbitrary constant.

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

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 equation of the curve passing
through (1, 2) is thereforey= 3 x−1.

Problem 1. Sketch the family of curves given

by the equation

dy

dx

= 4 xand determine the equa-

tion of one of these curves which passes through

the point (2, 3).

Integrating both sides of

dy

dx

= 4 xwith 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 andy= 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=2 and

y=3 are substituted into the equationy= 2 x^2 +c.

Figure 46.2

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.