SECOND ORDER DIFFERENTIAL EQUATIONS (HOMOGENEOUS) 477I
Whent=0,dy
dt= 3thus 3=(0+B)4
3e^0 +Ae^0i.e. 3=4
3B+Afrom which,A=−1, since
B=3.
Hence the particular solution isy=(−t+3)e4
3 t ory=(3−t)e4
3 tProblem 3. Solve the differential equation
d^2 y
dx^2+ 6dy
dx+ 13 y=0, given that when x=0,y=3 anddy
dx=7.Using the procedure of Section 50.2:
(a)d^2 y
dx^2+ 6dy
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)
2i.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+Bsin 2x)(d) Whenx=0,y=3, hence
3 =e^0 (Acos 0+Bsin 0), i.e.A=3.
Sincey=e−^3 x(Acos 2x+Bsin 2x)thendy
dx=e−^3 x(− 2 Asin 2x+ 2 Bcos 2x)−3e−^3 x(Acos 2x+Bsin 2x),
by the product rule,
=e−^3 x[(2B− 3 A) cos 2x
−(2A+ 3 B) sin 2x]Whenx=0,dy
dx=7,hence 7=e^0 [(2B− 3 A) cos 0−(2A+ 3 B) sin 0]
i.e. 7= 2 B− 3 A, from which,B=8, sinceA=3.
Hence the particular solution isy=e−^3 x(3 cos 2x+8 sin 2x)
Since, from Chapter 18, page 178,
acosωt+bsinωt=Rsin (ωt+α), where
R=√
(a^2 +b^2 ) andα=tan−^1a
bthen3 cos 2x+8 sin 2x=√
(3^2 + 82 ) sin (2x+tan−^138 )=√
73 sin(2x+ 20. 56 ◦)=√
73 sin(2x+ 0 .359)Thus the particular solution may also be
expressed asy=√
73 e−3xsin( 2 x+0.359)Now try the following exercise.Exercise 188 Further problems on differen-
tial equations of the formad^2 y
dx^2+bdy
dx+cy= 0In Problems 1 to 3, determine the general solu-
tion of the given differential equations.- 6
d^2 y
dt^2−dy
dt− 2 y= 0
[
y=Ae2
3 t+Be−1
2 t]- 4
d^2 θ
dt^2+ 4dθ
dt+θ= 0[
θ=(At+B)e−1
2 t]3.d^2 y
dx^2+ 2dy
dx+ 5 y= 0[y=e−x(Acos 2x+Bsin 2x)]In Problems 4 to 9, find the particular solu-
tion of the given differential equations for the
stated boundary conditions.