Page:Calculus Made Easy.pdf/259

 where skill and practice suggest a plan–of multiplying all the terms by $$\epsilon^{\frac{a}{b} t}$$, giving us:

which is the same as

and this being a perfect differential may be integrated thus:–since, if $$\dfrac{du}{dt} = \dfrac{dy}{dt} \epsilon^{\frac{a}{b} t} + y \dfrac{d(\epsilon^{\frac{a}{b} t})}{dt}$$,

or

The last term is obviously a term which will die out as $$t$$ increases, and may be omitted. The trouble now comes in to find the integral that appears as a factor. To tackle this we resort to the device (see p. 226) of integration by parts, the general formula for which is $$\int u dv = uv - \int v du$$. For this purpose write {{c|$$\begin{align} &\left\{ \begin{aligned} u &= \epsilon^{\frac{a}{b} t}; \\ dv &= \sin 2\pi nt \cdot dt. \end{aligned} \right. \\ \end{align}$$}} We shall then have {{c|$$ \begin{align} &\left\{ \begin{aligned} du &= \epsilon^{\frac{a}{b} t} \times \frac{a}{b}\, dt; \\ v &= - \frac{1}{2\pi n} \cos 2\pi nt. \end{aligned} \right. \end{align}$$}}