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) becomesU
dz
dx=9(8)
EliminatingUbetween (7) and (8)dz= 9 xexdxIntegratingz= 9∫
xexdx= 9 ex(x−1) (9)SubstitutingUandziny=Uz,y=9(x−1)
x1.65 d
(^2) y
dx^2
+
dy
dx− 2 y=2 cosh 2x (1)The complimentary solution is found fromd^2 y
dx^2+
dy
dx− 2 y= 0D^2 +D− 2 = 0
(D−1)(D+2)= 0
D= 1 ,− 2
Y=U=C 1 ex+C 2 e−^2 x (2)