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Solution Tutorial 2:
1(a):2)1()1( +=++ xy
dx
dyx
)1(
)1(
)1(
2
+
+=
++
x
x
x
y
dx
dy
)1()1(
+=+
+ xx
y
dx
dy
Linear differential equation 1)(,)1(
1)( +=
+= xxq
xxp
To find integrating factor:
)1ln(1
)( +=+
= xxdx
dxxp
Then, integrating factor is )1()( )1ln( +== + xex x
( ) 2)1()1)(1()()( +=++== xxxxxyxdx
d (1)
To find the general solution, integrate (1)
( ) += dxxdxyxdxd 2)1()(
Ax
++
=3
)1( 3
Ax
yx ++
=+3
)1()1(
3
)1(3
)1( 2
++
+=
x
Axy
(b):xxy
dx
dycoscot =+
xdxxdxxp sinlncot)( ==xx sin)( =
= xdxxxy cossinsin
A
xxdx +== 4
2cos2sin
2
1
The general solution is
Ax
xy +=4
2cossin
Bxxy =+ 2cossin4
(c):
2 2 5
2 2 5
6 2
( ) 6 and ( ) 2
dxt x t t
dt
p t t q t t t
+ = +
= = +
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( ) ( )
( )
( )
23
3 3
3 3 3
3
3
3 3
6 2
2 2 2 5
2 2 2 2 5
2
2 2
2 5 2 2 3
2
2 ----------(i)
* (**)
(*): (integrate by substitution)6
(**):2 2
t dt t
t t
t t t
t
t
t t
e e
de x e t t
dt
e x e t dt e t dt
ee t dt
e t dt e t t dt
= =
= +
= +
=
=
3
3
3
3
3 3
3 3 3
3
3 2 2
2
2
2
3 2 2
2 2
3
2 2 2
2 3
let '
' 36
12
6 2
3 6
From (i):
6 3 6
( )
t
t
t
t
t t
t t t
t
u t v e t
eu t v
et e t dt
e et
e e ee x t C
x t
= =
= =
=
=
= + +
33
2
3
ttCe
= +
(f)
2 3 4
4
3 4 2 3
3 4
4
32
3
cos ...............(1)
Divide eqn(1) with
cos ........(2)
Let ; 3
1or3
Substitute v and into eqn(2)
cos3
3 3cos
dy x y x y x
dx
y
dy x y x y x
dx
dv dyv y y
dx dx
dy dvydx dx
dy
dx
x dv x v x
dx
dvv x
dx x x
=
+ =
= =
=
+ =
+ =
3ln3)( x
x
dxdxxp ==
3)( xx =
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The general solution in v:3 cos 3sinvx xdx x A= = +
3
33sin
xx A
y= +
2.
(a)
( )
20
20 20
110 12
2
20 24
( ) 20, ( ) 24
20 20.
24
t
t t
dii
dt
dii
dt
p t q t
dt ti f e
de i e
dt
+ =
+ =
= =
=
=
=
20 20
20
24
6
5
t t
t
e i e dt
e C
=
= +
206( )5
ti t Ce = +
( )20
Impose the condition, 0, 0
6
5
6( ) 1
5
t
i t
C
i t e
= =
=
=
(b)
( )
5
5 5
2 10 20cos 5
5 10 cos 5
( ) 5, ( ) 10 cos 5
5 5
.
10 cos 5
t
t t
dii t
dt
di i tdt
p t q t t
dt t
i f e
de i e t
dt
+ =
+ =
= =
=
=
=
( )
5 5
5
10 cos5
cos5 sin 5
t t
t
e i e tdt
e t t C
=
= + +
5( ) cos5 sin 5 ti t t t Ce = + +
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5
Impose the condition, 0, 0
1
( ) cos5 sin 5 t
i t
C
i t t t e
= =
=
= +
3.(a) 0)()( =+++ dyxeedxyee
yxxy
xyxy eey
MyeeyxM +=
+=),(
yxyxee
x
NxeeyxN +=
+=),(
x
N
y
M
=
0)()( =+++ dyxeedxyee yxxy is an exact difftn. equation.
xy yee
x
u+=
(1)yx xee
y
u+=
(2)
To find the general solution ),( yxu , we integrate (1) with respect to x
+=
= dxyeedxx
uyxu xy )(),(
)(yyexexy ++= (3)
where )( y is some function in terms ofy only.
Differentiate (3) with respect to y,
)(' yexey
u xy++=
(4)
Compare (4) with (2), we have0)(' =y
ThereforeAy =)( , A a constant
The general solution is
)(),( yyexeyxu xy ++=kAyexe xy =++=
that is
kABByexeyxu xy ==++= ,0),(
(b) 0)(2 =+ dyyxydx
1),( =
=
y
MyyxM
1)(),( 2 =+=xNyxyxN
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x
N
y
M
0)( 2 =+ dyyxydx is not an exact diffrtn. equation.
(c)
( ) ( )
( )
2 2
2 2
2 2
2 2
2 2
2 0 , 1 2
!
2 2
2 ( )2
( )2
Rejecting the term in (ii) that already
dx
xt x t xdt
Exact
uM xt xt dx
x
x tx i
u N x t x tdt
t
x t
ii
= =
= =
=
= =
=
2 2
exist in (i) and then equate to the constant.
u(t,x) 22
x tx k = =
2 2
2 2
2 (where C any constant)2
Impose the condition:
42(2) 22
2 22
or
x tx C
C C
x tx
=
= =
=