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: 1Steps 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: 2Steps Involved: 5


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)


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)


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)


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)


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: 3Steps Involved: 3


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 .


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)


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.
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