Second order differential equations of the forma
d^2 y
dx^2 +b
dy
dx+cy=^0479
(d) Whent=0,y=3 hence 3=( 0 +B)e^0 ,i.e.B=3.
Sincey=(At+B)e
4
3 t
then
dy
dt
=(At+B)
(
4
3
e
4
3 t
)
+Ae
4
3 t,bythe
product rule.
When t= 0 ,
dy
dt
= 3
thus 3=( 0 +B)
4
3
e^0 +Ae^0
i.e. 3=
4
3
B+Afrom which,A=−1, since
B=3.
Hence the particular solution is
y=(−t+3)e
4
3 tory=(3−t)e
4
3 t
Problem 3. Solve the differential equation
d^2 y
dx^2
+ 6
dy
dx
+ 13 y=0, given that whenx=0,y= 3
and
dy
dx
=7.
Using the procedure of Section 50.2:
(a)
d^2 y
dx^2
+ 6
dy
dx
+ 13 y=0 in D-operator form is
(D^2 +6D+ 13 )y=0, where D≡
d
dx
(b) Substitutingmfor D gives the auxiliary equation
m^2 + 6 m+ 13 =0.
Using the quadratic formula:
m=
− 6 ±
√
[( 6 )^2 − 4 ( 1 )( 13 )]
2 ( 1 )
=
− 6 ±
√
(− 16 )
2
i.e.m=
− 6 ±j 4
2
=− 3 ±j 2
(c) Since the roots are complex,the general solu-
tion is
y=e−^3 x(Acos 2x+Bsin2x)
(d) Whenx=0,y=3, hence
3 =e^0 (Acos 0+Bsin0),i.e.A=3.
Sincey=e−^3 x(Acos 2x+Bsin2x)
then
dy
dx
=e−^3 x(− 2 Asin 2x+ 2 Bcos2x)
−3e−^3 x(Acos 2x+Bsin2x),
by the product rule,
=e−^3 x[( 2 B− 3 A)cos 2x
−( 2 A+ 3 B)sin2x]
Whenx=0,
dy
dx
=7,
hence 7=e^0 [( 2 B− 3 A)cos 0−( 2 A+ 3 B)sin0]
i.e. 7= 2 B− 3 A, from which,B=8, sinceA=3.
Hence the particular solution is
y=e−^3 x(3cos2x+8sin2x)
Since, from Chapter 17, page 165,
acosωt+bsinωt=Rsin(ωt+α),where
R=
√
(a^2 +b^2 )andα=tan−^1
a
b
then
3cos2x+8sin2x
=
√
( 32 + 82 )sin( 2 x+tan−^138 )
=
√
73sin( 2 x+ 20. 56 ◦)
=
√
73sin( 2 x+ 0. 359 )
Thus the particular solution may also be
expressed as
y=
√
73 e−3xsin( 2 x+0.359)
Now try the following exercise
Exercise 187 Further problemson
differential equations of the form
a
d^2 y
dx^2
+b
dy
dx
+cy= 0
In Problems 1 to 3, determine the general solution
of the given differential equations.
- 6
d^2 y
dt^2
−
dy
dt
− 2 y= 0
[
y=Ae
2
3 t+Be−
1
2 t
]
- 4
d^2 θ
dt^2
+ 4
dθ
dt
+θ= 0
[
θ=(At+B)e−
(^12) t]