Page:Calculus Made Easy.pdf/255

Now the mere inspection of this relation tells us that we have got to do with a case in which ${\dfrac {dy}{dx}}$ is proportional to $y$. If we think of the curve which will represent $y$ as a function of $x$, it will be such that its slope at any point will be proportional to the ordinate at that point, and will be a negative slope if $y$ is positive. So obviously the curve will be a die-away curve (p. 156), and the solution will contain $\epsilon ^{-x}$ as a factor. But, without presuming on this bit of sagacity, let us go to work.

As both $y$ and $dy$ occur in the equation and on opposite sides, we can do nothing until we get both $y$ and $dy$ to one side, and $dx$ to the other. To do this, we must split our usually inseparable companions $dy$ and $dx$ from one another.

${\frac {dy}{y}}=-{\frac {a}{b}}\,dx$.

Having done the deed, we now can see that both sides have got into a shape that is integrable, because we recognize ${\dfrac {dy}{y}}$, or ${\dfrac {1}{y}}\,dy$, as a differential that we have met with (p. 147) when differentiating logarithms. So we may at once write down the instructions to integrate,

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

and doing the two integrations, we have:

$\log _{\epsilon }y=-{\frac {a}{b}}x+\log _{\epsilon }C$,