Consider a family of
exponential curves (y = Ae^{x}), where A is an arbitrary constant for
different values of A, we get different members of the family. Differentiating
the relation (y = Ae^{x}) w.r.t.x, we get dy/dx = Ae^{x}

Eliminating the arbitrary
constant between y = Ae^{x} and dy/dx = Ae^{x}, we get
dy/dx = y. This is the differential equation of the family of curves
represented by y = Ae^{x}

Thus, by eliminating one
arbitrary constant, a differential equation of first order is obtained.

Now consider the family of
curves given by y = A cos 2x + B sin 2x ... (1)

Where A and B are
arbitrary consists.

Differentiating (1) w.r.t,
x we get dy/ dx = - 2Asin 2x + 2Bcos 2x ... (2)

Differentiating (2) w.r.t.
x we get d²y/ dx² = - 4Acos ax - 4Bsin 2x ... (3)

Eliminating A and B from
equations (1) and (2) (3), we get

d²y/ dx² = - 4y ⇒ d²y/ dx² + 4y = 0

Here we note that by
eliminating two arbitrary consists, a differential equation of second order is
obtained.

**Step
I:** write the given equation involving
independent variable x (say), dependent variable y (say) and the arbitrary
constants.

**Step
II:** obtained the numbers of a
arbitrary constants in step in step I. let there be n arbitrary consists.

**Step
III:** differentiate the relation in step
in times with respect to x.

**Step
IV:** eliminate arbitrary constants with
the help of n equations involving differential coefficient obtained in step III
and an equation in step I.

The equation so obtained
is the desired differential equation.

**Example:** Show that the differential
equation that represented all parabolas having their axis symmetry coincident
with the axis of x is yy₂ + y₁² = 0.

**Solution:** The equation that represents a
family of parabolas having their axis of symmetry coincident with the axis of x
is y² = 4a(x - h) ... (1)

This equation contains two
arbitrary constants, so we shall differentiate twice to obtain second order
differential equation.

Differentiating (1) w.r.t
x we get

2y dy/dx = 4a ⇒ y dy/ dx = 2a ... (2)

Differentiating (2) w.r.t,
x we get

y d²y/ dx² + (dy/ dx)² = 0 ⇒ yy₂ + y₁² = 0

Which is the required
differential equation.

**Example:** Find the differential
equation of all non-horizontal lines in a plane.

**Solution:** The equation of the family of all
non-horizontal line in a plane is given by

Ax + by = 1 ... (1)

Where a, b are arbitrary
constants such that (a ≠ 0)

Differentiating (1)
w.r.t, x we get

a dx/dy + b = 0

Differentiating this w.r.t
y we get

ad²x/ dy² = 0

⇒ d²x/
dy² = 0

Hence, the differential
equation of all non-horizontal lines in a plane is d²x/ dy² = 0.

**Example:** Find the differential equation of
all non- vertical lines in a plane.

**Solution:** The general equation of all
non-vertical lines in a plane is (ax + by = 1) where (b ≠ 0).

Now,

ax + by = 1

a + b dy/dx =
0 [differentiating
w.r.t.x]

b d²y/ dx =
0 [differentiating
w.r.t.x]

d²y/ dx² = 0 [∵ b ≠ 0]

Hence, the differential
equation is d²y/ dx² = 0

**Solution
of a differential equation:**** **The solution of a differential
equation is a relation between the variable involved which satisfies the
differential equation. Such a relation and the derivates obtained therefore
when substituted in the differential equation, makes left hand right hand sides
identically equal.

**General
solution: **The solution which
contains as many as arbitrary constants as the order of the differential
equations is called the general solution of the differential equation.

For example, y = Acos x +
Bsin x is the general solutions one arbitrary constant.

**Particular
solution: **Solution obtained by
giving particular values to the arbitrary constant in the general solution of a
differential equation is called a particular solution.

**Example:** Show that xy = ae^{x} +
be^{- x} + x² is a solution of the differential equation x
d²y/ dx² + 2 dy/dx - xy + x² - 2 = 0

**Solution:** We are given that xy =
ae^{x} + be^{- x} + x² ... (1)

Differentiating w.r.t.x,
we get x dy/ dx + y = ae^{x} - be^{- x} + 2x

Differentiating again
w.r.t.x, we get xd²y/ dx² + dy/dx + dy/dx = ae^{x} + be^{-
x} + 2

xd²y/ dx² + 2dy/dx = ae^{x} +
be^{- x} + 2 ... (2)

Now x d²y/ dx² + 2dy/dx -
xy + x2 - 2

= [ae^{x} + be^{- x} + 2] - [ae^{x} + be^{- x} + x²] + x² - 2

= 0 [using (1) and (2)]

Thus, (xy = ae^{x} + be^{- x} + x²) is a solution of the given differential equation.