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

#### Step-3

Using $(2)$ in $(1)$, we have
 $\dfrac{{xy'}}{{\sqrt {1 + {{(y')}^2}} }} - \dfrac{y}{{\sqrt {1 + {{(y')}^2}} }} = p$ $\Rightarrow {(xy' - y)^2} = {p^2}(1 + {(y')^2})$
As expected, this is a first order DE.