Download UNIT I ORDINARY DIFFERENTIAL EQUATIONS

UNIT I
ORDINARY DIFFERENTIAL EQUATIONS - FIRST ORDER
AND APPLICATIONS
1
A Differential equation is an equation which involves differential coefficients
or differentials.
Thus
(i) ex dx + ey dy = 0
d2 x
+ n2 x = 0
dt2
"
2 #3/2 2
dy
d y
=c
(iii) 1 +
dx
dx2
dx
dy
(iv)
− wy = a cos pt,
+ wx = a sin pt
dt
dt
dy
x
(v) y = x
+ dy
dx
/dx
2
∂2y
2∂ y
(vi) 2 = c
are all examples of differential equations.
∂t
∂x2
(ii)
An Ordinary differential equation is that in which all the differential coefficients have reference to a single independent variable. Thus equations (i) to
(v) are ordinary differential equations.
A Partial differential equation is that in which there are two are more independent variables and partial differential coefficients with respect to any of
them. Thus equation (vi) is a partial differential equation.
The Order of a differential equation is the order of the highest derivative appearing in it.
The Degree of a differential equation is the degree of the highest derivative appearing in it, after the equation has been expressed in a form free from radicals
and fractions as far as the derivatives are concerned.
Thus from the examples above,
(i) is of the first order and first degree; (ii) is of the second order and first degree;
(v) written as y
dy
=x
dx
dy
dx
2
+ x is clearly of the first order but of second
degree;
"
(iii) written as 1 +
dy
dx
2 # 3
=c
2
d2 y
dx2
degree.
2
2
is of the second order and second
Formation of a differential equation
An ordinary differential equation is formed in an attempt to eliminate certain
arbitrary constant from a relation in the variables and constants.
Example: Obtain the differential equation of all circles of radius a and centre
(h, k).
Solution Such a circle is (x − h)2 + (y − k)2 = a2
where h and k, the co-ordinates of the centre, and a are the constants.
Differentiating it twice, we have
2
dy
d2 y
dy
= 0 and 1 + (y − k) 2 +
=0
dx
dx
dx
2
dy
1+
dx
Then y − k = −
d2 y/dx2
"
2 #
dy
dy
1+
dx
dx
and x − h = −(y − k)dy/dx =
d2 y/dx2
h
i3
2
2
Substituting these in (i) and simplifying, we get 1 + (dy/dx)
= a2 (d2 y/dx2 )
as the required differential equation.
Example:
x − h + (y − k)
Obtain the differential equation of the coaxial circles of the system x2 + y 2 +
2ax + c2 = 0 where c is a constant and a is a variable.
Solution: We have x2 + y 2 + 2ax + c2 = 0.
dy
Differentiating w.r.t x, 2x + 2y
+ 2a = 0.
dx
dy
or 2a = −2 x + y
dx
dy
substituting in (i), x2 + y 2 − 2(x +
)x + c2 = 0
dx
dy
or 2xy
= y 2 − c2 which is the required differential equation.
dx
3
Solution of a differential equation
A solution of a differential equation is a relation between the variables which
satisfies the given differential equation.
Equations of the first order and first degree
Variable separable method If in an equation it is possible to collect all functions of x and dx on one side and all the functions of y and dy on the other
side, then the variables are said to be separable. Thus the general form of such
an equation is f (y)dy = φ(x)dx.
Integrating on both sides, we get
R
f (y)dy =
R
φ(x)dx + c as its solution.
dy
= e3x−2y + x2 e−2y
dx
dy
Solution: Given equation is
= e−2y e3x + x2 or e2y dy = (e3x + x2 )dx
dx
R
R 3x
Integrating both sides we get, e2y dy =
e + x2 dx + c
Example: Solve
e2y
e2x
x3
=
+
+ c or 3e2y = 2 e3x + x3 + 6c
2
3
3
dy
Example: Solve
= sin(x + y) + cos(x + y)
dx
Solution: Putting x + y = t so that
The given equation becomes
or
dy/dx
= dt/dx − 1
dt
− 1 = sin t + cos t
dx
dt
= 1 + sin t + cos t
dx
Z
dt
Integrating both sides, we get dx =
+c
1 + sin t + cos t
Z
2dθ
x=
+c
1 + sin 2θ + cos 2θ
Z
Z
2dθ
sec2 θ
=
+
c
=
dθ + c
2
2 cos θ + 2 sin θ cos θ
1 + tan θ
= log(1 + tan θ + c)
Hence the solution is x = log 1 + tan 12 (x + y) .
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Homogeneous equations
dy
f (x, y)
=
where f (x, y) and φ(x, y)
dx
φ(x, y)
are homogeneous functions of the same degree in x and y.
y
, is
Note: Any function f (x, y) which can be expressed in the form xn φ
x
called a homogeneous function of degree n in x and y.
Homogeneous equations are of the form
To solve a homogeneous equation
(i) Put y = vx, then
dy
dy
=v+x ,
dx
dx
(ii) Separate the variables v and x, and integrate.
Example: Solve (x2 − y 2 )dx − xydy = 0.
x2 − y 2
dy
=
which is homogeneous in x and y.
dx
xy
dy
dv
dv
1 − v2
Put y = vx, then
=v+x .
∴ (i) becomes v + x
=
dx
dx
dx
v
2
2
dv
1−v
1 − 2v
or
x
=
−v =
.
dx
v
v
v
dx
Separating the variables,
dv =
2
1 − 2v
x
Z
Z
dx
v
dv =
+c
Integrating both sides we get,
1 − 2v 2
x
Z
1
dx
1
−4v
or −1
dv =
+ c or − log(1 − 2v 2 ) = log x + c
4
1 − 2v 2
x
4
Solution: Given equation is
4 log x + log(1 − 2v 2 ) = −4c or log x4 (1 − 2v 2 ) = −4c
x4 (1 − 2y2/x2 ) = e−3c = c0
Hence the required solution is x2 (x2 − 2y 2 ) = c0 .
x
x
Example: Solve (1 + e /y )dx + e /y (1 − x/y)dy = 0.
Solution: The given equation may be rewritten as
x
dx
e /y (1 − x/y)
=−
x
dy
1 + e /y
which is a homogeneous equation. Putting x = vy so that (i) becomes
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ev (1 − v)
1 + ev
v + ev
dv
ev (1 − v)
=
−
or y
=−
dy
1 + ev
1 + ev
v+y =−
Separating the variables we get
−
dy
d(v + ev )
1 + ev
dv
=
=
y
v + ev
v + ev
Integrating on both sides, − log y = log(v + ev ) + c
x
or y(v + ev ) = e−c or x + ye /y = c0 (say) which is the required solution.
Equations reducible to homogeneous form
dy
ax + by + c
= 0
. . . (1) can be reduced to the
dx
a x + b0 y + c 0
homogeneous form as follows:
The equations of the form
Case I. When
a
b
6= 0
a0
b
Putting x = X +h, y = Y +k, (h, k being constants) so that dx = dX, dy = dY ,
(1) becomes
dY
aX + bY + (ah + bk + c)
. . . (2)
= 0
dX
a X + b0 Y + (a0 h + b0 k + c)
Choose h, k so that (2) may become homogeneous.
Put ah + bk + c = 0, and a0 h + b0 k + c = 0 so that
k
1
h
= 0
= 0
bc0 − b0 c
ca − c0 a
ab − ba0
bc0 − b0 c
ca0 − c0 a
or h = 0
,k= 0
0
ab − b a
ab − ba0
dY
aX + bY
= 0
which is homogeneous
dX
a X + b0 Y
in X, Y and can be solved by putting Y = vX.
Thus when ab0 − ba0 6= 0, (2) becomes
Case II. When
a
b
= 0
a0
b
i.e, ab0 − ba0 = 0, the above method fails as h and k become infinite or indeterminate.
Now
a
b
1
= 0 =
(say)
a0
b
m
6
∴ a0 = am, b00 = bm and (1) becomes
dy
(ax + by) + c
=
dx
m(ax + by) + c00
dy
dt
Put t = ax + by, so that a + b
=
dx
dx
dy
1 dt
1 dt
t+c
or
=
− a ∴ (4) becomes
−a =
dx
b dx
b dx
mt + c0
dt
(am + b)t + ac0 + bc
bt + bc
=
=a+
0
dx
mt + c
mt + c0
so that the variables are separable. In this solution, putting t = ax + by, we get
the required solution of (1).
dy
y+x−2
=
dx
y−x−4
y+x−2
dy
b
a
case 0 6= 0 . . . (i)
=
Solution. Given equation is
dx
y−x−4
a
b
Putting x = X +h, y = Y +k, (h, k being constants) so that dx = dX, dy = dY ,
(i) becomes
Example: Solve
Y + X + (k + h − 2)
dY
=
. . . (ii)
dX
Y − X + (k − h − 4)
Put k + h − 2 = 0 and k − h − 4 = 0 so that h = −1, k = 3.
y+X
dY
=
which is homogeneous in X and Y . . . (iii).
dX
y−X
dY
dv
∴ Put Y = vX, then
=v+X
dX
dX
dv
v+1
dv
v+1
1 + 2v − v 2
∴ (iii) becomes v + X
=
or X
=
−v =
dX
v−1
dX
v−1
v−1
v−1
dX
or
dv =
1 + 2v − v 2
X
Z
Z
1
2 − 2v
dX
Integrating both sides we get, −
dv =
+c
2
1 + 2v − v 2
X
1
or − log(1 + 2v − v 2 ) = log X + c
2
2Y
Y2
or log 1 +
− 2 + log X 2 = −2c
X
X
2
or log X + 2XY − Y 2 = −2c or X 2 + 2XY − Y 2 = e−2c = c0
∴ (ii) becomes
Putting X = x − h = x + 1, Y = y − k = y − 3, (iv) becomes
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(x + 1)2 + 2(x + 1)(y − 3) − (y − 3)2 = c0
or x2 + 2xy − y 2 − 4x + 8y − 14 = c0 which is the required solution.
Example: Solve (3y + 2x + 4)dx − (4x + 6y + 5)dy = 0.
dy
(2x + 3y) + 4
=
. . . (i)
dx
2(2x + 3y) + 5
dy
dt
1 dt
t+4
Putting 2x + 3y = t so that 2 + 3
=
∴ (i) becomes
−2 =
dx
dx
3 dx
2t + 5
dt
3t + 12
7t + 22
2t + 5
or
=2+
=
or
dt = dx
dx
2t + 5
2t + 5
7t + 22
Z
Z
2t + 5
dt = dx + c
Integrating both sides we get,
7t + 22
Z 2 9
1
2
9
or
− .
dt = x + c
or
t−
log(7t + 22) = x + c
7 7 7t + 22
7
49
Solution. Given equation is
Putting t = 2x + 3y, we have 14(2x + 3y) − 9 log(14x + 21y + 22) = 49x + 49c
21x − 42y + 9 log(14x + 21y + 22) = c0 which is the required solution.
Exact differential equations
A differential equation of the form M (x, y)dx + N (x, y)dy = 0 is said to be
exact if its left hand member is the exact differential of some function u(x, y)
i.e, du = M dx + N dy = 0. Its solution, therefore, is u(x, y) = c.
2
2
Example: Solve (y 2 exy + 4x3 )dx + (2xyexy − 3y 2 )dy = 0.
2
2
Solution. Here M = y 2 exy and N = 2xyexy − 3y 2
∴
∂N
∂M
2
= 2y 2 exy ;
= 2xy.
∂y
∂x
Thus the equation is exact and its solution is
R
R
M dx + (terms of N not containing x)dy = c
(y const)
R
2
2
(y 2 ey x + 4x3 )dx + (−3y 2 dy) = c or exy + x4 − y 3 = c
(y const)
R
Example: Solve (1 + 2xy cos x2 − 2xy)dx + (sin x2 − x2 )dy = 0
Solution. Here M = 1 + 2xy cos x2 − 2xy and N = sin x2 − x2 .
8
∂M
∂N
= 2x cos x2 − 2x =
∂y
∂x
thus the equation is exact and its solution is
R
R
M dx + (terms of N not containing x)dy = c
(y const)
R
yconst
(1 + 2xy cos x2 − 2xy)dx = c or x + y
R
cosx2 .2xdx −
R
2xdx = c
or x + y sin x2 − yx2 = c.
Equations reducible to exact equations
Sometimes a differential equation which is not exact, can be made so, on multiplication by a suitable factor called an integrating factor.
Example: Solve y(2xy + ex )dx = ex dy.
Solution. Rewriting the given equation (yex dx − ex dy) + 2xy 2 dx = 0.
1
1
is the integrating factor. Multiplying throughout by 2 , it follows
y2
y
x
e
yex dx − ex dy
+
2xdx
or
d
+ 2xdx = 0.
y2
y
ex
Integrating, we get
+ x2 = c which is the required solution.
y
Here
Example: Solve (1 + xy)ydx + (1 − xy)xdy = 0.
Solution. The given equation is of the form f1 (xy)ydx + f2 (xy)xdy = 0.
Here M = (1 + xy)y, N = (1 − xy)x.
∴ I.F=
1
1
1
=
= 2 2.
Mx − Ny
(1 + xy)yx − (1 − xy)xy
2x y
Multiplying throughout by 1/2x2 y2 , it becomes
1
1
1
1
+
dx
+
−
dy = 0 which is an exact equation.
2x2 y 2x
2xy 2
2y
Z
Z
∴ the solution is
M dx + (terms of N not containing x)dy = c
(yconst)
1
1
1
1
or
−
+ log x − log y = c.
2y
x
2
2
x
1
or log −
= c0 .
y
xy
9
Linear Equations
A differential equation is said to be linear if the dependent variable and its differential coefficients occur only in the first degree and not multiplied together.
Thus the standard form of a linear equation of the first order, commonly
known as Leibnitz’s linear equation is
dy
+ P y = Q where P , Q are the functions of x.
dx
dy
Example: Solve (x + 1)
− y = e3x (x + 1)2 .
dx
Solution. Dividing throughout by (x + 1), given equation becomes
y
dy
−
= e3x (x + 1) which is Leibnitz’s equation.
dx x + 1
Z
Z
dx
1
and P dx = −
= − log(x + 1) = log(x + 1)−1
Here P = −
x+1
x+1
Z
P dx
−1
1
I.F= e
= elog(x+1) =
x+1
Z
Thus the solution of (1) is y(I.F ) = [e3x (x + 1)](I.F )dx + c
Z
1
1 3x
y
or
= e3x dx + c = e3x + c or y =
e + c (x + 1)
x+1
3
3
Example: Solve y(log y)dx + (x − log y)dy = 0.
Solution. We have
dx
x
1
+
= . . . (i)
dy
y log y
y
which is a Leibnitz equation in x.
Z
1
dy
y
log
y
I.F= e
= elog(log y) = log y
Z
1
Thus the solution of (i) is x(I.F ) =
(I.F )dy + c
y
Z
1
1
x log y =
log ydy + c = (log y)2 + c
y
2
1
x = log y + c(log y)−1
2
10
Example: Solve (1 + y 2 )dx = (tan−1 y − x)dy.
Solution. This equation contains y 2 and tan−1 y and is , therefore, not a linear
in y; but since only x occurs, it can be written as
tan−1 y
dx
dx
x
=
which is a Leibnitz’s equa= tan−1 y − x or
+
2
dy
dy 1 + y
1 + y2
tion in x.
Z
1
R
−1
2
P dy
∴ I.F = e
= e 1 + y dy = etan y
(1 + y 2 )
Z
tan−1 y
Thus the solution is x(I.F ) =
(I.F )dy + c
1 + y2
Z
tan−1 y tan−1 y
−1
or xetan y =
.e
dy + c
1 + y2
R
R
= tet dt + c = t.et − 1.et dt + c
[Put tan−1 y = t, ∴
−1
= t.et − et + c = (tan−1 y − 1)etan
−1
x = tan−1 y − 1 + ce− tan
y
+c
y
11
dy
= dt]
1 + y2
Applications
Orthogonal Trajectories The families of curves such that every member of
either family cuts each member of the other family at right angles are called
orthogonal trajectories.
The concept of the orthogonal trajectories is of wide use in applied mathematics especially in field problems. For instance, in an electric field, the paths
along which the current flows are the orthogonal trajectories of the equipotential curves and vice versa.
To find the orthogonal trajectories of the family of curves F (x, y, c) =
0.
dy/dx)
= 0 by eliminating c.
− dx/dy ,
(so that the product
(i) Form its differential equation in the form f (x, y,
(ii) Replace, in this differential equation, dy/dx by
of their slopes at eac point of intersection is −1).
Solve the differential equation of the orthogonal trajectories i.e, f (x, y, −
dx/dy ) = 0.
Example: Find the orthogonal trajectories of the family of confocal conics
y2
x2
+ 2
, where λ is the parameter.
2
a
b +λ
2x
2y dy
=0
Solution. Differentiating the given equation, we get 2 + 2
a
b + λ dx
y
x
y2
−xy
or 2
= − 2 dy
or 2
= 2 dy
b +λ
a ( /dx)
b +λ
a ( /dx)
Substituting in the given equation, we get
x2
xy
dy
− 2 dy
= 1 or (x2 − a2 )
= xy . . . (i)
2
a
a ( /dx)
dx
which is the differential equation of the given family.
Changing dy/dx to − dx/dy in (i), we get (a2 − x2 )dx/dy = xy as the differential
equation of the orthogonal trajectories.
Separating the variables and integrating, we obtain
Z
Z 2
Z
a − x2
1 2
1
ydy =
dx + c
or
y = a2 log x − x2 + c
x
2
2
or
x2 + y 2 = 2a2 log x + c0
[c0 = 2c]
which is the equation of the required orthogonal trajectories
12
Example: Find the orthogonal trajectories of a system of confocal and coaxial
parabolas.
Solution. The equation of the family of confocal parabolas having x−axis as
their axis, is of the form
y 2 = 4a(x + a). . . (i)
Differentiating, y
dy
= 2a . . . (ii)
dx
1 dy
dy
x+ y
Substituting the value of a from (ii) in (i), we get y = 2y
dx
2 dx
2
dy
dy
+ 2x
i.e, y
− y = 0 as the differential equation of the family. . . . (iii)
dx
dx
2
dy
dx
dx
dx
Replacing
by −
in (iii), we obtain y
− 2x
−y =0
dx
dy
dy
dy
2
dy
dy
y
+ 2x
− y = 0 which is the same as (iii).
dx
dx
2
Thus we see that a system of confocal and coaxial parabolas is selforthogonal,
i.e each member of the family (i) cuts every other member of the same family
orthogonally.
To find the orthogonal trajectories of the curves F (r, θ, c) = 0.
dr
(i) Form its differential equation in the form f (r, θ,
) = 0 by eliminating c.
dθ
dr
dθ
(ii) Replace in this differential equation,
by −r2 .
dθ
dr
(iii) Solve the differential equation of the orthogonal trajectories f (r, θ, −
r2 dθ/dr)
Example: Find the orthogonal trajectory of the cardioids r = a(1 − cos θ)
Solution. Differentiating r = a(1 − cos θ) . . . (i)
with respect to θ we get
dr
= a sin θ . . . (ii)
dθ
eliminating a from (i) and (ii), we obtain
dr 1
sinθ
θ
. =
= cot which is the differential equation of the given family.
dθ r
1 − cos θ
2
Replacing
dr/dθ
by −r2 dθ/dr, we obtain
13
1
r
−r2
dθ
dr
= cot
θ
dr
θ
or
+ tan dθ = 0
2
r
2
as the differential equation of orthogonal trajectories. It can be rewritten as
θ
sin
dθ
dr
2
=−
θ
r
cos
2
Integrating, log r = 2 log cos θ/2 + log c
or r = c cos2 θ/2 =
1
c(1 + cos θ) or r = a0 (1 + cos θ)
2
which is the required orthogonal trajectory.
Example: Find the orthogonal trajectory of the family of curves rn = a sin nθ.
Solution. We have n log r = log a + log sin nθ
Differentiating w.r.t θ, we have
n cos nθ
1 dr
n dr
=
or
= cot nθ
r dθ
sin nθ
r dθ
dr
dθ
Replacing
by −r2 , we obtain
dθ
dr
1
dθ
dr
= 0.
−r2
= cot nθ or tan nθ.dθ −
r
dr
r
Z
Z
dr
sin nθ
Integrating,
+
dθ = c,
r
cos nθ
n r
1
= nc = log b. (say)
i.e, log r − cos nθ = c or log
n
cos nθ
or rn = b cos nθ, which is the required orthogonal trajectory.
14