- #1
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Homework Statement
My DE book claims that
[itex]\frac{d\mu(t)/dt}{\mu(t)}[/itex]
is equivalent to
[itex]\frac{d}{dt}ln|\mu(t)|[/itex]
But says nothing to justify this claim.
Can you?
Why would it? Most Authors would assume that if a student is studying differential equations, they have already mastered the basics of single variable calculus, such as computing derivatives and anti-derivatives.But says nothing to justify this claim.
Just compute the anti-derivative:I see how from the bottom, the top follows by direct application of the chain rule.
But given the top, I don't know how to come to the bottom expression.
This all seems kind of pointless. Integrating ##\frac{d\mu(t)/dt}{\mu(t)}## to get ##\ln|\mu(t)|##, you've used the fact you're trying to show, so why bother? Also, the righthand side is only equal to qt if q is a constant, which it most likely isn't.That's not really my issue. I just like to see it written out even if it is recognizable - I can integrate and differentiate until the cows come home.
Nonetheless:
[itex]\frac{d\mu(t)/dt}{\mu(t)} = q[/itex]
Integrating both sides
[itex]ln|\mu(t)| = qt[/itex]
But you defined [itex]q[/itex] as being equal to [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex], so it is only constant if [itex]\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}[/itex] is constant, which it isn't, in general.q is constant, I just put it there so I could visualize it thoroughly.
Try computing [itex]\frac{d}{dx}\ln(x)[/itex] when [itex]x<0[/itex]. There's a reason that calculus books give the antiderivative of [itex]\frac{1}{x}[/itex] as [itex]\ln|x|+C[/itex] instead of [itex]\ln(x)+C[/itex].though I don't know why they dropped the absolute value