Page:Calculus Made Easy.pdf/257

curve y becomes either a maximum or a minimum, so that ${\dfrac {dy}{dx}}=0$, then $y$ will have the value $={\dfrac {g}{a}}$. But let us go to work as before, separating the differentials and trying to transform the thing into some integrable shape.

${\begin{aligned}b{\frac {dy}{dx}}&=g-ay;\\{\frac {dy}{dx}}&={\frac {a}{b}}\left({\frac {g}{a}}-y\right);\\{\frac {dy}{y-{\dfrac {g}{a}}}}&=-{\frac {a}{b}}\,dx.\end{aligned}}$

Now we have done our best to get nothing but $y$ and $dy$ on one side, and nothing but $dx$ on the other. But is the result on the left side integrable?

It is of the same form as the result on p. 148; so, writing the instructions to integrate, we have:

$\int {\frac {dy}{y-{\dfrac {g}{a}}}}=-\int {{\frac {a}{b}}\,dx}$;

and, doing the integration, and adding the appropriate constant,

${\begin{aligned}\log _{\epsilon }\left(y-{\frac {g}{a}}\right)&=-{\frac {a}{b}}x+\log _{\epsilon }C;\\{\text{whence}}\quad \quad \quad y-{\frac {g}{a}}&=C\epsilon ^{-{\frac {a}{b}}x};\\{\text{and finally,}}\quad \quad \quad y&={\frac {g}{a}}+C\epsilon ^{-{\frac {a}{b}}x},\end{aligned}}$

which is the solution.