Page:Calculus Made Easy.pdf/252

called ^[1] its “solution”; and it is quite astonishing how in many cases the solution looks as if it had no relation to the differential equation of which it is the integrated form. The solution often seems as different from the original expression as a butterfly does from the caterpillar that it was. Who would have supposed that such an innocent thing as

${\dfrac {dy}{dx}}={\dfrac {1}{a^{2}-x^{2}}}$

could blossom out into

$y={\dfrac {1}{2a}}\log _{\epsilon }{\dfrac {a+x}{a-x}}+C?$

yet the latter is the solution of the former.

As a last example, let us work out the above together.

By partial fractions,

${\begin{aligned}{\frac {1}{a^{2}-x^{2}}}&={\frac {1}{2a(a+x)}}+{\frac {1}{2a(a-x)}},\\dy&={\frac {dx}{2a(a+x)}}+{\frac {dx}{2a(a-x)}},\\y&={\frac {1}{2a}}\left(\int {\frac {dx}{a+x}}+\int {\frac {dx}{a-x}}\right)\\&={\frac {1}{2a}}\left(\log _{\epsilon }(a+x)-\log _{\epsilon }(a-x)\right)\\&={\frac {1}{2a}}\log _{\epsilon }{\frac {a+x}{a-x}}+C.\end{aligned}}$

1. This means that the actual result of solving it is called its “solution.” But many mathematicians would say, with Professor Forsyth, “every differential equation is considered as solved when the value of the dependent variable is expressed as a function of the independent variable by means either of known functions, or of integrals, whether the integrations in the latter can or cannot be expressed in terms of functions already known.”