Page:Calculus Made Easy.pdf/258

If the condition is laid down that $y=0$ when $x=0$ we can find $C$; for then the exponential becomes $=1$; and we have

$0={\frac {g}{a}}+C$,

or

$C=-{\frac {g}{a}}$.

Putting in this value, the solution becomes

$y={\frac {g}{a}}(1-\epsilon ^{-{\frac {a}{b}}x})$.

But further, if $x$ grows indefinitely, $y$ will grow to a maximum; for when $x=\infty$, the exponential $=0$, giving $y_{\text{max.}}={\dfrac {g}{a}}$. Substituting this, we get finally

$y=y_{\text{max.}}(1-\epsilon ^{-{\frac {a}{b}}x})$.

This result is also of importance in physical science.

 

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Example 3.
Let $\quad \quad \quad \quad ay+b{\frac {dy}{dt}}=g\cdot \sin 2\pi nt$.

We shall find this much less tractable than the preceding. First divide through by $b$.

${\frac {dy}{dt}}+{\frac {a}{b}}y={\frac {g}{b}}\sin 2\pi nt$.

Now, as it stands, the left side is not integrable. But it can be made so by the artifice–and this