62 1 Mathematical Physics
dy
dx
+Py=Q (2)
dy
dx
+
y(x+1)
x
=9(3)
Lety=Uz (4)
dy
dx
=
Udz
dx
+
zdU
dx
(5)
Substituting (4) and (5) in (3)
Udz
dx
+
(
dU
dx
+
U(x+1)
x
)
z=9(6)
Now to determineU, we place the coefficients ofzequal to zero. This gives
dU
dx
+
U(x+1)
x
= 0
dU
U
=−
(
1 +
1
x
)
dx
Integrating, lnU=−x−lnxor
U=e−x/x (7)
As the term inzdrops off, Eq. (6) becomes
U
dz
dx
=9(8)
EliminatingUbetween (7) and (8)
dz= 9 xexdx
Integratingz= 9
∫
xexdx= 9 ex(x−1) (9)
SubstitutingUandziny=Uz,
y=
9(x−1)
x
1.65 d
(^2) y
dx^2
+
dy
dx
− 2 y=2 cosh 2x (1)
The complimentary solution is found from
d^2 y
dx^2
+
dy
dx
− 2 y= 0
D^2 +D− 2 = 0
(D−1)(D+2)= 0
D= 1 ,− 2
Y=U=C 1 ex+C 2 e−^2 x (2)