Math 333 Logarithms Worksheet With Answers Page 11
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1
2
3
4
5
6
7
8
9
10
11
12
13so that
2
2
2
x
y
3xy
+ 4y
= x
(2)
3x(2x) + 4x
1
1
1
2
2
2
= 2x
6x
+ 4x
= 0.
This shows that y
is a solution to the corresponding homogeneous equation. Similarly,
1
we have
2
y
= x
ln x
2
y
= 2x ln x + x
2
y
= 2 ln x + 3,
2
so that
2
2
2
x
y
3xy
+ 4y
= x
(2 ln x + 3)
3x(2x ln x + x) + 4(x
ln x)
2
2
2
2
2
2
2
2
= (2x
6x
+ 4x
) ln x + 3x
3x
= 0.
To solve the non-homogeneous differential equation, we use variation of parameters.
We let
2
2
y = u
x
+ u
x
ln x.
(4)
1
2
Then we have
2
2
y
= u
x
+ 2xu
+ u
x
ln x + (2x ln x + x)u
1
2
1
2
2
2
= 2xu
+ (2x ln x + x)u
+ x
u
+ (x
ln x)u
.
1
2
1
2
We impose the condition
2
2
x
u
+ (x
ln x)u
= 0
(5)
1
2
so that we have
y
= 2xu
+ (2x ln x + x)u
,
1
2
from which it follows that
y
= 2u
+ 2xu
+ (2 ln x + 3)u
+ (2x ln x + x)u
.
1
2
1
2
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