Tuesday, 5 August 2014

CHAPTER 3 - General DEs and Arbitrary Constants

Finally, an n^{th} linear DE (degree one) is an equation of the form
{a_0}\dfrac{{{d^n}y}}{{d{x^n}}} + {a_1}\dfrac{{{d^{n - 1}}y}}{{d{x^{n - 1}}}} + \ldots + {a_{n - 1}}\dfrac{{dy}}{{dx}} + {a_n}y = b
where the {a_i}^\prime s and b are functions of x.
Solving an n^{th} order DE to evaluate the unknown function will essentially consist of doing n integrations on the DE. Each integration step will introduce an arbitrary constant. Thus, you can expect in general that the solution of an n^{th} order DE will contain n independent arbitrary constants.
By n independent constants, we mean to say that the most general solution of the DE cannot be expressed in fewer than n constants. As an example, the second order DE
\dfrac{{{d^2}y}}{{d{x^2}}} + y = 0
has its most general solution of the form
y = A\cos x + B\sin x
(verify that this is a solution by explicit substitution).
Thus, two arbitrary and independent constants must be included in the general solution. We cannot reduce (1) to a relation containing only one arbitrary constant. On the other hand, it can be verified that the function
y = a{e^{x + b}}
is a solution to the second-order DE
\dfrac{{{d^2}y}}{{d{x^2}}} = y
but even through it (seems to)contain two arbitrary constants, it is not the general solution to this DE. This is because it can be reduced to a relation involving only one arbitrary constant:
y = a{e^{x + b}}
 = a{e^x} \cdot {e^b}
 = c{e^x} (where c = a \cdot {e^b})
Let us summarise what we’ve seen till now : the most general solution of an n^{th} order DE will consist of n arbitrary constants; conversely, from a functional relation involving n arbitrary constants, an n^{th} order DE can be generated (we’ll soon see how to do this). We are generally interested in solutions of the DE satisfying some particular constraints (say, some initial values).Since the most general solution of the DE involves n arbitrary constants, we see that the maximum number of independent conditions which can be imposed on a solution of the DE is n.
As a first example, consider the functional relation
y = {x^2} + {c_1}{e^{2x}} + {c_2}{e^{3x}}\ldots (1)
This curve’s equation contains two arbitrary constants; as we vary {c_1} and {c_2} we obtain different curves; those curves constitute a family of curves. All members of this family will satisfy the DE that we can generate from this general relation; this DE will be second order since the relation contains two arbitrary constants.
We now see how to generate the DE. Differentiate the given relation twice to obtain
y' = 2x + 2{c_1}{e^{2x}} + 3{c_2}{e^{3x}}\ldots (2)
y'' = 2 + 4{c_1}{e^{2x}} + 9{c_2}{e^{3x}}\ldots (3)
From (1),(2),(3){c_1} and {c_2} can be eliminated to obtain
\left| {\begin{array}{*{20}{c}}  {{e^{2x}}}&{{e^{3x}}}&{{x^2} - y}\\  {2{e^{2x}}}&{3{e^{3x}}}&{2x - y'}\\  {4{e^{2x}}}&{9{e^{3x}}}&{2 - y''}  \end{array}} \right| = 0
 \Rightarrow \left| {\begin{array}{*{20}{c}}  1&1&{{x^2} - y}\\  2&3&{2x - y'}\\  4&9&{2 - y''}  \end{array}} \right| = 0
 \Rightarrow 6 - 3y'' - 18x + 9y' + 8x - 4y' - 4 + 2y'' + 6{x^2} - 6y = 0
 \Rightarrow y'' - 5y' + 6y = 6{x^2} - 10x + 2\ldots (4)
This is the required DE; it corresponds to the family of curves given by (1). Differently put, the most general solution of this DE is given by (1).
As an exercise for the reader, show that the DE corresponding to the general equation
y = A{e^{2x}} + B{e^x} + C
where A, B, C are arbitrary constants, is
y''' - 3y'' + 2y' = 0
As expected, the three arbitrary constants cause the DE to be third order.
Post a Comment