DIFFERENTIAL EQUATIONS 379
He who seeks for methods without having a definite problem in mind
seeks for the most part in vain. – D. HILBERT
9.1 Introduction
In Class XI and in Chapter 5 of the present book, we
discussed how to differentiate a given function f with respect
to an independent variable, i.e., how to find f ′(x) for a given
function f at each x in its domain of definition. Further, in
the chapter on Integral Calculus, we discussed how to find
a function f whose derivative is the function g, which may
also be formulated as follows:
For a given function g, find a function f such that
dy
dx
= g(x), where y = f (x) ... (1)
An equation of the form (1) is known as a differential
equation. A formal definition will be given later.
These equations arise in a variety of applications, may it be in Physics, Chemistry,
Biology, Anthropology, Geology, Economics etc. Hence, an indepth study of differential
equations has assumed prime importance in all modern scientific investigations.
In this chapter, we will study some basic concepts related to differential equation,
general and particular solutions of a differential equation, formation of differential
equations, some methods to solve a first order - first degree differential equation and
some applications of differential equations in different areas.
9.2 Basic Concepts
We are already familiar with the equations of the type:
x
2
– 3x + 3 = 0 ... (1)
sin x + cos x = 0 ... (2)
x + y = 7 ... (3)
Chapter
9
DIFFERENTIAL EQUATIONS
Henri Poincare
(1854-1912 )
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MATHEMATICS380
Let us consider the equation:
dy
xy
dx
+
= 0 ... (4)
We see that equations (1), (2) and (3) involve independent and/or dependent variable
(variables) only but equation (4) involves variables as well as derivative of the dependent
variable y with respect to the independent variable x. Such an equation is called a
differential equation.
In general, an equation involving derivative (derivatives) of the dependent variable
with respect to independent variable (variables) is called a differential equation.
A differential equation involving derivatives of the dependent variable with respect
to only one independent variable is called an ordinary differential equation, e.g.,
3
2
2
2
d y dy
dx
dx
+
= 0 is an ordinary differential equation .... (5)
Of course, there are differential equations involving derivatives with respect to
more than one independent variables, called partial differential equations but at this
stage we shall confine ourselves to the study of ordinary differential equations only.
Now onward, we will use the term ‘differential equation’ for ‘ordinary differential
equation’.
Note
1. We shall prefer to use the following notations for derivatives:
2 3
2 3
,,
dy d y d y
y yy
dx
dx dx
′ ′′ ′′′
= ==
2. For derivatives of higher order, it will be inconvenient to use so many dashes
as supersuffix therefore, we use the notation y
n
for nth order derivative
n
n
dy
dx
.
9.2.1. Order of a differential equation
Order of a differential equation is defined as the order of the highest order derivative of
the dependent variable with respect to the independent variable involved in the given
differential equation.
Consider the following differential equations:
dy
dx
= e
x
... (6)
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2
2
dy
y
dx
+
= 0 ... (7)
3
3 2
2
3 2
dy dy
x
dx dx
+
= 0 ... (8)
The equations (6), (7) and (8) involve the highest derivative of first, second and
third order respectively. Therefore, the order of these equations are 1, 2 and 3 respectively.
9.2.2 Degree of a differential equation
To study the degree of a differential equation, the key point is that the differential
equation must be a polynomial equation in derivatives, i.e., y′, y″, y″′ etc. Consider the
following differential equations:
2
32
32
2
d y d y dy
y
dx
dx dx
+ −+
= 0 ... (9)
2
2
sin
dy dy
y
dx dx
+−
= 0 ... (10)
sin
dy dy
dx dx
+
= 0 ... (11)
We observe that equation (9) is a polynomial equation in y″′, y″ and y′, equation (10)
is a polynomial equation in y′ (not a polynomial in y though). Degree of such differential
equations can be defined. But equation (11) is not a polynomial equation in y′ and
degree of such a differential equation can not be defined.
By the degree of a differential equation, when it is a polynomial equation in
derivatives, we mean the highest power (positive integral index) of the highest order
derivative involved in the given differential equation.
In view of the above definition, one may observe that differential equations (6), (7),
(8) and (9) each are of degree one, equation (10) is of degree two while the degree of
differential equation (11) is not defined.
Note Order and degree (if defined) of a differential equation are always
positive integers.
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Example 1 Find the order and degree, if defined, of each of the following differential
equations:
(i)
cos 0
dy
x
dx
−=
(ii)
2
2
2
0
d y dy dy
xy x y
dx dx
dx
+ −=
(iii)
2
0
y
yye
′
′′′
++=
Solution
(i) The highest order derivative present in the differential equation is
dy
dx
, so its
order is one. It is a polynomial equation in y′ and the highest power raised to
dy
dx
is one, so its degree is one.
(ii) The highest order derivative present in the given differential equation is
2
2
dy
dx
, so
its order is two. It is a polynomial equation in
2
2
dy
dx
and
dy
dx
and the highest
power raised to
2
2
dy
dx
is one, so its degree is one.
(iii) The highest order derivative present in the differential equation is
y
′′′
, so its
order is three. The given differential equation is not a polynomial equation in its
derivatives and so its degree is not defined.
EXERCISE 9.1
Determine order and degree (if defined) of differential equations given in Exercises
1 to 10.
1.
4
4
sin( ) 0
dy
y
dx
′′′
+=
2. y′ + 5y = 0 3.
4
2
2
30
ds d s
s
dt
dt
+=
4.
2
2
2
cos 0
d y dy
dx
dx
+=
5.
2
2
cos3 sin 3
dy
xx
dx
=+
6.
2
()
y
′′′
+ (y″)
3
+ (y′)
4
+ y
5
= 0 7.
y
′′′
+ 2y″ + y′ = 0
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8. y′ + y = e
x
9. y″ + (y′)
2
+ 2y = 0 10. y″ + 2y′ + sin y = 0
11. The degree of the differential equation
3
2
2
2
sin 1 0
d y dy dy
dx dx
dx
+ + +=
is
(A) 3 (B) 2 (C) 1 (D) not defined
12. The order of the differential equation
2
2
2
230
d y dy
xy
dx
dx
− +=
is
(A) 2 (B) 1 (C) 0 (D) not defined
9.3. General and Particular Solutions of a Differential Equation
In earlier Classes, we have solved the equations of the type:
x
2
+ 1 = 0 ... (1)
sin
2
x – cos x = 0 ... (2)
Solution of equations (1) and (2) are numbers, real or complex, that will satisfy the
given equation i.e., when that number is substituted for the unknown x in the given
equation, L.H.S. becomes equal to the R.H.S..
Now consider the differential equation
2
2
0
dy
y
dx
+=
... (3)
In contrast to the first two equations, the solution of this differential equation is a
function φ that will satisfy it i.e., when the function φ is substituted for the unknown y
(dependent variable) in the given differential equation, L.H.S. becomes equal to R.H.S..
The curve y = φ (x) is called the solution curve (integral curve) of the given
differential equation. Consider the function given by
y = φ (x) = a sin (x + b), ... (4)
where a, b ∈ R. When this function and its derivative are substituted in equation (3),
L.H.S. = R.H.S.. So it is a solution of the differential equation (3).
Let a and b be given some particular values say a = 2 and
4
b
π
=
, then we get a
function y = φ
1
(x) =
2sin
4
x
π
+
... (5)
When this function and its derivative are substituted in equation (3) again
L.H.S. = R.H.S.. Therefore φ
1
is also a solution of equation (3).
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Function φ consists of two arbitrary constants (parameters) a, b and it is called
general solution of the given differential equation. Whereas function φ
1
contains no
arbitrary constants but only the particular values of the parameters a and b and hence
is called a particular solution of the given differential equation.
The solution which contains arbitrary constants is called the general solution
(primitive) of the differential equation.
The solution free from arbitrary constants i.e., the solution obtained from the general
solution by giving particular values to the arbitrary constants is called a particular
solution of the differential equation.
Example 2 Verify that the function y = e
– 3x
is a solution of the differential equation
2
2
60
d y dy
y
dx
dx
+−=
Solution Given function is y = e
– 3x
. Differentiating both sides of equation with respect
to x , we get
3
3
x
dy
e
dx
−
=−
... (1)
Now, differentiating (1) with respect to x, we have
2
2
dy
dx
= 9 e
– 3x
Substituting the values of
2
2
,
d y dy
dx
dx
and y in the given differential equation, we get
L.H.S. = 9 e
– 3x
+ (–3e
– 3x
) – 6.e
– 3x
= 9 e
– 3x
– 9 e
– 3x
= 0 = R.H.S..
Therefore, the given function is a solution of the given differential equation.
Example 3 Verify that the function y = a cos x + b sin x, where, a, b ∈ R is a solution
of the differential equation
2
2
0
dy
y
dx
+=
Solution The given function is
y = a cos x + b sin x ... (1)
Differentiating both sides of equation (1) with respect to x, successively, we get
dy
dx
= – a sin x + b cos x
2
2
dy
dx
= – a cos x – b sin x
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Substituting the values of
2
2
dy
dx
and y in the given differential equation, we get
L.H.S. = (– a cos x – b sin x) + (a cos x + b sin x) = 0 = R.H.S.
Therefore, the given function is a solution of the given differential equation.
EXERCISE 9.2
In each of the Exercises 1 to 10 verify that the given functions (explicit or implicit) is a
solution of the corresponding differential equation:
1. y = e
x
+ 1 : y″ – y′ = 0
2. y = x
2
+ 2x + C : y′ – 2x – 2 = 0
3. y = cos x + C : y′ + sin x = 0
4. y =
2
1
x
+
: y′ =
2
1
xy
x
+
5. y = Ax : xy′ = y (x ≠ 0)
6. y = x sin x : xy′ = y + x
22
xy
−
(x ≠ 0 and x > y or x < – y)
7. xy = log y + C : y′ =
2
1
y
xy
−
(xy ≠ 1)
8. y – cos y = x : (y sin y + cos y + x) y′ = y
9. x + y = tan
–1
y : y
2
y′ + y
2
+ 1 = 0
10. y =
22
ax
−
x ∈ (–a, a) : x + y
dy
dx
= 0 (y ≠ 0)
11. The number of arbitrary constants in the general solution of a differential equation
of fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
12. The number of arbitrary constants in the particular solution of a differential equation
of third order are:
(A) 3 (B) 2 (C) 1 (D) 0
9.4 Formation of a Differential Equation whose General Solution is given
We know that the equation
x
2
+ y
2
+ 2x – 4y + 4 = 0 ... (1)
represents a circle having centre at (–1, 2) and radius 1 unit.
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Differentiating equation (1) with respect to x, we get
dy
dx
=
1
2
x
y
+
−
(y ≠ 2) ... (2)
which is a differential equation. You will find later on [See (example 9 section 9.5.1.)]
that this equation represents the family of circles and one member of the family is the
circle given in equation (1).
Let us consider the equation
x
2
+ y
2
= r
2
... (3)
By giving different values to r, we get different members of the family e.g.
x
2
+ y
2
= 1, x
2
+ y
2
= 4, x
2
+ y
2
= 9 etc. (see Fig 9.1).
Thus, equation (3) represents a family of concentric
circles centered at the origin and having different radii.
We are interested in finding a differential equation
that is satisfied by each member of the family. The
differential equation must be free from r because r is
different for different members of the family. This
equation is obtained by differentiating equation (3) with
respect to x, i.e.,
2x + 2y
dy
dx
= 0 or x + y
dy
dx
= 0 ... (4)
which represents the family of concentric circles given by equation (3).
Again, let us consider the equation
y = mx + c ... (5)
By giving different values to the parameters m and c, we get different members of
the family, e.g.,
y = x (m = 1, c = 0)
y =
3
x
(m =
3
, c = 0)
y = x + 1 (m = 1, c = 1)
y = – x (m = – 1, c = 0)
y = – x – 1 (m = – 1, c = – 1) etc. ( see Fig 9.2).
Thus, equation (5) represents the family of straight lines, where m, c are parameters.
We are now interested in finding a differential equation that is satisfied by each
member of the family. Further, the equation must be free from m and c because m and
Fig 9.1
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X
X’
Y
Y’
y =
x+
1
y = x
y = –x
y = –x–1
y = x3
O
c are different for different members of the family.
This is obtained by differentiating equation (5) with
respect to x, successively we get
dy
m
dx
=
, and
2
2
0
dy
dx
=
... (6)
The equation (6) represents the family of straight
lines given by equation (5).
Note that equations (3) and (5) are the general
solutions of equations (4) and (6) respectively.
9.4.1 Procedure to form a differential equation that will represent a given
family of curves
(a) If the given family F
1
of curves depends on only one parameter then it is
represented by an equation of the form
F
1
(x, y, a) = 0 ... (1)
For example, the family of parabolas y
2
= ax can be represented by an equation
of the form f (x, y, a) : y
2
= ax.
Differentiating equation (1) with respect to x, we get an equation involving
y′, y, x, and a, i.e.,
g (x, y, y′, a) = 0 ... (2)
The required differential equation is then obtained by eliminating a from equations
(1) and (2) as
F (x, y, y′) = 0 ... (3)
(b) If the given family F
2
of curves depends on the parameters a, b (say) then it is
represented by an equation of the from
F
2
(x, y, a, b) = 0 ... (4)
Differentiating equation (4) with respect to x, we get an equation involving
y′, x, y, a, b, i.e.,
g (x, y, y′, a, b) = 0 ... (5)
But it is not possible to eliminate two parameters a and b from the two equations
and so, we need a third equation. This equation is obtained by differentiating
equation (5), with respect to x, to obtain a relation of the form
h (x, y, y′, y″, a, b) = 0 ... (6)
Fig 9.2
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MATHEMATICS388
The required differential equation is then obtained by eliminating a and b from
equations (4), (5) and (6) as
F (x, y, y′, y″) = 0 ... (7)
Note The order of a differential equation representing a family of curves is
same as the number of arbitrary constants present in the equation corresponding to
the family of curves.
Example 4 Form the differential equation representing the family of curves y = mx,
where, m is arbitrary constant.
Solution We have
y = mx ... (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx
= m
Substituting the value of m in equation (1) we get
dy
yx
dx
=⋅
or
dy
x
dx
– y = 0
which is free from the parameter m and hence this is the required differential equation.
Example 5 Form the differential equation representing the family of curves
y = a sin (x + b), where a, b are arbitrary constants.
Solution We have
y = a sin (x + b) ... (1)
Differentiating both sides of equation (1) with respect to x, successively we get
dy
dx
= a cos (x + b) ... (2)
2
2
dy
dx
= – a sin (x + b) ... (3)
Eliminating a and b from equations (1), (2) and (3), we get
2
2
dy
y
dx
+
= 0 ... (4)
which is free from the arbitrary constants a and b and hence this the required differential
equation.
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Example 6 Form the differential equation
representing the family of ellipses having foci on
x-axis and centre at the origin.
Solution We know that the equation of said family
of ellipses (see Fig 9.3) is
22
22
xy
ab
+
= 1 ... (1)
Differentiating equation (1) with respect to x, we get
22
22
0
x y dy
dx
ab
+=
or
y dy
x dx
=
2
2
b
a
−
... (2)
Differentiating both sides of equation (2) with respect to x, we get
y
x
dy
dx
x
dy
dx
y
x
dy
dx
+
−
2
2 2
= 0
or
= 0 ... (3)
which is the required differential equation.
Example 7 Form the differential equation of the family
of circles touching the x-axis at origin.
Solution Let C denote the family of circles touching
x-axis at origin. Let (0, a) be the coordinates of the
centre of any member of the family (see Fig 9.4).
Therefore, equation of family C is
x
2
+ (y – a)
2
= a
2
or
x
2
+ y
2
= 2ay ... (1)
where, a is an arbitrary constant. Differentiating both
sides of equation (1) with respect to x,we get
22
dy
xy
dx
+
=
2
dy
a
dx
Fig 9.3
Fig 9.4
X
X
’
Y’
Y
O
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MATHEMATICS390
or
dy
xy
dx
+
=
dy
a
dx
or a =
dy
xy
dx
dy
dx
+
... (2)
Substituting the value of a from equation (2) in equation (1), we get
x
2
+ y
2
=
2y
xy
dy
dx
dy
dx
+
or
22
()
dy
xy
dx
+
=
2
22
dy
xy y
dx
+
or
dy
dx
=
22
2
–
xy
xy
This is the required differential equation of the given family of circles.
Example 8 Form the differential equation representing the family of parabolas having
vertex at origin and axis along positive direction of x-axis.
Solution Let P denote the family of above said parabolas (see Fig 9.5) and let (a, 0) be the
focus of a member of the given family, where a is an arbitrary constant. Therefore, equation
of family P is
y
2
= 4ax ... (1)
Differentiating both sides of equation (1) with respect to x, we get
2
dy
y
dx
= 4a ... (2)
Substituting the value of 4a from equation (2)
in equation (1), we get
y
2
=
2 ()
dy
yx
dx
or
2
2
dy
y xy
dx
−
= 0
which is the differential equation of the given family
of parabolas.
Fig 9.5
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EXERCISE 9.3
In each of the Exercises 1 to 5, form a differential equation representing the given
family of curves by eliminating arbitrary constants a and b.
1.
1
xy
ab
+=
2. y
2
= a (b
2
– x
2
) 3. y = a e
3x
+ b e
– 2x
4. y = e
2x
(a + bx) 5. y = e
x
(a cos x + b sin x)
6. Form the differential equation of the family of circles touching the y-axis at
origin.
7. Form the differential equation of the family of parabolas having vertex at origin
and axis along positive y-axis.
8. Form the differential equation of the family of ellipses having foci on y-axis and
centre at origin.
9. Form the differential equation of the family of hyperbolas having foci on x-axis
and centre at origin.
10. Form the differential equation of the family of circles having centre on y-axis
and radius 3 units.
11. Which of the following differential equations has y = c
1
e
x
+ c
2
e
–x
as the general
solution?
(A)
2
2
0
dy
y
dx
+=
(B)
2
2
0
dy
y
dx
−=
(C)
2
2
10
dy
dx
+=
(D)
2
2
10
dy
dx
−=
12. Which of the following differential equations has y = x as one of its particular
solution?
(A)
2
2
2
d y dy
x xy x
dx
dx
− +=
(B)
2
2
d y dy
x xy x
dx
dx
+ +=
(C)
2
2
2
0
d y dy
x xy
dx
dx
− +=
(D)
2
2
0
d y dy
x xy
dx
dx
+ +=
9.5. Methods of Solving First Order, First Degree Differential Equations
In this section we shall discuss three methods of solving first order first degree differential
equations.
9.5.1 Differential equations with variables separable
A first order-first degree differential equation is of the form
dy
dx
= F (x, y) ... (1)
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If F (x, y) can be expressed as a product g (x) h(y), where, g(x) is a function of x
and h(y) is a function of y, then the differential equation (1) is said to be of variable
separable type. The differential equation (1) then has the form
dy
dx
= h (y) . g (x) ... (2)
If h (y) ≠ 0, separating the variables, (2) can be rewritten as
1
()
hy
dy = g(x) dx ... (3)
Integrating both sides of (3), we get
1
()
dy
hy
∫
=
()
g x dx
∫
... (4)
Thus, (4) provides the solutions of given differential equation in the form
H(y) = G (x) + C
Here, H (y) and G (x) are the anti derivatives of
1
()
hy
and g (x) respectively and
C is the arbitrary constant.
Example 9 Find the general solution of the differential equation
1
2
dy x
dx y
+
=
−
, (y ≠ 2)
Solution We have
dy
dx
=
1
2
x
y
+
−
... (1)
Separating the variables in equation (1), we get
(2 – y) dy = (x + 1) dx ... (2)
Integrating both sides of equation (2), we get
(2 )
y dy−
∫
=
( 1)
x dx
+
∫
or
2
2
2
y
y −
=
2
1
C
2
x
x
++
or x
2
+ y
2
+ 2x – 4y + 2 C
1
= 0
or x
2
+ y
2
+ 2x – 4y + C = 0, where C = 2C
1
which is the general solution of equation (1).
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Example 10 Find the general solution of the differential equation
2
2
1
1
dy y
dx
x
+
=
+
.
Solution Since 1 + y
2
≠ 0, therefore separating the variables, the given differential
equation can be written as
2
1
dy
y
+
=
2
1
dx
x
+
... (1)
Integrating both sides of equation (1), we get
2
1
dy
y
+
∫
=
2
1
dx
x
+
∫
or tan
–1
y = tan
–1
x + C
which is the general solution of equation (1).
Example 11 Find the particular solution of the differential equation
2
4
dy
xy
dx
=−
given
that y = 1, when x = 0.
Solution If y ≠ 0, the given differential equation can be written as
2
dy
y
= – 4x dx ... (1)
Integrating both sides of equation (1), we get
2
dy
y
∫
=
4
x dx
−
∫
or
1
y
−
= – 2x
2
+ C
or y =
2
1
2C
x
−
... (2)
Substituting y = 1 and x = 0 in equation (2), we get, C = – 1.
Now substituting the value of C in equation (2), we get the particular solution of the
given differential equation as
2
1
21
y
x
=
+
.
Example 12 Find the equation of the curve passing through the point (1, 1) whose
differential equation is x dy = (2x
2
+ 1) dx (x ≠ 0).
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Solution The given differential equation can be expressed as
dy* =
or dy =
1
2
x dx
x
+
... (1)
Integrating both sides of equation (1), we get
dy
=
1
2
x dx
x
+
or y = x
2
+ log |x | + C ... (2)
Equation (2) represents the family of solution curves of the given differential equation
but we are interested in finding the equation of a particular member of the family which
passes through the point (1, 1). Therefore substituting x = 1, y = 1 in equation (2), we
get C = 0.
Now substituting the value of C in equation (2) we get the equation of the required
curve as y = x
2
+ log | x |.
Example 13 Find the equation of a curve passing through the point (–2, 3), given that
the slope of the tangent to the curve at any point (x, y) is
2
2
x
y
.
Solution We know that the slope of the tangent to a curve is given by
dy
dx
.
so,
dy
dx
=
2
2x
y
... (1)
Separating the variables, equation (1) can be written as
y
2
dy = 2x dx ... (2)
Integrating both sides of equation (2), we get
2
y dy
=
2
x dx
or
3
3
y
= x
2
+ C ... (3)
* The notation
dy
dx
due to Leibnitz is extremely flexible and useful in many calculation and formal
transformations, where, we can deal with symbols dy and dx exactly as if they were ordinary numbers. By
treating dx and dy like separate entities, we can give neater expressions to many calculations.
Refer: Introduction to Calculus and Analysis, volume-I page 172, By Richard Courant,
Fritz John Spinger – Verlog New York.
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DIFFERENTIAL EQUATIONS 395
Substituting x = –2, y = 3 in equation (3), we get C = 5.
Substituting the value of C in equation (3), we get the equation of the required curve as
3
2
5
3
y
x
=+
or
1
2
3
(3 15)
yx
=+
Example 14 In a bank, principal increases continuously at the rate of 5% per year. In
how many years Rs 1000 double itself?
Solution Let P be the principal at any time t. According to the given problem,
dp
dt
=
5
P
100
×
or
dp
dt
=
P
20
... (1)
separating the variables in equation (1), we get
P
dp
=
20
dt
... (2)
Integrating both sides of equation (2), we get
log P =
1
C
20
t
+
or P =
1
C
20
t
ee
⋅
or P =
20
C
t
e
(where
1
C
C
e
=
) ... (3)
Now P = 1000, when t = 0
Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3),
gives
P = 1000
20
t
e
Let t years be the time required to double the principal. Then
2000 = 1000
20
t
e
⇒ t = 20 log
e
2
EXERCISE 9.4
For each of the differential equations in Exercises 1 to 10, find the general solution:
1.
1 cos
1 cos
dy x
dx x
−
=
+
2.
2
4 ( 2 2)
dy
yy
dx
= − −< <
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MATHEMATICS396
3.
1 ( 1)
dy
yy
dx
+= ≠
4. sec
2
x tan y dx + sec
2
y tan x dy = 0
5. (e
x
+ e
–x
) dy – (e
x
– e
–x
) dx = 0 6.
22
(1 ) (1 )
dy
xy
dx
=+ +
7. y log y dx – x dy = 0 8.
5 5
dy
xy
dx
=−
9.
1
sin
dy
x
dx
−
=
10. e
x
tan y dx + (1 – e
x
) sec
2
y dy = 0
For each of the differential equations in Exercises 11 to 14, find a particular solution
satisfying the given condition:
11.
32
(
1)
dy
xxx
dx
+ ++
= 2x
2
+ x; y = 1 when x = 0
12.
2
( 1) 1
dy
xx
dx
−=
; y = 0 when x = 2
13.
cos
dy
a
dx
=
(a ∈ R); y = 1 when x = 0
14.
tan
dy
yx
dx
=
; y = 1 when x = 0
15. Find the equation of a curve passing through the point (0, 0) and whose differential
equation is y′ = e
x
sin x.
16. For the differential equation
( 2) ( 2)
dy
xy x y
dx
=+ +
, find the solution curve
passing through the point (1, –1).
17. Find the equation of a curve passing through the point (0, –2) given that at any
point (x, y) on the curve, the product of the slope of its tangent and y coordinate
of the point is equal to the x coordinate of the point.
18. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the
line segment joining the point of contact to the point (– 4, –3). Find the equation
of the curve given that it passes through (–2, 1).
19. The volume of spherical balloon being inflated changes at a constant rate. If
initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of
balloon after t seconds.
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DIFFERENTIAL EQUATIONS 397
20. In a bank, principal increases continuously at the rate of r% per year. Find the
value of r if Rs 100 double itself in 10 years (log
e
2 = 0.6931).
21. In a bank, principal increases continuously at the rate of 5% per year. An amount
of Rs 1000 is deposited with this bank, how much will it worth after 10 years
(e
0.5
= 1.648).
22. In a culture, the bacteria count is 1,00,000. The number is increased by 10% in 2
hours. In how many hours will the count reach 2,00,000, if the rate of growth of
bacteria is proportional to the number present?
23. The general solution of the differential equation
xy
dy
e
dx
+
=
is
(A) e
x
+ e
–y
= C (B) e
x
+ e
y
= C
(C) e
–x
+ e
y
= C (D) e
–x
+ e
–y
= C
9.5.2 Homogeneous differential equations
Consider the following functions in x and y
F
1
(x, y) = y
2
+ 2xy,F
2
(x, y) = 2x – 3y,
F
3
(x, y) =
cos
y
x
,F
4
(x, y) = sin x + cos y
If we replace x and y by λx and λy respectively in the above functions, for any nonzero
constant λ, we get
F
1
(λx, λy) = λ
2
(y
2
+ 2xy) = λ
2
F
1
(x, y)
F
2
(λx, λy) = λ (2x – 3y) = λ F
2
(x, y)
F
3
(λx, λy) =
cos cos
yy
xx
λ
=
λ
= λ
0
F
3
(x, y)
F
4
(λx, λy) = sin λx + cos λy ≠ λ
n
F
4
(x, y), for any n ∈ N
Here, we observe that the functions F
1
, F
2
, F
3
can be written in the form
F(λx, λy) = λ
n
F
(x, y) but F
4
can not be written in this form. This leads to the following
definition:
A function F(x, y) is said to be homogeneous function of degree n if
F(λx, λy) = λ
n
F(x, y) for any nonzero constant λ.
We note that in the above examples, F
1
, F
2
, F
3
are homogeneous functions of
degree 2, 1, 0 respectively but F
4
is not a homogeneous function.
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MATHEMATICS398
We also observe that
F
1
(x, y) =
2
22
1
2
2
yy y
x xh
xx
x
+=
or F
1
(x, y) =
2 2
2
2
1
xx
y yh
yy
+=
F
2
(x, y) =
1 1
3
3
2
yy
x xh
xx
−=
or F
2
(x, y) =
1 1
4
23
x x
y yh
y y
−=
F
3
(x, y) =
0 0
5
cos
yy
x xh
xx
=
F
4
(x, y) ≠
6
n
y
xh
x
, for any n ∈ N
or F
4
(x, y) ≠
7
n
x
yh
y
, for any n ∈ N
Therefore, a function F (x, y) is a homogeneous function of degree n if
F (x, y) =
or
nn
y x
xg yh
x
y
A differential equation of the form
dy
dx
= F (x, y) is said to be homogenous if
F(x, y) is a homogenous function of degree zero.
To solve a homogeneous differential equation of the type
()
F,
dy
xy
dx
=
=
y
g
x
... (1)
We make the substitution y = v . x ... (2)
Differentiating equation (2) with respect to x, we get
dy
dx
=
dv
vx
dx
+
... (3)
Substituting the value of
dy
dx
from equation (3) in equation (1), we get
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DIFFERENTIAL EQUATIONS 399
dv
vx
dx
+
= g (v)
or
dv
x
dx
= g (v) – v ... (4)
Separating the variables in equation (4), we get
()
dv
gv v
−
=
dx
x
... (5)
Integrating both sides of equation (5), we get
()
dv
gv v
−
=
1
C
dx
x
+
... (6)
Equation (6) gives general solution (primitive) of the differential equation (1) when
we replace v by
y
x
.
Note If the homogeneous differential equation is in the form
F( , )
dx
xy
dy
=
where, F (x, y) is homogenous function of degree zero, then we make substitution
x
v
y
=
i.e., x = vy and we proceed further to find the general solution as discussed
above by writing
F( , ) .
dx x
xy h
dy y
==
Example 15 Show that the differential equation (x – y)
dy
dx
= x + 2y is homogeneous
and solve it.
Solution The given differential equation can be expressed as
dy
dx
=
2
xy
xy
+
−
... (1)
Let F (x, y) =
2
xy
xy
+
−
Now F(λx, λy) =
0
( 2)
(, )
()
xy
f xy
xy
λ+
=λ ⋅
λ−
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MATHEMATICS400
Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential
equation is a homogenous differential equation.
Alternatively,
2
1
1
y
dy
x
y
dx
x
+
=
−
=
y
g
x
... (2)
R.H.S. of differential equation (2) is of the form
g
y
x
and so it is a homogeneous
function of degree zero. Therefore, equation (1) is a homogeneous differential equation.
To solve it we make the substitution
y = vx ... (3)
Differentiating equation (3) with respect to, x we get
dy
dx
=
dv
vx
dx
+
... (4)
Substituting the value of y and
dy
dx
in equation (1) we get
dv
vx
dx
+
=
12
1
v
v
+
−
or
dv
x
dx
=
12
1
v
v
v
+
−
−
or
dv
x
dx
=
2
1
1
vv
v
++
−
or
2
1
1
v
dv
vv
−
++
=
dx
x
−
Integrating both sides of equation (5), we get
=
or = – log | x| + C
1
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DIFFERENTIAL EQUATIONS 401
or
or
or
or (Why?)
Replacing v by
y
x
, we get
or
or
2
21
1
2
12
log 1 3 tan C
2
3
y y yx
x
x
x
x
−
+
++ = +
or
2 2 1
1
2
log ( ) 2 3 tan 2C
3
yx
y xy x
x
−
+
++ = +
or
2 2 1
2
log ( ) 2 3 tan
C
3
−
+
++ = +
xy
x xy y
x
which is the general solution of the differential equation (1)
Example 16 Show that the differential equation
cos cos
y dy y
x
yx
x dx x
=+
is
homogeneous and solve it.
Solution The given differential equation can be written as
dy
dx
=
cos
cos
y
yx
x
y
x
x
+
... (1)
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MATHEMATICS402
It is a differential equation of the form
F( , )
dy
xy
dx
=
.
Here F (x, y) =
cos
cos
y
yx
x
y
x
x
+
Replacing x by λx and y by λy, we get
F (λx, λy) =
0
[ cos ]
[F( , )]
cos
y
yx
x
xy
y
x
x
λ+
=λ
λ
Thus, F(x, y) is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation.
To solve it we make the substitution
y = vx ... (2)
Differentiating equation (2) with respect to x, we get
dy
dx
=
dv
vx
dx
+
... (3)
Substituting the value of y and
dy
dx
in equation (1), we get
dv
vx
dx
+
=
cos 1
cos
vv
v
+
or
dv
x
dx
=
cos 1
cos
vv
v
v
+
−
or
dv
x
dx
=
1
cos
v
or cos v dv =
dx
x
Therefore
cos
v dv
∫
=
1
dx
x
∫
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DIFFERENTIAL EQUATIONS 403
or sin v = log | x | + log |C|
or sin v = log | Cx |
Replacing v by
y
x
, we get
sin
y
x
= log | Cx |
which is the general solution of the differential equation (1).
Example 17 Show that the differential equation
2 20
x x
y y
y e dx y x e dy
+− =
is
homogeneous and find its particular solution, given that, x = 0 when y = 1.
Solution The given differential equation can be written as
dx
dy
=
2
2
x
y
x
y
xe y
ye
−
... (1)
Let F(x, y) =
2
2
x
y
x
y
xe y
ye
−
Then F (λx, λy) =
0
2
[F( , )]
2
x
y
x
y
xe y
xy
ye
λ−
=λ
λ
Thus, F(x, y) is a homogeneous function of degree zero. Therefore, the given
differential equation is a homogeneous differential equation.
To solve it, we make the substitution
x = vy ... (2)
Differentiating equation (2) with respect to y, we get
dx
dy
=
+
dv
vy
dy
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MATHEMATICS404
Substituting the value of
and
dx
x
dy
in equation (1), we get
dv
vy
dy
+
=
21
2
v
v
ve
e
−
or
dv
y
dy
=
21
2
v
v
ve
v
e
−
−
or
dv
y
dy
=
1
2
v
e
−
or 2e
v
dv =
dy
y
−
or
2
v
e dv
⋅
∫
=
dy
y
−
∫
or 2 e
v
= – log |y| + C
and replacing v by
x
y
, we get
2
x
y
e
+ log | y | = C ... (3)
Substituting x = 0 and y = 1 in equation (3), we get
2 e
0
+ log | 1| = C ⇒ C = 2
Substituting the value of C in equation (3), we get
2
x
y
e
+ log | y | = 2
which is the particular solution of the given differential equation.
Example 18 Show that the family of curves for which the slope of the tangent at any
point (x, y) on it is
22
2
xy
xy
+
, is given by x
2
– y
2
= cx.
Solution We know that the slope of the tangent at any point on a curve is
dy
dx
.
Therefore,
dy
dx
=
22
2
xy
xy
+
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DIFFERENTIAL EQUATIONS 405
or
dy
dx
=
2
2
1
2
y
x
y
x
+
... (1)
Clearly, (1) is a homogenous differential equation. To solve it we make substitution
y = vx
Differentiating y = vx with respect to x, we get
dy
dx
=
dv
vx
dx
+
or
dv
vx
dx
+
=
2
1
2
v
v
+
or
dv
x
dx
=
2
1
2
v
v
−
2
2
1
v
dv
v
−
=
dx
x
or
2
2
1
v
dv
v
−
=
dx
x
−
Therefore
2
2
1
v
dv
v
−
∫
=
1
dx
x
−
∫
or log | v
2
– 1 | = – log | x | + log | C
1
|
or log | (v
2
– 1) (x) | = log |C
1
|
or (v
2
– 1) x = ± C
1
Replacing v by
y
x
, we get
2
2
1
y
x
x
−
= ± C
1
or (y
2
– x
2
) = ± C
1
x or x
2
– y
2
= Cx
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EXERCISE 9.5
In each of the Exercises 1 to 10, show that the given differential equation is homogeneous
and solve each of them.
1. (x
2
+ xy) dy = (x
2
+ y
2
) dx 2.
xy
y
x
+
′
=
3. (x – y) dy – (x + y) dx = 0 4. (x
2
– y
2
) dx + 2xy dy = 0
5.
2 22
2
dy
x x y xy
dx
=− +
6. x dy – y dx =
22
x y dx
+
7.
cos sin sin cos
yy
yy
x y y dx y x x dy
xx
xx
+ =−
8.
sin 0
dy y
x yx
dx x
−+ =
9.
log 2 0
y
y dx x dy x dy
x
+ −=
10.
For each of the differential equations in Exercises from 11 to 15, find the particular
solution satisfying the given condition:
11. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1
12. x
2
dy + (xy + y
2
) dx = 0; y = 1 when x = 1
13.
when x = 1
14.
cosec 0
dy y y
dx x x
−+ =
; y = 0 when x = 1
15.
22
2 20
dy
xy y x
dx
+− =
; y = 2 when x = 1
16. A homogeneous differential equation of the from
dx x
h
dy y
=
can be solved by
making the substitution.
(A) y = vx (B) v = yx (C) x = vy (D) x = v
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17. Which of the following is a homogeneous differential equation?
(A) (4x + 6y + 5) dy – (3y + 2x + 4) dx = 0
(B) (xy) dx – (x
3
+ y
3
) dy = 0
(C) (x
3
+ 2y
2
) dx + 2xy dy = 0
(D) y
2
dx + (x
2
– xy – y
2
) dy = 0
9.5.3 Linear differential equations
A differential equation of the from
P
dy
y
dx
+
= Q
where, P and Q are constants or functions of x only, is known as a first order linear
differential equation. Some examples of the first order linear differential equation are
dy
y
dx
+
= sin x
1
dy
y
dx x
+
= e
x
log
dy y
dx x x
+
=
1
x
Another form of first order linear differential equation is
1
P
dx
x
dy
+
= Q
1
where, P
1
and Q
1
are constants or functions of y only. Some examples of this type of
differential equation are
dx
x
dy
+
= cos y
2
dx x
dy y
−
+
= y
2
e
– y
To solve the first order linear differential equation of the type
P
dy
y
dx
+
= Q ... (1)
Multiply both sides of the equation by a function of x say g (x) to get
g(x)
dy
dx
+ P.(g(x)) y = Q.g (x) ... (2)
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MATHEMATICS408
Choose g (x) in such a way that R.H.S. becomes a derivative of y . g (x).
i.e. g (x)
dy
dx
+ P. g (x) y =
d
dx
[y . g (x)]
or g (x)
dy
dx
+ P. g(x) y = g (x)
dy
dx
+ y g′ (x)
⇒ P.g (x) = g′ (x)
or P =
()
()
gx
gx
′
Integrating both sides with respect to x, we get
P
dx
=
()
()
gx
dx
gx
′
or
P
dx
⋅
= log (g (x))
or g (x) =
P
dx
e
On multiplying the equation (1) by g(x) =
P
dx
e
, the L.H.S. becomes the derivative
of some function of x and y. This function g(x) =
P
dx
e
is called Integrating Factor
(I.F.) of the given differential equation.
Substituting the value of g (x) in equation (2), we get
=
or
d
dx
ye
dxP
∫
=
Integrating both sides with respect to x, we get
=
Q
P
.e dx
dx
∫
∫
or y =
e e dx
dx dx−
∫∫
+
∫
PP
QC
..
which is the general solution of the differential equation.
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Steps involved to solve first order linear differential equation:
(i) Write the given differential equation in the form
PQ
dy
y
dx
+=
where P, Q are
constants or functions of x only.
(ii) Find the Integrating Factor (I.F) =
.
(iii) Write the solution of the given differential equation as
y (I.F) =
In case, the first order linear differential equation is in the form
11
PQ
dx
x
dy
+=
,
where, P
1
and Q
1
are constants or functions of y only. Then I.F =
1
P
dy
e
and the
solution of the differential equation is given by
x . (I.F) =
(
)
1
Q × I.F C
dy
+
Example 19 Find the general solution of the differential equation
cos
dy
yx
dx
−=
.
Solution Given differential equation is of the form
PQ
dy
y
dx
+=
, where P = –1 and Q = cos x
Therefore I.F =
Multiplying both sides of equation by I.F, we get
= e
–x
cos x
or
()
x
dy
ye
dx
−
= e
–x
cos x
On integrating both sides with respect to x, we get
ye
– x
=
cos C
x
e x dx
−
+
... (1)
Let I =
cos
x
e x dx
−
=
cos ( sin ) ( )
1
x
x
e
x x e dx
−
−
−− −
−
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MATHEMATICS410
=
cos sin
x x
xe xe dx
− −
−−
∫
=
cos sin (– ) cos ( )
xx x
x e x e x e dx
− − −
−− −−
∫
=
cos sin cos
x x x
xe xe xe dx
− − −
−+−
∫
or I = – e
–x
cos x + sin x e
–x
– I
or 2I = (sin x – cos x) e
–x
or I =
(sin cos )
2
x
x xe
−
−
Substituting the value of I in equation (1), we get
ye
– x
=
sin cos
C
2
x
xx
e
−
−
+
or y =
sin cos
C
2
x
xx
e
−
+
which is the general solution of the given differential equation.
Example 20 Find the general solution of the differential equation
2
2 ( 0)
dy
x yxx
dx
+= ≠
.
Solution The given differential equation is
2
dy
xy
dx
+
= x
2
... (1)
Dividing both sides of equation (1) by x, we get
2
dy
y
dx x
+
= x
which is a linear differential equation of the type
PQ
dy
y
dx
+=
, where
2
P
x
=
and Q = x.
So I.F =
2
dx
x
e
∫
= e
2 log x
=
2
log 2
x
ex
=
log ( )
[ ( )]
fx
as e f x
=
Therefore, solution of the given equation is given by
y . x
2
=
2
( )( ) C
x x dx
+
∫
=
3
C
x dx
+
∫
or y =
2
2
C
4
x
x
−
+
which is the general solution of the given differential equation.
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DIFFERENTIAL EQUATIONS 411
Example 21 Find the general solution of the differential equation y dx – (x + 2y
2
) dy = 0.
Solution The given differential equation can be written as
dx x
dy y
−
= 2y
This is a linear differential equation of the type
11
PQ
dx
x
dy
+=
, where
1
1
P
y
=−
and
Q
1
= 2y. Therefore
1
1
log log ( )
1
I.F
dy
yy
y
e ee
y
−
−
−
= == =
Hence, the solution of the given differential equation is
1
x
y
=
1
(2 ) C
y dy
y
+
or
x
y
=
(2 ) C
dy
+
or
x
y
= 2y + C
or x = 2y
2
+ Cy
which is a general solution of the given differential equation.
Example 22 Find the particular solution of the differential equation
cot
+
dy
yx
dx
= 2x + x
2
cot x (x ≠ 0)
given that y = 0 when
2
x
π
=
.
Solution The given equation is a linear differential equation of the type
PQ
dy
y
dx
+=
,
where P = cot x and Q = 2x + x
2
cot x. Therefore
I.F =
ee x
x dx
x
cot
log sin
sin
∫
==
Hence, the solution of the differential equation is given by
y . sin x = ∫ (2x + x
2
cot x) sin x dx + C
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MATHEMATICS412
or y sin x = ∫2x sin x dx + ∫x
2
cos x dx + C
or y sin x =
2 2
2
2 2
sin cos cos C
2 2
x x
x x dx x x dx
− + +
∫∫
or y sin x =
2 2 2
sin cos cos C
x x x x dx x x dx
− + +
∫∫
or y sin x = x
2
sin x + C ... (1)
Substituting y = 0 and
2
x
π
=
in equation (1), we get
0 =
2
sin C
22
ππ
+
or C =
2
4
−π
Substituting the value of C in equation (1), we get
y sin x =
2
2
sin
4
xx
π
−
or y =
2
2
(sin 0)
4 sin
xx
x
π
− ≠
which is the particular solution of the given differential equation.
Example 23 Find the equation of a curve passing through the point (0, 1). If the slope
of the tangent to the curve at any point (x, y) is equal to the sum of the x coordinate
(abscissa) and the product of the x coordinate and y coordinate (ordinate) of that point.
Solution We know that the slope of the tangent to the curve is
dy
dx
.
Therefore,
dy
dx
= x + xy
or
dy
xy
dx
−
= x ... (1)
This is a linear differential equation of the type
PQ
dy
y
dx
+=
, where P = – x and Q = x.
Therefore, I.F =
2
2
x
x dx
ee
−
−
∫
=
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Hence, the solution of equation is given by
2
2
x
ye
−
⋅
=
(
)
2
2
() C
x
x dx
e
−
+
∫
... (2)
Let I =
2
2
()
x
x dx
e
−
∫
Let
2
2
x
t
−
=
, then – x dx = dt or x dx = – dt.
Therefore, I =
2
2
–
x
tt
e dt e
e
−
− =− =
∫
Substituting the value of I in equation (2), we get
2
2
x
y
e
−
=
2
2
+C
−
−
x
e
or y =
2
2
1C
x
e
−+
... (3)
Now (3) represents the equation of family of curves. But we are interested in
finding a particular member of the family passing through (0, 1). Substituting x = 0 and
y = 1 in equation (3) we get
1 = – 1 + C . e
0
or C = 2
Substituting the value of C in equation (3), we get
y =
2
2
12
x
e
−+
which is the equation of the required curve.
EXERCISE 9.6
For each of the differential equations given in Exercises 1 to 12, find the general solution:
1.
2 sin
dy
yx
dx
+=
2.
2
3
x
dy
ye
dx
−
+=
3.
2
dy y
x
dx x
+=
4.
(sec ) tan 0
2
dy
xy x x
dx
π
+ = ≤<
5.
2
cos tan
dy
xy x
dx
+=
0
2
x
π
≤<
6.
2
2 log
dy
x yx x
dx
+=
7.
2
log log
dy
xx y x
dx x
+=
8. (1 + x
2
) dy + 2xy dx = cot x dx (x ≠ 0)
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MATHEMATICS414
9.
cot 0 ( 0)
dy
x y x xy x x
dx
+−+ = ≠
10.
()1
dy
xy
dx
+=
11. y dx + (x – y
2
) dy = 0 12.
2
( 3 ) ( 0)
dy
x y yy
dx
+ =>
.
For each of the differential equations given in Exercises 13 to 15, find a particular
solution satisfying the given condition:
13.
2 tan sin ; 0 when
3
dy
y x xy x
dx
π
+ == =
14.
2
2
1
(1 ) 2 ; 0 when 1
1
dy
x xy y x
dx
x
+ += = =
+
15.
3 cot sin 2 ; 2 when
2
dy
y x xy x
dx
π
−= = =
16. Find the equation of a curve passing through the origin given that the slope of the
tangent to the curve at any point (x, y) is equal to the sum of the coordinates of
the point.
17. Find the equation of a curve passing through the point (0, 2) given that the sum of
the coordinates of any point on the curve exceeds the magnitude of the slope of
the tangent to the curve at that point by 5.
18. The Integrating Factor of the differential equation
2
2
dy
x yx
dx
−=
is
(A) e
–x
(B) e
–y
(C)
1
x
(D) x
19. The Integrating Factor of the differential equation
2
(1 )
dx
y yx
dy
−+
=
( 1 1)
−< <
ay y
is
(A)
2
1
1
y
−
(B)
2
1
1
y
−
(C)
2
1
1
y
−
(D)
2
1
1
y
−
Miscellaneous Examples
Example 24 Verify that the function y = c
1
e
ax
cos bx + c
2
e
ax
sin bx, where c
1
, c
2
are
arbitrary constants is a solution of the differential equation
()
2
22
2
2
0
d y dy
a a by
dx
dx
− ++ =
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DIFFERENTIAL EQUATIONS 415
Solution The given function is
y = e
ax
[c
1
cosbx + c
2
sinbx] ... (1)
Differentiating both sides of equation (1) with respect to x, we get
dy
dx
=
[
]
[
]
1 2 1 2
– sin cos cos sin
ax ax
e bc bx b c bx c bx c bx e a
+ ++ ⋅
or
dy
dx
=
21 21
[( )cos ( )sin ]
ax
e bc ac bx ac bc bx
+ +−
... (2)
Differentiating both sides of equation (2) with respect to x, we get
2
2
dy
dx
=
21 21
[( ) ( sin ) ( ) ( cos )]
ax
e bc ac b bx ac bc b bx
+ − +−
+
21 21
[( ) cos ( ) sin ] .
ax
bc ac bx ac bc bx e a
+ +−
=
2 2 2 2
2 12 1 21
[(2)sin(2)cos]
ax
e a c ab c b c bx a c ab c b c bx
− − ++ −
Substituting the values of
2
2
,
d y dy
dx
dx
and y in the given differential equation, we get
L.H.S. =
2 2 2 2
2 12 1 21
[2)sin(2)cos]
ax
e a c abc b c bx a c abc b c bx
− − ++ −
21 21
2 [( )cos ( )sin ]
ax
ae bc ac bx ac bc bx
− + +−
22
12
( ) [ cos sin ]
ax
a b e c bx c bx
++ +
=
(
)
2 22 22
2122122
2 2 222
1 21 2 111
2 22 sin
(2 22 )cos
ax
a c abc b c a c abc a c b c bx
e
a c abc b c abc a c a c b c bx
−−−+++
+ + −− − ++
=
[0 sin 0 cos ]
ax
e bx bx
×+
= e
ax
× 0 = 0 = R.H.S.
Hence, the given function is a solution of the given differential equation.
Example 25 Form the differential equation of the family of circles in the second
quadrant and touching the coordinate axes.
Solution Let C denote the family of circles in the second quadrant and touching the
coordinate axes. Let (–a, a) be the coordinate of the centre of any member of
this family (see Fig 9.6).
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MATHEMATICS416
X
X
’
Y
Y
’
(– , )aa
O
Equation representing the family C is
(x + a)
2
+ (y – a)
2
= a
2
... (1)
or x
2
+ y
2
+ 2ax – 2ay + a
2
= 0 ... (2)
Differentiating equation (2) with respect to x, we get
22 22
dy dy
x y aa
dx dx
+ +−
= 0
or
dy
xy
dx
+
=
1
dy
a
dx
−
or a =
1
x yy
y
′
+
′
−
Substituting the value of a in equation (1), we get
2
2
1 1
x yy x yy
x y
y y
′ ′
+ +
+ +−
′ ′
− −
=
2
1
x yy
y
′
+
′
−
or [xy′ – x + x + y y′]
2
+ [y y′ – y – x – y y′]
2
= [x + y y′]
2
or (x + y)
2
y′
2
+ [x + y]
2
= [x + y y′]
2
or (x + y)
2
[(y′)
2
+ 1] = [x + y y′]
2
which is the differential equation representing the given family of circles.
Example 26 Find the particular solution of the differential equation
log 3 4
dy
xy
dx
=+
given that y = 0 when x = 0.
Solution The given differential equation can be written as
dy
dx
= e
(3x + 4y)
or
dy
dx
= e
3x
. e
4y
... (1)
Separating the variables, we get
4
y
dy
e
= e
3x
dx
Therefore
4 y
e dy
−
∫
=
3x
e dx
∫
Fig 9.6
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DIFFERENTIAL EQUATIONS 417
or
4
4
y
e
−
−
=
3
C
3
x
e
+
or 4 e
3x
+ 3 e
– 4y
+ 12 C = 0 ... (2)
Substituting x = 0 and y = 0 in (2), we get
4 + 3 + 12 C = 0 or C =
7
12
−
Substituting the value of C in equation (2), we get
4 e
3x
+ 3 e
– 4y
– 7 = 0,
which is a particular solution of the given differential equation.
Example 27 Solve the differential equation
(x dy – y dx) y sin
y
x
= (y dx + x dy) x cos
y
x
.
Solution The given differential equation can be written as
2 2
sin cos cos sin
yy
yy
x y x dy xy y dx
xx
xx
− = +
or
dy
dx
=
2
2
cos sin
sin cos
yy
xy y
xx
yy
xy x
xx
+
−
Dividing numerator and denominator on RHS by x
2
, we get
dy
dx
=
2
2
cos sin
sin cos
y yy y
xx x
x
yy y
xx x
+
−
... (1)
Clearly, equation (1) is a homogeneous differential equation of the form
dy y
g
dx x
=
.
To solve it, we make the substitution
y = vx ... (2)
or
dy
dx
=
dv
vx
dx
+
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MATHEMATICS418
or
dv
vx
dx
+
=
2
cos sin
sin cos
v vv v
vv v
+
−
(using (1) and (2))
or
dv
x
dx
=
2 cos
sin cos
vv
vv v
−
or
sin cos
cos
vv v
dv
vv
−
=
2
dx
x
Therefore
sin cos
cos
vv v
dv
vv
−
∫
=
1
2
dx
x
∫
or
1
tan
v dv dv
v
−
∫∫
=
1
2
dx
x
∫
or
log sec log | |
vv
−
=
1
2log | | log | C |
x
+
or
2
sec
log
v
vx
= log | C
1
|
or
2
sec
v
vx
= ± C
1
... (3)
Replacing v by
y
x
in equation (3), we get
2
sec
()
y
x
y
x
x
= C where, C = ± C
1
or
sec
y
x
= C xy
which is the general solution of the given differential equation.
Example 28 Solve the differential equation
(tan
–1
y
– x) dy = (1 + y
2
) dx.
Solution The given differential equation can be written as
2
1
dx x
dy
y
+
+
=
1
2
tan
1
y
y
−
+
... (1)
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DIFFERENTIAL EQUATIONS 419
Now (1) is a linear differential equation of the form
1
P
dx
dy
+
x = Q
1
,
where, P
1
=
2
1
1
y
+
and
1
1
2
tan
Q
1
y
y
−
=
+
.
Therefore, I . F =
1
2
1
tan
1
dy
y
y
ee
−
+
∫
=
Thus, the solution of the given differential equation is
1
tan
y
xe
−
=
1
1
tan
2
tan
C
1
y
y
e dy
y
−
−
+
+
∫
... (2)
Let I =
1
1
tan
2
tan
1
y
y
e dy
y
−
−
+
∫
Substituting tan
–1
y = t so that
2
1
1
dy dt
y
=
+
, we get
I =
t
t e dt
∫
= t e
t
– ∫1 . e
t
dt = t e
t
– e
t
= e
t
(t – 1)
or I =
1
tan
y
e
−
(tan
–1
y –1)
Substituting the value of I in equation (2), we get
1 1
tan tan 1
. (tan 1) C
yy
xe e y
−−
−
= −+
or x =
1
1 tan
(tan 1) C
y
ye
−
−−
−+
which is the general solution of the given differential equation.
Miscellaneous Exercise on Chapter 9
1. For each of the differential equations given below, indicate its order and degree
(if defined).
(i)
2
2
2
5 6 log
d y dy
x yx
dx
dx
+ −=
(ii)
3 2
4 7 sin
dy dy
yx
dx dx
− +=
(iii)
4 3
4 3
sin 0
dy dy
dx dx
−=
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MATHEMATICS420
2. For each of the exercises given below, verify that the given function (implicit or
explicit) is a solution of the corresponding differential equation.
(i) xy = a e
x
+ b e
–x
+ x
2
:
2
2
2
2 20
d y dy
x xy x
dx
dx
+ − + −=
(ii) y = e
x
(a cos x + b sin x) :
2
2
2 20
d y dy
y
dx
dx
− +=
(iii) y = x sin 3x :
2
2
9 6cos3 0
dy
yx
dx
+− =
(iv) x
2
= 2y
2
log y :
22
() 0
dy
x y xy
dx
+ −=
3. Form the differential equation representing the family of curves given by
(x – a)
2
+ 2y
2
= a
2
, where a is an arbitrary constant.
4. Prove that x
2
– y
2
= c (x
2
+ y
2
)
2
is the general solution of differential equation
(x
3
– 3x y
2
) dx = (y
3
– 3x
2
y) dy, where c is a parameter.
5. Form the differential equation of the family of circles in the first quadrant which
touch the coordinate axes.
6. Find the general solution of the differential equation
2
2
1
0
1
dy y
dx
x
−
+=
−
.
7. Show that the general solution of the differential equation
2
2
1
0
1
dy y y
dx
xx
++
+=
++
is
given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter.
8. Find the equation of the curve passing through the point
0,
4
π
whose differential
equation is sin x cos y dx + cos x sin y dy = 0.
9. Find the particular solution of the differential equation
(1 + e
2x
) dy + (1 + y
2
) e
x
dx = 0, given that y = 1 when x = 0.
10. Solve the differential equation
2
( 0)
x x
y y
y e dx x e y dy y
=+ ≠
.
11. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy,
given that y = –1, when x = 0. (Hint: put x – y = t)
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DIFFERENTIAL EQUATIONS 421
12. Solve the differential equation
2
1( 0)
x
e y dx
x
dy
xx
−
− =≠
.
13. Find a particular solution of the differential equation
cot
dy
yx
dx
+
= 4x cosec x
(x ≠ 0), given that y = 0 when
2
x
π
=
.
14. Find a particular solution of the differential equation (x + 1)
dy
dx
= 2 e
–y
– 1, given
that y = 0 when x = 0.
15. The population of a village increases continuously at the rate proportional to the
number of its inhabitants present at any time. If the population of the village was
20, 000 in 1999 and 25000 in the year 2004, what will be the population of the
village in 2009?
16. The general solution of the differential equation
0
y dx x dy
y
−
=
is
(A) xy = C (B) x = Cy
2
(C) y = Cx (D) y = Cx
2
17. The general solution of a differential equation of the type
11
PQ
dx
x
dy
+=
is
(A)
(
)
1 1
P P
1
Q C
dy dy
y e e dy
∫∫
= +
∫
(B)
(
)
1 1
P P
1
.Q C
dx dx
y e e dx
∫∫
= +
∫
(C)
(
)
1 1
P P
1
QC
dy dy
x e e dy
∫∫
=
+
∫
(D)
(
)
1 1
P P
1
Q C
dx dx
x e e dx
∫∫
= +
∫
18. The general solution of the differential equation e
x
dy + (y e
x
+ 2x) dx = 0 is
(A) x e
y
+ x
2
= C (B) x e
y
+ y
2
= C
(C) y e
x
+ x
2
= C (D) y e
y
+ x
2
= C
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MATHEMATICS422
Summary
An equation involving derivatives of the dependent variable with respect to
independent variable (variables) is known as a differential equation.
Order of a differential equation is the order of the highest order derivative
occurring in the differential equation.
Degree of a differential equation is defined if it is a polynomial equation in its
derivatives.
Degree (when defined) of a differential equation is the highest power (positive
integer only) of the highest order derivative in it.
A function which satisfies the given differential equation is called its solution.
The solution which contains as many arbitrary constants as the order of the
differential equation is called a general solution and the solution free from
arbitrary constants is called particular solution.
To form a differential equation from a given function we differentiate the
function successively as many times as the number of arbitrary constants in
the given function and then eliminate the arbitrary constants.
Variable separable method is used to solve such an equation in which variables
can be separated completely i.e. terms containing y should remain with dy
and terms containing x should remain with dx.
A differential equation which can be expressed in the form
(,)or (,)
dy dx
f xy gxy
dx dy
= =
where, f (x, y) and g(x, y) are homogenous
functions of degree zero is called a homogeneous differential equation.
A differential equation of the form
+P Q
dy
y
dx
=
, where P and Q are constants
or functions of x only is called a first order linear differential equation.
Historical Note
One of the principal languages of Science is that of differential equations.
Interestingly, the date of birth of differential equations is taken to be November,
11,1675, when Gottfried Wilthelm Freiherr Leibnitz (1646 - 1716) first put in black
and white the identity
2
1
2
y dy y
=
∫
, thereby introducing both the symbols ∫ and dy.
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DIFFERENTIAL EQUATIONS 423
Leibnitz was actually interested in the problem of finding a curve whose tangents
were prescribed. This led him to discover the ‘method of separation of variables’
1691. A year later he formulated the ‘method of solving the homogeneous
differential equations of the first order’. He went further in a very short time
to the discovery of the ‘method of solving a linear differential equation of the
first-order’. How surprising is it that all these methods came from a single man
and that too within 25 years of the birth of differential equations!
In the old days, what we now call the ‘solution’ of a differential equation,
was used to be referred to as ‘integral’ of the differential equation, the word
being coined by James Bernoulli (1654 - 1705) in 1690. The word ‘solution was
first used by Joseph Louis Lagrange (1736 - 1813) in 1774, which was almost
hundred years since the birth of differential equations. It was Jules Henri Poincare
(1854 - 1912) who strongly advocated the use of the word ‘solution’ and thus the
word ‘solution’ has found its deserved place in modern terminology. The name of
the ‘method of separation of variables’ is due to John Bernoulli (1667 - 1748),
a younger brother of James Bernoulli.
Application to geometric problems were also considered. It was again John
Bernoulli who first brought into light the intricate nature of differential equations.
In a letter to Leibnitz, dated May 20, 1715, he revealed the solutions of the
differential equation
x
2
y″ = 2y,
which led to three types of curves, viz., parabolas, hyperbolas and a class of
cubic curves. This shows how varied the solutions of such innocent looking
differential equation can be. From the second half of the twentieth century attention
has been drawn to the investigation of this complicated nature of the solutions of
differential equations, under the heading ‘qualitative analysis of differential
equations’. Now-a-days, this has acquired prime importance being absolutely
necessary in almost all investigations.
—
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