Given a family of functions parameterized by some constants, a differential equation can be formed by eliminating those constants of this family.

**Formation of
Differential Equations from Geometrical Problems**

Given a
family of functions parameterized by some constants, a differential equation
can be formed by eliminating those constants of this family. For instance, the
elimination of constants A and B from *y*
=
A*e ^{x}* +
B

Consider
an equation of a family of curves, which contains *n* arbitrary constants. To form a differential equation not
containing any of these constants, let us proceed as follows:

Differentiate
the given equation successively *n*
times, getting *n* differential
equations. Then eliminate *n* arbitrary
constants from (*n* +1)
equations made up of the given equation and *n*
newly obtained equations arising from *n*
successive differentiations. The result of elimination gives the required
differential equation which must contain a derivative of the *n*th order.

** **

**Example 10.2**

Find the
differential equation for the family of all straight lines passing through the
origin.

**Solution**

The
family of straight lines passing through the origin is *y* = *mx* , where *m* is an arbitrary constant. … (1)

Differentiating
both sides with respect to *x*, we get

From (1)
and (2), we get *y *=* x* *dy*/*dx*. This is the required differential
equation.

Observe
that the given equation *y* =
*mx* contains only one arbitrary constant
and thus we get the differential equation of order one.

** **

**Example 10.3**

Form the
differential equation by eliminating the arbitrary constants A and B from *y *=* *A cos* x *+* *Bsin* x *.

**Solution**

Given
that

*
y *= A cos *x* + Bsin *x *... (1)

Differentiating
(1) twice successively, we get

Substituting
(1) in (3), we get *d*^{2}*y/dx*^{2} +
*y* = 0 as the required differential
equation.

Find the
differential equation of the family of circles passing through the points ( *a*, 0) and ( −*a*, 0) .

**Solution**

A circle
passing through the points ( *a*,
0)
and (
−*a*, 0) has its centre on *y* - axis.

Let (
0, *b*) be the centre of the circle. So,
the radius of the circle is √[*a*^{2}* *+* b*^{2}].

Therefore
the equation of the family of circles passing through the points (
*a*, 0) and ( −*a*, 0) is *x*^{2}* *+* *(* y *−* b*)^{2}* *=* a*^{2}* *+* b*^{2}* *, *b *is an arbitrary constant.

Differentiating
both sides of (1) with respect to *x*,
we get

Substituting
the value of *b* in equation (1), we
get

⇒ ( *x*^{2} − *y*^{2} − *a*^{2} ) *dy/dx* − 2*xy* = 0 , which is the required differential
equation.

** **

Find the
differential equation of the family of parabolas *y*^{2} = 4*ax* , where *a* is an
arbitrary constant.

The
equation of the family of parabolas is given by *y*^{2} = 4*ax* , *a* is an arbitrary
constant. ... (1)

Differentiating
both sides of (1) with respect to *x* ,
we get 2*y dy/dx *= 4 *a* ⇒ *a* = *y/2 dy*/*dx*

Substituting
the value of *a* in (1) and
simplifying, we get *dy*/*dx* = *y/2x*
as the required differential equation.

** **

Find the
differential equation of the family of all ellipses having foci on the *x* -axis and centre at the origin.

**Solution**

The
equation of the family of all ellipses having foci on the *x *-axis and centre at the origin is given by

where *a* and *b* are arbitrary constants.

Differentiating
equation (1) with respect to *x*, we
get

Substituting
the value of 1/a^{2} in equation (2) and simplifying, we get

which is
the required differential equation.

The
result of eliminating one arbitrary constant yields a first order differential
equation and that of eliminating two arbitrary constants leads to a second
order differential equation and so on.

Tags : Mathematics , 12th Maths : UNIT 10 : Ordinary Differential Equations

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12th Maths : UNIT 10 : Ordinary Differential Equations : Formation of Differential Equations from Geometrical Problems | Mathematics

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