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A tutorial (page 17) on ODE scaling takes the equation $m\ddot{u}+c\dot{u}+ku=F_0sin(\omega t)$, with $u$ a function of $t$, and substitutes $\eta=u/a$, $\tau=t/b$ to get:

$$\frac{ma}{b^2}\ddot{\eta}+\frac{ca}{b}\dot{\eta}+ka\eta=F_0sin(\omega\tau)$$

Now, when I differentiate $\eta=u/a$, I get $\dot{\eta}=t\dot{u}/a=\tau b\dot{u}/a$. Substituting, the final result is:

$$\frac{ma}{\tau^2b^2}\ddot{\eta}+\frac{ca}{\tau b}\dot{\eta}+ka\eta=F_0sin(\omega\tau b)$$

That's not the same. What am I missing?

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2 Answers 2

up vote 2 down vote accepted

Gerry has already explained the differences on the left-hand side. On the right-hand side, this is a typo in the book. The next equation has $\sin\omega'\tau$ instead of $\sin\omega\tau$, with $\omega'=\omega\sqrt{m/k}$ and $b=\sqrt{m/k}$, so this is $\sin\omega\tau b$ as you expected. So it should be either $\sin\omega t$ or $\sin\omega'\tau$ on the right-hand side.

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Differentiating $\eta=u/a$ gives $\eta'=u'/a$, not $\eta'=tu'/a$, no?

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Heh. Of course, you are right. I would be much obliged, if you coud explain how to get their result. –  Don Reba Aug 24 '11 at 8:53
    
@Don, I'm a little puzzled myself as to how they get $\omega\tau$ on the right side, where your $\omega\tau b$ looks better to me. Sorry, don't have the time to work through it now, but I'm sure someone will be along to help out. –  Gerry Myerson Aug 24 '11 at 9:22

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