Math 333 Logarithms Worksheet With Answers Page 12
ADVERTISEMENT
1
2
3
4
5
6
7
8
9
10
11
12
13Substituting our expressions for y, y , and y into the original differential equation, we
get
2
2
x
ln x = x
(2u
+ 2xu
+ (2 ln x + 3)u
+ (2x ln x + x)u
)
1
2
1
2
2
2
3x(2xu
+ (2x ln x + x)u
) + 4(x
u
+ (x
ln x)u
)
1
2
1
2
3
3
3
= 2x
u
+ (2x
ln x + x
)u
1
2
2
2
2
2
2
2
2
2
+(2x
6x
+ 4x
)u
+ (2x
ln x + 3x
6x
ln x
3x
+ 4x
ln x)u
1
2
3
3
3
= 2x
u
+ (2x
ln x + x
)u
.
1
2
The coefficients of u
and u
are both zero, as expected.
1
2
Next we recall condition (5) and solve the system
2
2
x
u
+ (x
ln x)u
= 0
1
2
3
3
3
2
2x
u
+ (2x
ln x + x
)u
= x
ln x.
1
2
Since we are assuming x > 0, we may divide through by suitable powers of x to get
the somewhat simpler system
u
+ (ln x)u
= 0
(6)
1
2
ln x
2u
+ (2 ln x + 1)u
=
.
(7)
1
2
x
From (6), we get u
=
(ln x)u
. Making this substitution in (7) yields
1
2
ln x
2 ln xu
+ 2 ln xu
+ u
=
2
2
2
x
ln x
so that u
=
. From this we get
2
x
u
=
(ln x)u
1
2
2
(ln x)
=
.
x
We integrate u
to get
1
3
(ln x)
u
=
+ c
1
1
3
and u
to get
2
2
(ln x)
u
=
+ c
.
2
2
2
ADVERTISEMENT
0 votes
Related Articles
Related forms
Related Categories
Parent category: Education