02 first order differential equations
DESCRIPTION
first order differential equationTRANSCRIPT
FIRST ORDER DIFFERENTIAL EQUATIONS
In this lecture we discuss various methods of solving first order differential equations.
These include:
• Variables separable
• Homogeneous equations
• Exact equations
• Equations that can be made exact by multiplying by an integrating factor
( , )y f x y
Linear Non-linear
Integrating Factor
Separable Homogeneous Exact
IntegratingFactor
Transform to ExactTransform to separable
The first-order differential equation
,dy
f x ydx
is called separable provided that f(x,y) can be written as the product of a function of x and a function of y.
(1)
Variables Separable
Suppose we can write the above equation as
)()( yhxgdx
dy
We then say we have “ separated ” the variables. By taking h(y) to the LHS, the equation becomes
1( )
( )dy g x dx
h y
Integrating, we get the solution as
1( )
( )dy g x dx c
h y
where c is an arbitrary constant.
Example 1. Consider the DE ydx
dy
Separating the variables, we get
dxdyy
1
Integrating we get the solution as
kxy ||ln
or ccey x , an arbitrary constant.
Example 2. Consider the DE3
23 ln 0x
x y dx dyy
Separating the variables, we get
dxx
dyyy
3
ln
1
Integrating we get the solution askxy ln3|ln|ln
or cx
cy ,ln
3 an arbitrary constant.
Homogeneous equations
Definition A function f(x, y) is said to be homogeneous of degree n in x, y if
),(),( yxfttytxf n for all t, x, y
Examples 22 2),( yxyxyxf is homogeneous of degree
)sin(),(x
xy
x
yyxf
is homogeneous of degree
2.
0.
A first order DE 0),(),( dyyxNdxyxM
is called homogeneous if ),(,),( yxNyxM
are homogeneous functions of x and y of the same degree.
This DE can be written in the form
),( yxfdx
dy
where ),(/),(),( yxNyxMyxf is
clearly homogeneous of degree 0.
The substitution y = z x converts the given equation into “variables separable” form and hence can be solved. (Note that z is also a (new) variable,) We illustrate by means of examples.
Example 3. Solve the DE
02)3( 22 dxxydyyx
Let y = z x. Hence we get
That is )3(
222 yx
xy
dx
dy
)3(
2
)3(
22222
2
z
z
xzx
xz
dx
dzxz
)3()3(
22
3
2 z
zzz
z
z
dx
dzx
or
Separating the variables, we get
x
dxdz
zz
z
3
2 )3(
Integrating we get
x
dxdz
zz
z3
2 )3(
We express the LHS integral by partial fractions. We get
cx
dxdz
zzzln
1
1
1
13
or ccxzzz ,)1)(1(3 an arbitrary constant.
Noting z = y/x, the solution is:
,))(( 3ycxyxy c an arbitrary constant
or ,322 ycyx c an arbitrary constant
Working Rule to solve a HDE:
1. Put the given equation in the form
.Let .3 zxy
)1(0),(),( dyyxNdxyxM
2. Check M and N are Homogeneous function of the same degree.
5. Put this value of dy/dx into (1) and solve the equation for z by separating the variables.
6. Replace z by y/x and simplify.
dx
dzxz
dx
dy
4. Differentiate y = z x to get
Example 4. Solve the DE
)sin(x
xy
x
yy
Let y = z x. Hence we get
zzdx
dzxz sin or z
dx
dzx sin
Separating the variables, we getx
dxdz
z
sin
1
Integrating we get
where
cosec z – cot z = c x
and c an arbitrary constant.y
zx
We shall now see that some equations can be brought to homogeneous form by appropriate substitution.
Non-homogeneous equations
Example 5 Solve the DE
0)23()12( dyyxdxyx
That is )23(
)12(
yx
yx
dx
dy
We shall now put x = u+h, y = v+k
where h, k are constants ( to be chosen).
Hence the given DE becomes
)233(
)122(
khvu
khvu
du
dv
We now choose h, k such that
023
012
kh
kh Hence
5
3,
5
1 kh
Hence the DE becomes
)3(
)2(
vu
vu
du
dv
which is homogeneous in u and v.
Let v = z u. Hence we get
)31(
)2(
z
z
du
dzuz
)31(
)322(
)31(
)2( 2
z
zzz
z
z
du
dzu
or
Separating the variables, we get
u
dudz
zz
z
2322
)31(
Integrating we get czzu )322( 22
cvuvu 22 322i.e.
or 2 21 3 32( 1/ 5) 2( )( ) 3( )
5 5 5x x y y c
Example 6 Solve the DE
0)223()346( dyyxdxyx
)223(
)346(
yx
yx
dx
dyThat is
Now the previous method does not work as the lines
0223
0346
yx
yx
are parallel. We now put u = 3x + 2y.
The given DE becomes
)2(
)32()3(
2
1
u
u
dx
du
or2)2(
)64(3
u
u
u
u
dx
du
Separating the variables, we get
dxduu
u
2
Integrating, we get
dxduu
u
2
cxuu 2ln
i.e.
cxyxyx 2)23(ln23
EXACT DIFFERENTIAL EQUATIONS
A first order DE 0),(),( dyyxNdxyxM
is called an exact DE if there exits a function f(x, y) such that
( , ) ( , )df M x y dx N x y dy
Here df is the ‘total differential’ of f(x, y) and equals f f
dx dyx y
Hence the given DE becomes df = 0
Integrating, we get the solution as
f(x, y) = c, c an arbitrary constant
Thus the solution curves of the given DE are the ‘level curves’ of the function f(x, y) .
Example 8 The DE 0y dx x dy is exact as it is d (xy) = 0
Hence the solution is: x y = c
Example 7 The DE 0x dx y dy
is exact as it is d (x2+ y2) = 0
Hence the solution is: x2+ y2 = c
Example 9 The DE 2
1sin( ) sin( ) 0
x x xdx dy
y y y y
is exact as it is (cos( ) ) 0x
dy
Hence the solution is: cos( )x
cy
Test for exactnessSuppose 0),(),( dyyxNdxyxM
is exact. Hence there exists a function f(x, y)
such that
( , ) ( , )f f
M x y dx N x y dy df dx dyx y
Hence ,
f fM N
x y
Assuming all the 2nd order mixed derivatives of f(x, y) are continuous, we get
2 2M f f N
y y x x y x
Thus a necessary condition for exactness is
M N
y x
We saw a necessary condition for exactness is
M N
y x
We now show that the above condition is also sufficient for M dx + N dy = 0 to be exact. That is, if
,f f
M Nx y
then there exists a function f(x, y) such that
M N
y x
Integrating f
Mx
partially w.r.t. x, we get
( , ) ( )x
f x y M dx g y where g(y) is a function of y alone
We know that for this f(x, y), f
Ny
……. (*)
Differentiating (*) partially w.r.t. y, we get
( )x
fM dx g y
y y
= N gives
or ( )x
g y N M dxy
… (**)
We now show that the R.H.S. of (**) is independent of x and thus g(y) (and so f(x, y)) can be found by integrating (**) w.r.t. y.
x
N M dxx y
x
NM dx
x x y
x
NM dx
x y x
NM
x y
= 0
Q.E.D.
Note (1) The solution of the exact DE d f = 0 is f(x, y) = c.
Note (2) When the given DE is exact, the solution f(x, y) = c is found as we did in the previous theorem. That is, we integrate M partially w.r.t. x to get ( , ) ( )
x
f x y M dx g y
The following examples will help you in understanding this.
We now differentiate this partially w.r.t. y and equating to N, find g (y) and hence g(y).
Example 8 Test whether the following DE is exact. If exact, solve it. 2
( ) 0x dy y dxy
Here 2, ( )M y N x
y
1M N
y x
Hence exact.
Now ( , ) ( )x x
f x y M dx y dx xy g y
Differentiating partially w.r.t. y, we get2
( )f
x g y N xy y
Hence 2( )g y
y
Integrating, we get ( ) 2ln | |g y y
(Note that we have NOT put the arb constant )
Hence ( , ) ( ) 2ln | |f x y xy g y xy y Thus the solution of the given d.e. is
( , )f x y c 2ln | | ,xy y c or c an arb const.
Example 9 Test whether the following DE is exact. If exact, solve it.
4 2 3(2 sin ) (4 cos ) 0x y y dx x y x y dy
Here 4 2 32 sin , 4 cosM xy y N x y x y
38 cosM N
xy yy x
Hence exact.
Now 4( , ) (2 sin )x x
f x y M dx xy y dx 2 4 sin ( )x y x y g y
Differentiating partially w.r.t. y, we get2 34 cos ( )
fx y x y g y
y
Hence ( ) 0g y
Integrating, we get ( ) 0g y (Note that we have NOT put the arb constant )Hence
2 4 2 4( , ) sin ( ) sinf x y x y x y g y x y x y Thus the solution of the given d.e. is
( , )f x y c 2 4 sin ,x y x y c or c an arb const.
2 34 cosN x y x y
In the above problems, we found f(x, y) by integrating M partially w.r.t. x and then
.f
to Ny
We can reverse the roles of x and y. That is we can find f(x, y) by integrating N partially
.f
to Mx
The following problem illustrates this.
w.r.t. y and then equate
equated
Example 10 Test whether the following DE is exact. If exact, solve it.
2 2(1 sin 2 ) 2 cos 0y x dx y x dy
Here M
2 sin 2 ;M
y xy
Hence exact.Now 2( , ) 2 cos
y y
f x y N dy y x dy 2 2cos ( )y x g x
N 21 sin 2 ,y x 22 cosy x
4 cos sin 2 sin 2N
y x x y xx
Differentiating partially w.r.t. x, we get
22 cos sin ( )f
y x x g xx
gives ( ) 1g x
Integrating, we get ( )g x x(Note that we have NOT put the arb constant )Hence
2 2 2 2( , ) cos ( ) cosf x y y x g x y x x Thus the solution of the given d.e. is
( , )f x y c 2 2cos ,x y x c or c an arb const.
21 sin 2M y x
2 sin 2 ( )y x g x
The DE 2( 1) ( ) 0x y dx x xy dy
is NOT exact but becomes exact when
multiplied by 1
x 1( ) ( ) 0y dx x y dy
x
i.e. 21ln 0
2d xy x y
We say1
x
as it becomes
is an Integrating Factor of the given DE
Integrating Factors
Definition If on multiplying by (x, y), the DE
0M dx N dy
becomes an exact DE, we say that (x, y) is an Integrating Factor of the above DE
2 2
1 1 1, ,
xy x yare all integrating factors of
the non-exact DE 0y dx x dy
We give some methods of finding integrating factors of an non-exact DE
Problem Under what conditions will the DE
0M dx N dy have an integrating factor that is a function of x alone ?
Solution. Suppose = h(x) is an I.F.
Multiplying by h(x) the above d.e. becomes
( ) ( ) 0 .......(*)h x M dx h x N dy Since (*) is an exact DE, we have
( ( ) ) ( ( ) )h x M h x Ny x
i.e. ( ) ( ) ( ) ( )M N
h x M h x h x N h xy y x x
( ) 0 ( ) ( )M N
h x M h x N h xy x
or
( )( ) ( )M N
h x N h xy x
( )
( )
M Nh xy x
N h x
or
or
( )
M Ny x
g xN
Hence if
is a function of x alone, then
( )g x dxe
is an integrating factor of the given DE
0M dx N dy
Rule 2: Consider the DE 0M dx N dy
If ( )
M Ny x
h yM
, a function of y alone,
then( )h y dy
e is an integrating factor of the given DE
Problem Under what conditions will the DE
0M dx N dy have an integrating factor that is a function of the product z = x y ?
Solution. Suppose = h(z) is an I.F.
Multiplying by h(z) the above d.e. becomes
( ) ( ) 0 .......(*)h z M dx h z N dy Since (*) is an exact DE, we have
( ( ) ) ( ( ) )h z M h z Ny x
i.e. ( ) ( ) ( ) ( )M N
h z M h z h z N h zy y x x
( ) ( ) ( ) ( )M N
h z M h z x h z N h z yy x
or
( )( ) ( )( )M N
h z h z Ny Mxy x
( )
( )
M Nh zy x
Ny Mx h z
or
or
( )
M Ny x
g zNy Mx
Hence if
is a function of z = x y alone, then
( )g z dze
is an integrating factor of the given DE
0M dx N dy
Example 11 Find an I.F. for the following DE and hence solve it.
2( 3 ) 2 0x y dx xy dy Here
6 2M N
y yy x
2( 3 ); 2M x y N xy
Hence the given DE is not exact.
M Ny x
N
Now
6 2
2
y y
xy
2( ),g x
x
a function of x alone. Hence
( )g x dxe
22dx
xe x is an integrating factor of the given DE
Multiplying by x2, the given DE becomes
3 2 2 3( 3 ) 2 0x x y dx x y dy
which is of the form 0M dx N dy
Note that now 3 2 2 3( 3 ); 2M x x y N x y
Integrating, we easily see that the solution is4
3 2 ,4
xx y c c an arbitrary constant.
Example 12 Find an I.F. for the following DE and hence solve it.
( cot 2 csc ) 0x xe dx e y y y dy Here
0 cotxM Ne y
y x
; cot 2 cscx xM e N e y y y
Hence the given DE is not exact.
M Ny x
M
Now0 cotx
x
e y
e
cot ( ),y h y
a function of y alone. Hence
( )h y dye
cotsin
y dye y
is an integrating factor of the given DE
Multiplying by sin y, the given DE becomes
sin ( cos 2 ) 0x xe y dx e y y dy
which is of the form 0M dx N dy
Note that now sin ; cos 2x xM e y N e y y
Integrating, we easily see that the solution is
c an arbitrary constant.2sin ,xe y y c
Example 13 Find an I.F. for the following DE and hence solve it.
2 3( 2 ) 0y dx x x y dy Here
31 1 4M N
xyy x
2 3; 2M y N x x y
Hence the given DE is not exact.
M Ny x
Ny Mx
Now 3
2 4
1 (1 4 )
( 2 )
xy
xy x y xy
2 2
( ),g zxy z
a function of z =x y alone. Hence
( )g z dze
2
2 2 2
1 1dzze
z x y
is an integrating factor of the given DE
2 2
1 1( 2 ) 0d x y d y
x y xy
which is of the form 0M dx N dy
Integrating, we easily see that the solution is
c an arbitrary constant.21,y c
xy
Multiplying by2 2
1,
x y the given DE becomes
Problem Under what conditions will the DE
0M dx N dy have an integrating factor that is a function of the sum z = x + y ?
Solution. Suppose = h(z) is an I.F.
Multiplying by h(z) the above DE becomes
( ) ( ) 0 .......(*)h z M dx h z N dy Since (*) is an exact DE, we have
( ( ) ) ( ( ) )h z M h z Ny x
i.e. ( ) ( ) ( ) ( )M N
h z M h z h z N h zy y x x
( ) ( ) ( ) ( )M N
h z M h z h z N h zy x
or
( )( ) ( )( )M N
h z h z N My x
( )
( )
M Nh zy x
N M h z
or
or
( )
M Ny x
g zN M
Hence if
is a function of z = x + y alone, then ( )g z dz
e
is an integrating factor of the given DE
0M dx N dy
Linear EquationsLinear Equations
A linear first order equation is an equation that A linear first order equation is an equation that can be expressed in the formcan be expressed in the form
1 0( ) ( ) ( ), (1)dy
a x a x y b xdx
where a1(x), a0(x), and b(x) depend only on the independent variable x, not on y.
We assume that the function a1(x), a0(x), and b(x) are continuous on an interval and that a1(x) 0on that interval. Then, on dividing by a1(x), we can rewrite equation (1) in the standard form
( ) ( ) (2)dy
P x y Q xdx
where P(x), Q(x) are continuous functions on the interval.
Let’s express equation (2) in the differential form ( ) ( ) 0 (3)P x y Q x dx dy
If we test this equation for exactness, we find
( ) 0M N
P x andy x
Consequently, equation(3) is exact only when P(x) = 0. It turns out that an integrating factor , which depends only on x, can easily obtained the general solution of (3).
Multiply (3) by a function (x) and try to determine (x) so that the resulting equation
( ) ( ) ( ) ( ) ( ) 0 (4)x P x y x Q x dx x dy
is exact. ( ) ( ) ( )M N d
x P x and xy x dx
We see that (4) is exact if satisfies the DE
( ) ( ) ( ) (5)d
x P x xdx
( ) exp ( ) (6)x P x dx Which is our desired IF
In (2), we multiply by (x) defined in (6) to obtain
( ) ( ) ( ) ( ) ( ) (7)dy
x P x x y x Q xdx
We know from (5) ( ) ( )d
P x xdx
and so (7) can be written in the form
( ) ( ) ( ) (8)
( ) ( ) ( ) ( )
dx y x Q x
dx
dy dx x y x Q x
dx dx
Integrating (8) w.r.t. x gives
( ) ( ) ( )x y x Q x dx C and solving for y yields
exp ( ) ( ) ( )y P x dx x Q x dx C
Working Rule to solve a LDE:
1. Write the equation in the standard form
( ) ( )dy
P x y Q xdx
2. Calculate the IF (x) by the formula
( ) exp ( )x P x dx 3. Multiply the equation in standard form
by (x) and recalling that the LHS is just ( ) ,
dx y
dx obtain
( ) ( ) ( )d
x y x Q xdx
4. Integrate the last equation and solve for y by dividing by (x).
Ex 1. Solve
1cossincos
xxxydx
dyxx
Solution :- Dividing by x cos x, throughout, we get
x
xy
xx
dx
dy sec1tan
1tan( )( )
x dxP x dx xx e e
log sec log loglog sec( ) sece e ex x xxx e e e x x
secsec sec
d x
dx xy x x x x
2sec sec tany x x dx C x C
Multiply by secx x yields
Integrate both side we get
Problem (2g p. 62): Find the general solution of the equation
.0'cot xyxxyxy
Ans.: .cossinsin cxxxxxy
The usual notation implies that x is independent variable & y the dependent variable. Sometimes it is helpful to replace x by y and y by x & work on the resulting equation.
* When diff equation is of the form
dx
dy
yQxyPdy
dx .
&
P y dyIF e solution is x IF Q IF dy c
Q. 4 (b) Solve
dx
dyeyxy y2
yeyxdy
dxy 2
yeyxydy
dx 1
2' ' yy xy y y e
yeeIF y
dyy 1ln
1
dyedyy
eyy
x yy 11
cey
x y
cyeyx y