1. KNF1023
Engineering
Mathematics II
First Order ODEs
Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 2008/2009
2. Learning Objectives
Demonstrate how to find integrating
factor for non-exact differential equation
Demonstrate the solution of
Homogeneous 1st order ODE in linear form
Demonstrate the solution of
inhomogeneous 1st order ODE in linear form
3. Integrating Factor
If a function u ( x, y ) has continuous partial
derivatives, its total or exact differential is
∂u ∂u
du = dx + dy
∂x ∂y
From this it follows that if u ( x, y ) =c=const,
then du = 0
A first-order differential equation of the form
M ( x, y )dx + N ( x, y )dy = 0 − −− → (1)
Is called exact if its left side is the total or exact
differential
4. Exact Differential Equation
∂u ∂u
du = dx + dy − −− → (2)
∂x ∂y
of some function u ( x, y ) . Then the differential
equation (1) can be written
du = 0
By integration we immediately obtain the
general solution of (1) in the form
u ( x, y ) = C − −− → (3)
5. Exact Differential Equation
Comparing (1) and (2), we see that (1) is
exact if there is some function u ( x, y ) such that
∂u ∂u
a) =M b) =N (4)
∂x ∂y
∂M ∂u 2
∂N ∂ 2u
= =
∂y ∂y∂x ∂x ∂x∂y
By the assumption of continuity the two second
derivatives are equal. Thus
∂M ∂N
= − −− → (5)
∂y ∂x
6. Exact Differential Equation
This condition is not only necessary but also
sufficient for Mdx + Ndy to be an exact
differential.
If (1) is exact, the function u ( x, y ) can be
found by guessing or in the following
systematic way. From 4(a) we have by
integration with respect to x
u = ∫ Mdx + k ( y ) − −− → ( 6 )
7. 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∂u / ∂y from
(6), use (4b) to get ∂k / ∂y and integrate ∂k / ∂y 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 = ∫ Ndy + l ( x) − −− → (*6 )
8. Exact Differential Equation
To determine l(x) we derive ∂u / ∂x from (6*),
use (4a) to get ∂l / ∂x and integrate it to get l .
9. Example 1: An exact equation
3 2 2 3
Solve ( x + 3 xy )dx + (3x y + y )dy = 0 (7)
Solution:
1st step: Test for exactness.
Our equation is of the form (1) with
M = x 3 + 3 xy 2
N = 3x 2 y + y 3
∂M
= 6 xy
∂y
∂N
= 6 xy
∂x exact solution
10. Continue…
2nd Step: Implicit solution.
From (6) we obtain
u = ∫ Mdx + k ( y ) = ∫ ( x 3 + 3 xy 2 )dx + k ( y )
1 4 3 2 2
= x + x y + k ( y) (8)
4 2
To find k(y), we differentiate this formula
with respect to y and use formula (4b),
obtaining
∂u 2 dk
= 3x y + = N = 3x 2 y + y3
∂y dy
11. Continue…
dk 3 y4 *
Hence dy = y , so that k = 4 + c . Inserting
this into (8) we get the answer
1 4
u ( x, y ) = ( x + 6 x 2 y 2 + y 4 ) = c (9)
4
3rd step: Checking.
For checking, we can differentiate u ( x, y ) = c
implicitly and see whether this leads to dy = − M
dx N
or Mdx+ Ndy= 0 , the given equation.
12. Continue…
In the present case, differentiating (9)
implicitly with respect to x, we obtain
1
(4 x3 + 12 xy 2 + 12 x 2 yy '+ 4 y 3 y ') = 0
4
Collecting terms, we see that this equals M +
Ny’=0 with M and N as in (7); thus Mdx + Ndy = 0
.This completes the check.
13. Example 2
( )
Solve cos( x + y)dx + 3y 2 + 2 y + cos(x + y) dy = 0
Solution:
M = cos( x + y )
N = 3 y 2 + 2 y + cos( x + y )
Thus
∂M
= − sin( x + y )
∂y
exact
∂N
= − sin( x + y )
∂x
14. Continue…
Step 2: Implicit general solution.
u = ∫ Mdx + k ( y ) = ∫ cos( x + y )dx + k ( y ) = sin( x + y ) + k ( y )
To find k(y), we differentiate this formula
with respect to y and obtain
∂u dk
= cos( x + y ) + = N = 3 y 2 + 2 y + cos( x + y )
∂y dy
2
Hence dk / dy = 3 y + 2 y
By integration, k = y 3 + y 2 + c * . Inserting this result
u ( x, y ) = sin( x + y ) + y 3 + y 2 = c
15. 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:
∂u ∂u
du = dx + dy = cos( x + y )dx + (cos( x + y ) + 3 y 2 + 2 y )dy = 0
∂x ∂y
This completes the check.
17. Identify whether the ODE is
homogeneous or not???
dy
Differential equation ( x , y ) is call
= f
dx
Homogeneous equation if f ( λ x, λ y ) = f ( x, y )
for every real value of λ
Example 3: Identify whether dy = y − x is
dx y + x
homogeneous or not.
18. Continue…
y−x
f ( x, y ) =
y+x
λ y − λx
f ( λ x, λ y ) =
λ y + λx
y−x
=
y+x
= f ( x, y )
This equation is homogeneous
19. Example 4
Check whether the equation given is
homogeneous or not?
dy
= x− y
dx
f ( x, y ) = x − y
f ( λ x, λ y ) = λ x − λ y
= λ ( x − y)
= λ f ( x, y )
This equation is not homogeneous.
20. Example 5
dy x + 3 y
Solve =
dx 2x
Solution:
The above equation is a homogeneous
y = ux
dy du
=u+x
dx dx
du x + 3(ux) x(1 + 3u ) 1 + 3u
u+x = = =
dx 2x 2x 2
21. Continue…
dv 1 + 3u
u+x =
dx 2
du 1 + 3u 1 + 3u − 2u 1 + u
x = −u = =
dx 2 2 2
du 1 + u 1
= = (1 + u )( )
dx 2x 2x
du dx
=
u +1 2x
du 1 dx
∫ u +1 = 2 ∫ x
22. Continue…
1
ln ( u + 1) = ln x + ln c
2
1
ln ( u + 1) = ln x + ln c
2
1
ln ( u + 1) = ln( x c)
2
1
u + 1 = x 2c
24. Example 6
dy xy
Solve = 2
dx x + y 2
subject y(0)=2
Let y = xu , rewrite this equation become
du x ( xu )
u+x = 2
dx x + ( xu )2
u
=
1+ u2
du u
x = 2
−u
dx 1 + u
u3
=−
1+ u2
25. Continue…
1+ u2 dx
3 du = −
u x
1 1 dx
3 + du = −
u u x
1 1 dx
∫ u 3 + u du = − ∫ x
1
− 2 + ln u = − ln x + c
2u
26. Continue…
1
ln u + ln x = c + 2
2u
1
ln xu = c + 2
2u
y
Substitute u=
x
x2
ln y = c +
2y2
x2
2 y2
y = Ae
28. Summary on solving Homogeneous
equations
1. Identify whether the equation is
homogeneous or not.
dy du
2. Use substitution of y = xu and dx = u + x dx
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.
y
5. Substitute back u = .
x
6. If the equation has subject to any
value, substitute it to get the
constant value A.
29. Prepared By
Annie ak Joseph
Prepared By
Annie ak Joseph Session 2008/2009
