Page:Calculus Made Easy.pdf/259

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FINDING SOLUTIONS

239

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

${\frac {dy}{dt}}\epsilon ^{{\frac {a}{b}}t}+{\frac {a}{b}}y\epsilon ^{{\frac {a}{b}}t}={\frac {g}{b}}\epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt$ ,

which is the same as

${\frac {dy}{dt}}\epsilon ^{{\frac {a}{b}}t}+y{\frac {d(\epsilon ^{{\frac {a}{b}}t})}{dt}}={\frac {g}{b}}\epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt$ ;

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

$y\epsilon ^{{\frac {a}{b}}t}={\frac {g}{b}}\int \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt\cdot dt+C$ ,

or

$y={\frac {g}{b}}\epsilon ^{-{\frac {a}{b}}t}\int \epsilon ^{{\frac {a}{b}}t}\cdot \sin 2\pi nt\cdot dt+C\epsilon ^{-{\frac {a}{b}}t}..............{\text{[A]}}$

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 $\int udv=uv-\int vdu$ . For this purpose write

${\begin{aligned}&\left\{{\begin{aligned}u&=\epsilon ^{{\frac {a}{b}}t};\\dv&=\sin 2\pi nt\cdot dt.\end{aligned}}\right.\\\end{aligned}}$

We shall then have

${\begin{aligned}&\left\{{\begin{aligned}du&=\epsilon ^{{\frac {a}{b}}t}\times {\frac {a}{b}}\,dt;\\v&=-{\frac {1}{2\pi n}}\cos 2\pi nt.\end{aligned}}\right.\end{aligned}}$

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