week 3 [compatibility mode]
DESCRIPTION
KNF1023TRANSCRIPT
Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2008/2009
KNF1023Engineering
Mathematics II
First Order ODEs
Learning Objectives
Demonstrate the solution of
inhomogeneous 1st order ODE in linear form
Demonstrate the solution of
Homogeneous 1st order ODE in linear form
Demonstrate how to find integrating
factor for non-exact differential equation
Integrating Factor
� If a function has continuous partial derivatives, its total or exact differential is
� From this it follows that if =c=const, then
� A first-order differential equation of the form
Is called exact if its left side is the total or exact differential
dyy
udx
x
udu
∂
∂+
∂
∂=
),( yxu
),( yxu0=du
( , ) ( , ) 0 (1)M x y dx N x y dy+ = − −− →
Exact Differential Equation
of some function . Then the differential equation (1) can be written
By integration we immediately obtain the general solution of (1) in the form
(2)u u
du dx dyx y
∂ ∂= + − −− →
∂ ∂
),( yxu
0=du
( , ) (3)u x y C= − −− →
Exact Differential Equation
� Comparing (1) and (2), we see that (1) is exact if there is some function such that
By the assumption of continuity the two second
derivatives are equal. Thus
)u
a Mx
∂=
∂
),( yxu
) (4)u
b Ny
∂=
∂
2M u
y y x
∂ ∂=
∂ ∂ ∂
2N u
x x y
∂ ∂=
∂ ∂ ∂
(5)M N
y x
∂ ∂= − −− →
∂ ∂
Exact Differential Equation
� This condition is not only necessary but also sufficient for to be an exact differential.
� If (1) is exact, the function can be found by guessing or in the following systematic way. From 4(a) we have by integration with respect to x
NdyMdx +
),( yxu
( )( ) 6u Mdx k y= + − −− →∫
Exact Differential Equation
� in this integration, y is to be regarded as a constant, and k(y) plays the role of a “constant” of integration. To determine k(y), we derive from (6), use (4b) to get and integrate to get k.
� Formula (6) was obtained from (4a). Instead of 4(a) we may equally well use (4b). Then instead of (6) we first have
/u y∂ ∂/k y∂ ∂ /k y∂ ∂
( )( ) *6u Ndy l x= + − −− →∫
Exact Differential Equation
� To determine l(x) we derive from (6*), use (4a) to get and integrate it to get .
/u x∂ ∂
/l x∂ ∂ l
Example 1: An exact equation
Solve
Solution:
1st step: Test for exactness.
Our equation is of the form (1) with
exact solution
3 2 2 3( 3 ) (3 ) 0 (7)x xy dx x y y dy+ + + =
3 2
2 3
3
3
6
6
M x xy
N x y y
Mxy
y
Nxy
x
= +
= +
∂=
∂
∂=
∂
Continue…
2nd Step: Implicit solution.
From (6) we obtain
To find k(y), we differentiate this formula
with respect to y and use formula (4b),
obtaining
( )3 2
4 2 2
( ) 3 ( )
1 3( ) (8)
4 2
u Mdx k y x xy dx k y
x x y k y
= + = + +
= + +
∫ ∫
2 2 33 3
u dkx y N x y y
y dy
∂= + = = +
∂
Continue…
� Hence , so that . Inserting this into (8) we get the answer
3rd step: Checking.
For checking, we can differentiate
implicitly and see whether this leads to
or , the given equation.
3ydy
dk= *
4
4c
yk +
=
4 2 2 41( , ) ( 6 ) (9)
4u x y x x y y c= + + =
cyxu =),(
NM
dxdy
−=
0=+ NdyMdx
Continue…
� In the present case, differentiating (9) implicitly with respect to x, we obtain
� Collecting terms, we see that this equals M + Ny’=0 with M and N as in (7); thus .This completes the check.
3 2 2 31(4 12 12 ' 4 ') 0
4x xy x yy y y+ + + =
0=+ NdyMdx
Example 2
� Solve
Solution:
Thus
exact
)cos(23
)cos(
2 yxyyN
yxM
+++=
+=
( ) ( )( ) 0cos23cos2 =+++++ dyyxyydxyx
sin( )
sin( )
Mx y
y
Nx y
x
∂= − +
∂
∂= − +
∂
Continue…
Step 2: Implicit general solution.
To find k(y), we differentiate this formula
with respect to y and obtain
Hence
By integration, . Inserting this result
( ) cos( ) ( ) sin( ) ( )u Mdx k y x y dx k y x y k y= + = + + = + +∫ ∫
2cos( ) 3 2 cos( )
u dkx y N y y x y
y dy
∂= + + = = + + +
∂
2/ 3 2dk dy y y= +
*23cyyk ++=
3 2( , ) sin( )u x y x y y y c= + + + =
Continue…
Step 3: Checking an implicit solution.
We can check by differentiating the implicit solution
u(x,y) = c implicitly and see whether this leads to
the given ODE:
This completes the check.
2cos( ) (cos( ) 3 2 ) 0
u udu dx dy x y dx x y y y dy
x y
∂ ∂= + = + + + + + =
∂ ∂
Homogeneous Equation
Identify whether the ODE is homogeneous or not???
�Differential equation is call Homogeneous equation if
for every real value of
Example 3: Identify whether is
homogeneous or not.
( ),dy
f x ydx
=
( ) ( ), ,f x y f x yλ λ =
λ
dy y x
dx y x
−=
+
Continue…
This equation is homogeneous
( )
( )
,
,
y xf x y
y x
y xf x y
y x
λ λλ λ
λ λ
−=
+
−=
+
y x
y x
−=
+
( ),f x y=
Example 4
� Check whether the equation given is homogeneous or not?
This equation is not homogeneous.
dyx y
dx= −
( )
( )
,
,
f x y x y
f x y x yλ λ λ λ
= −
= −
( )
( ),
x y
f x y
λ
λ
= −
=
Example 5
Solve
Solution:
The above equation is a homogeneous
x
yx
dx
dy
2
3+=
dy duu x
dx dx= +
y ux=
3( ) (1 3 ) 1 3
2 2 2
du x ux x u uu x
dx x x
+ + ++ = = =
Continue…
1 3
2
dv uu x
dx
++ =
1 3 1 3 2 1
2 2 2
du u u u ux u
dx
+ + − += − = =
1 1(1 )( )
2 2
du uu
dx x x
+= = +
1 2
du dx
u x=
+
1
1 2
du dx
u x=
+∫ ∫
Continue…
( )1
ln 1 ln ln2
u x c+ = +
( )1
2ln 1 ln lnu x c+ = +
( )1
2ln 1 ln( )u x c+ =
121u x c+ =
Continue…
cxxy 23
=+
cxx
y2
1
1 =+
Example 6
Solve subject y(0)=2
Let , rewrite this equation become
22 yx
xy
dx
dy
+=
y xu=
( )
( )22
21
x xuduu x
dx x xu
u
u
+ =+
=+
21
du ux u
dx u= −
+
3
21
u
u= −
+
Continue…
2
3
1 u dxdu
u x
+= −
3
1 1 dxdu
u u x
+ = −
3
1 1 dxdu
u u x
+ = −
∫ ∫
2
1ln ln
2u x c
u− + = − +
Continue…
Substitute
2
1ln ln
2u x c
u+ = +
2
1ln
2xu c
u= +
yu
x=
2
2
2ln
y
xcy +=
2
2
2 y
x
Aey =
Continue…
2)0( =y
2
2
2
0
2
2
y
x
ey
Ae
=
=
Summary on solving Homogeneous equations
1. Identify whether the equation is homogeneous or not.
2. Use substitution of and
in the original equation.
3. Separate the variable x and u in 2.
4. Integral both sides with respect to the related variables and put only one constant, say A.
5. Substitute back .
6. If the equation has subject to any value, substitute it to get the constant value A.
y xu= dy duu x
dx dx= +
yu
x=
Prepared By
Annie ak Joseph
Prepared ByAnnie ak Joseph Session 2008/2009