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.

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,

and doing the two integrations, we have: