Page:Calculus Made Easy.pdf/266

Subtracting (2) from (1) and dividing by $2$, we then have

$y={\frac {1}{2}}C\epsilon ^{nx}-{\frac {1}{2}}\,{\frac {c^{2}}{C}}\epsilon ^{-nx},$

which is more conveniently written

$y=A\epsilon ^{nx}+B\epsilon ^{-nx}.$

Or, the solution, which at first sight does not look as if it had anything to do with the original equation, shows that $y$ consists of two terms, one of which grows logarithmically as $x$ increases, and of a second term which dies away as $x$ increases.

 

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Example 7.

Let

$b{\frac {d^{2}y}{dt^{2}}}+a{\frac {dy}{dt}}+gy=0$.

Examination of this expression will show that, if $b=0$, it has the form of Example 1, the solution of which was a negative exponential. On the other hand, if $a=0$, its form becomes the same as that of Example 6, the solution of which is the sum of a positive and a negative exponential. It is therefore not very surprising to find that the solution of the present example is

$y=(\epsilon ^{-mt})(A\epsilon ^{nt}+B\epsilon ^{-nt})$,

where

$m={\frac {a}{2b}}\quad {\text{and}}\quad n={\sqrt {\frac {a^{2}}{4b^{2}}}}-{\frac {g}{b}}$.

The steps by which this solution is reached are not given here; they may be found in advanced treatises.

 

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