62
1 Mathematical Physics
dy
dx
+ Py = Q
(2)
dy
dx
+
y(x + 1)
x
= 9
( 3 )
Let y = U z
(4)
dy
dx
=
U dz
dx
+
zdU
dx
(5)
Substituting (4) and (5) in (3)
U dz
dx
+
dU
dx
+
U (x + 1)
x
z = 9
( 6 )
Now to determine U , we place the coefficients of z equal to zero. This gives
dU
dx
+
U (x + 1)
x
= 0
dU
U
= −
1 +
1
x
dx
Integrating, ln U = −x − ln x or
U = e
−x
/x
(7)
As the term in z drops off, Eq. (6) becomes
U
dz
dx
= 9
( 8 )
Eliminating U between (7) and (8)
dz = 9x e
x dx
Integrating z = 9
xe
x dx = 9e
x (x − 1)
(9)
Substituting U and z in y = U z,
y =
9(x − 1)
x
1.65 d
2 y
dx 2 +
dy
dx
− 2y = 2 cosh 2x
(1)
The complimentary solution is found from
d
2 y
dx 2 +
dy
dx
− 2y = 0
D
2
+ D − 2 = 0
(D − 1)(D + 2) = 0
D = 1, −2
Y = U = C 1 e
x
+ C 2 e
−2x
(2)
1 Mathematical Physics
dy
dx
+ Py = Q
(2)
dy
dx
+
y(x + 1)
x
= 9
( 3 )
Let y = U z
(4)
dy
dx
=
U dz
dx
+
zdU
dx
(5)
Substituting (4) and (5) in (3)
U dz
dx
+
dU
dx
+
U (x + 1)
x
z = 9
( 6 )
Now to determine U , we place the coefficients of z equal to zero. This gives
dU
dx
+
U (x + 1)
x
= 0
dU
U
= −
1 +
1
x
dx
Integrating, ln U = −x − ln x or
U = e
−x
/x
(7)
As the term in z drops off, Eq. (6) becomes
U
dz
dx
= 9
( 8 )
Eliminating U between (7) and (8)
dz = 9x e
x dx
Integrating z = 9
xe
x dx = 9e
x (x − 1)
(9)
Substituting U and z in y = U z,
y =
9(x − 1)
x
1.65 d
2 y
dx 2 +
dy
dx
− 2y = 2 cosh 2x
(1)
The complimentary solution is found from
d
2 y
dx 2 +
dy
dx
− 2y = 0
D
2
+ D − 2 = 0
(D − 1)(D + 2) = 0
D = 1, −2
Y = U = C 1 e
x
+ C 2 e
−2x
(2)