446 DIFFERENTIAL EQUATIONSFind the equation of the curve if it passes
through the point(
1,^13)
.
[
y=3
2x^2 −x^3
6− 1]- The acceleration,a, of a body is equal to its
rate of change of velocity,dv
dt. Find an equa-
tion forvin terms oft, given that whent=0,
velocityv=u.[v=u+at]
8. An object is thrown vertically upwards with
an initial velocity,u, of 20 m/s. The motion
of the object follows the differential equation
ds
dt
=u−gt, wheresis the height of the objectin metres at timetseconds andg= 9 .8m/s^2.
Determine the height of the object after 3
seconds ifs=0 whent=0. [15.9 m]46.4 The solution of equations of the
form
dy
dx
=f(y)
A differential equation of the form
dy
dx=f(y)isinitially rearranged to give dx=dy
f(y)and then thesolution is obtained by direct integration,i.e.∫
dx=∫
dy
f(y)Problem 6. Find the general solution of
dy
dx= 3 + 2 y.Rearrangingdy
dx= 3 + 2 ygives:dx=dy
3 + 2 yIntegrating both sides gives:∫
dx=∫
dy
3 + 2 yThus, by using the substitutionu=(3+ 2 y) — see
Chapter 39,x=^12 ln( 3 + 2 y)+c (1)It is possible to give the general solution of a differ-
ential equation in a different form. For example, if
c=lnk, wherekis a constant, then:x=^12 ln(3+ 2 y)+lnk,i.e. x=ln(3+ 2 y)1(^2) +lnk
or x=ln [k
√
(3+ 2 y)] (2)
by the laws of logarithms, from which,
ex=k
√
(3+ 2 y) (3)
Equations (1), (2) and (3) are all acceptable general
solutions of the differential equation
dy
dx
= 3 + 2 y
Problem 7. Determine the particular solu-
tion of (y^2 −1)
dy
dx
= 3 ygiven thaty=1 when
x= 2
1
6
Rearranging gives:
dx=
(
y^2 − 1
3 y
)
dy=
(
y
3
−
1
3 y
)
dy
Integrating gives:
∫
dx=
∫ (
y
3
−
1
3 y
)
dy
i.e. x=
y^2
6
−
1
3
lny+c,
which is the general solution.
Wheny=1,x= 216 , thus 2^16 =^16 −^13 ln 1+c, from
which,c=2.
Hence the particular solution is:
x=
y^2
6
−
1
3
lny+ 2
Problem 8. (a) The variation of resistance,
R ohms, of an aluminium conductor with
temperatureθ◦C is given by
dR
dθ
=αR, where