Solution of first order differential equations by separation of variables 447
5.1
ex+ 2 =x− 3dy
dx,giveny=1whenx=0.
[
y=1
6(
x^2 − 4 x+2
ex+ 4)]- The gradient of a curve is given by:
dy
dx
+x^2
2= 3 xFind the equation of the curve if it passes
through the point(
1 ,^13)
.
[
y=3
2x^2 −x^3
6− 1]- Theacceleration,a,ofabodyisequal toitsrate
of change of velocity,
dv
dt. Findan equationfor
vin terms oft, given that whent=0, velocity
v=u.[v=u+at]
8. An object is thrown vertically upwards with
an initial velocity,u, of 20m/s. The motion
of the object follows the differential equation
ds
dt
=u−gt,wheresis theheight of the object
in metres at timetseconds andg= 9 .8m/s^2.
Determine the height of the object after 3
seconds ifs=0whent=0. [15.9m]46.4 The solution of equations of the
form
dy
dx
=f(y)
Adifferentialequationoftheform
dy
dx=f(y)isinitiallyrearranged to give dx=dy
f(y)and then the solution isobtained by direct integration,
i.e.
∫
dx=∫
dy
f(y)Problem 6. Find the general solution of
dy
dx= 3 + 2 y.Rearranging
dy
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 solutionof a differential
equation in a different form. For example, ifc=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 solution of
(y^2 − 1 )
dy
dx
= 3 ygiven thaty=1whenx= 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.
When y=1, x= 216 , thus 2^16 =^16 −^13 ln1+c, from
which,c=2.
Hence the particular solution is:
x=
y^2
6
−
1
3
lny+ 2