Math 333 Logarithms Worksheet With Answers Page 10
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Adding these two equations, we get 2u
= 1, from which u
=
. Substituting this
2
2
2
value for u
back into equation (2) gives
2
sin 2t
u
cos 2t +
= 0
1
2
which we solve to get
tan 2t
u
=
.
1
2
Integrating u
and u
, we get
1
2
ln(sec 2t)
u
=
+ c
1
1
4
t
u
=
+ c
.
2
2
2
Substituting these functions back into our original solution, we get the general solution
ln(sec 2t)
t sin 2t
y = c
cos 2t
cos 2t + c
sin 2t +
.
1
2
4
2
2
2
C2. (B & D, 3.7, problem 17) Verify that y
= x
and y
= x
ln x (with x > 0) are
1
2
solutions to the differential equation
2
x
y
3xy + 4y = 0.
2
2
Find a particular solution the differential equation x
y
3xy + 4y = x
ln x.
Solution: Given the differential equation
2
2
x
y
3xy + 4y = x
ln x,
2
2
we first check that the given functions y
= x
and y
= x
ln x are indeed solutions to
1
2
the corresponding homogeneous equation.
We have
2
y
= x
1
y
= 2x
1
y
= 2,
1
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