ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 4

Tuesday, 5 August 2014

CHAPTER 4 - Worked Out Examples

Example: 1

Find the DE corresponding to the family of rectangular hyperbolas $xy=c^2$

Solution: 1	Steps Involved: 1

Since the equation for a rectangular hyperbola contains only one arbitrary constants, the corresponding DE for the family of rectangular hyperbolas will be first order and can be obtained by differentiation once.

	$xy = {c^2}$
	$\Rightarrow xy' + y = 0$

This is the required DE.

Example: 2

Find the DE associated with the family of circles of a fixed radius $r$ .

Solution: 2	Steps Involved: 5

Step-1

The circles are of a fixed radius but their centres are not. Let the centre be denoted by the variable point $(h,k)$

Then the equation of an arbitrary circle of the family is

${(x - h)^2} + {(y - k)^2} = {r^2}$

$\ldots (1)$

Step-2

This contains two arbitrary constants and therefore will give rise to a second-order DE. Differentiating $(1)$ , we have

$(x - h) + (y - k)\dfrac{{dy}}{{dx}} = 0$

$\ldots (2)$

Step-3

Differentiating $(2)$ again, we have

	$1 + (y - k)\dfrac{{{d^2}y}}{{d{x^2}}} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} = 0$	$\ldots (3)$
	$\Rightarrow (y - k) = - \dfrac{{1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}}}{{\dfrac{{{d^2}y}}{{d{x^2}}}}}$	$\ldots (4)$

Step-4

Using $(4)$ in $(2)$ , we have

$(x - h) = \dfrac{{\left\{ {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right\}\dfrac{{dy}}{{dx}}}}{{\dfrac{{{d^2}y}}{{d{x^2}}}}}$

$\ldots (5)$

Step-5

Using $(4)$ and $(5)$ in $(1)$ , and simplifying, we have the required DE as

${\left[ {1 + {{\left( {\dfrac{{dy}}{{dx}}} \right)}^2}} \right]^3} = {r^2}{\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^2}$

which as expected, is second order.

If all this talk about arbitrary constants and solutions to DEs, confuses you, lets view the whole discussion from a slightly different perspective.

Referring to example -1, suppose we are given the DE

$xy' + y = 0$

$\ldots (1)$

This is a first-order DE and its most general solution will contain one arbitrary constant; in fact, the most general solution of this DE is

$xy = \lambda$

where $\lambda$ is an arbitrary constant. Now suppose we are told that the curve satisfying the DE in $(1)$ passes through $(2, 2)$ . This additional constraint enables us to determine the value of for this particular curve:

$(2)(2) = \lambda \Rightarrow \lambda = 4$

and thus the curve $xy=4$ is a particular solution of the DE in $(1)$ .

To emphasize once more,

	$xy=\lambda$ is a general solution to $(1)$
	while $xy=4$ is a particular solution to $(1)$

which was obtained from the general solution by using the fact that the curve passes through $(2,2)$ .

As another example, consider the DE obtained in example – 2.

${\left( {1 + {{\left( {y'} \right)}^2}} \right)^3} = {r^2}{\left( {y''} \right)^2}$

This is a second-order DE and its most general solution will contain two arbitrary constants; the most general solution can be found to be

${\left( {x - \alpha } \right)^2} + {\left( {y - \beta } \right)^2} = {r^2}$

where $\alpha$ and $\beta$ are arbitrary constants.

To determine a particular solution to this DE, we need two additional constraints which can enable us to evaluate $\alpha$ and $\beta$

Example: 3

Find the DE associated with the family of straight lines, each of which is at a constant distance $p$ from the origin.

Solution: 3	Steps Involved: 3

Step-1

Any such line has the equation

$x\cos \alpha + y\sin \alpha = p$

$\ldots (1)$

where $\alpha$ is a variable. Different values of $\alpha$ give different lines belonging to this family. Since the equation representing this family contains only one arbitrary constant, its corresponding DE will be first order .

Step-2

Differentiating $(1)$ , we have

	$\cos \alpha + y'\sin \alpha = 0$
	$\Rightarrow \tan \alpha = - \dfrac{1}{{y'}}$
	$\Rightarrow \sin \alpha = \dfrac{{ - 1}}{{\sqrt {1 + {{(y')}^2}} }},\cos \alpha = \dfrac{{y'}}{{\sqrt {1 + {{(y')}^2}} }}$	$\ldots (2)$