Second order differential equations of the forma
d^2 y
dx^2 +b
dy
dx+cy=f(x)^491
(v) Substitutingvinto
(D^2 +D− 6 )v= 12 x−50sinxgives:
(D^2 +D− 6 )(ax+b+csinx+dcosx)
= 12 x−50sinx
D(ax+b+csinx+dcosx)
=a+ccosx−dsinx
D^2 (ax+b+csinx+dcosx)
=−csinx−dcosx
Hence (D^2 +D− 6 )(v)
=(−csinx−dcosx)+(a+ccosx
−dsinx)− 6 (ax+b+csinx+dcosx)
= 12 x−50sinx
Equating constant terms gives:
a− 6 b=0(1)
Equating coefficients ofxgives:− 6 a=12, from
which,a=−2.
Hence, from (1),b=−^13
Equating the coefficients of cosxgives:
−d+c− 6 d= 0
i.e. c− 7 d= 0
(2)
Equating the coefficients of sinxgives:
−c−d− 6 c=− 50
i.e. − 7 c−d=− 50
(3)
Solving equations (2) and (3) gives:c=7and
d=1.
Hence the P.I.,
υ=− 2 x−^13 +7sinx+cosx
(vi) The general solution,y=u+v,
i.e. y=Ae^2 x+Be−^3 x− 2 x
−^13 +7sinx+cosx
Problem 10. Solve the differential equation
d^2 y
dx^2
− 2
dy
dx
+ 2 y=3excos2x, given that when
x=0,y=2and
dy
dx
=3.
Using the procedure of Section 51.2:
(i)
d^2 y
dx^2
− 2
dy
dx
+ 2 y=3excos2x in D-operator
form is
(D^2 −2D+ 2 )y=3excos2x
(ii) The auxiliary equation ism^2 − 2 m+ 2 = 0
Using the quadratic formula,
m=
2 ±
√
[4− 4 ( 1 )( 2 )]
2
=
2 ±
√
− 4
2
=
2 ±j 2
2
i.e.m= 1 ±j 1.
(iii) Since the roots are complex, the C.F.,
u=ex(Acosx+Bsinx).
(iv) Since the right hand side of the given dif-
ferential equation is a product of an expo-
nential and a cosine function, let the P.I.,
v=ex(Csin2x+Dcos 2x)(see Table 51.1(f) —
again, constantsCandDare used sinceAandB
have already been used for the C.F.).
(v) Substitutingvinto(D^2 −2D+ 2 )v=3excos2x
gives:
(D^2 −2D+ 2 )[ex(Csin2x+Dcos 2x)]
=3excos2x
D(v)=ex( 2 Ccos2x− 2 Dsin 2x)
+ex(Csin2x+Dcos2x)
(≡ex{( 2 C+D)cos 2x
+(C− 2 D)sin2x})
D^2 (v)=ex(− 4 Csin2x− 4 Dcos2x)
+ex( 2 Ccos2x− 2 Dsin2x)
+ex( 2 Ccos2x− 2 Dsin2x)
+ex(Csin2x+Dcos 2x)
≡ex{(− 3 C− 4 D)sin2x
+( 4 C− 3 D)cos2x}
Hence (D^2 −2D+ 2 )v
=ex{(− 3 C− 4 D)sin2x
+( 4 C− 3 D)cos2x}
−2ex{( 2 C+D)cos2x
+(C− 2 D)sin2x}
+2ex(Csin2x+Dcos 2x)
=3excos2x