B
˙
ILKENT UNIVERSITY
Department of Mathematics
MATH 225,
LINEAR ALGEBRA and DIFFERENTIAL EQUATIONS,
Solution of Homework set
1
# 3
U. Mu˘gan
June 20, 2008
Homework problems from the 2
nd
Edition, SECTION 1.5
5
(5)
)
2
Linear first order D.E. which can be solved by introducing the integrating factor
ρ
(
x
)
ρ
= exp
•Z
2
x
dx
‚
=
e
2 ln
x
=
x
2
.
The D.E. can be written as
d
dx
(
yx
2
) = 3
x
2
Integration yields
y
(
x
) =
x
+
c
x
2
,
c
= constant
I.C. implies that
c
= 4, so the solution of the I.V.P. is
y
(
x
) =
x
+
4
x
2
.
10
(10)
)
Integrating factor is
ρ
(
x
) = exp
•Z

3
2
x
dx
‚
=
e
(

3 ln
x
)
/
2
=
x

3
/
2
.
Then the D.E. can be written as
d
dx
(
yx

3
/
2
) =
9
2
x
1
/
2
Therefore the general solution of the D.E. is
y
(
x
) = 3
x
3
+
cx
3
/
2
,
c
= constant
.
14
(14)
)
Integrating factor is
ρ
(
x
) = exp
•Z

3
x
dx
‚
=
e

3 ln
x
=
x

3
.
Then the D.E. can be written as
d
dx
(
yx

3
) =
1
x
1
I made every effort to avoid the calculation errors and/or typos while I prepared the solution set.
You are
responsible to check all the solutions and correct the errors if there is any.
If you find any errors and/or
misprints, please notify me.
2
The number in the parenthesis denotes the problem number in the International Edition
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Therefore the general solution of the D.E. is
y
(
x
) =
x
3
ln
x
+
cx
3
,
c
= constant
.
I.C. implies that
c
= 10, so the solution of the I.V.P. is
y
(
x
) =
x
3
ln
x
+ 10
x
3
.
20
(20)
)
Integrating factor is
ρ
(
x
) = exp
•Z
(

1

x
)
dx
‚
=
e

x

x
2
2
.
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 Trigraph, general solution, I.C., BILKENT UNIVERSITY Department of Mathematics MATH

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