DE book does not justify this claim

[1MileCrash]

Homework Statement

My DE book claims that

$\frac{d\mu(t)/dt}{\mu(t)}$

is equivalent to

$\frac{d}{dt}ln|\mu(t)|$

But says nothing to justify this claim.

Can you?

 

[gabbagabbahey]

But says nothing to justify this claim.

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.

This is just an application of the chain rule, combined with the fact that $\frac{d}{dx} \ln|x| = \frac{1}{x}$.

 

[1MileCrash]

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.

 

[Bacle2]

But that seems just like the chain rule , (though I don't know why they dropped the

absolute value): d/dt f(g(t))= f'(g(t))g'(t) , where f(t)=ln|t| g(t)=u(t). Or

maybe you can do a triple chain rule f(g(h)) , with g(t)=|t|, h(t)=u(t), and

f(t)=ln(t).

 

[gabbagabbahey]

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.

Just compute the anti-derivative:

$$\int \frac{ \frac{d\mu}{dt} }{ \mu(t) }dt = \int \frac{d\mu}{\mu} = \ln|\mu| + C$$

More importantly though, given how common a function the natural logarithm is, you need to be able to recognize its derivative when you see it. A big part of problem solving with differential equations requires you to have experience with calculating common derivatives (which is why most calculus textbooks have all those practice problems in them ) and applying that experience, by recognizing derivatives of common functions when you see them.

 

[Nessdude14]

Remember that:
$\large \frac{δ}{δt}ln|x|=\frac{1}{x}$

but since you have a another function inside the log, you need to apply the chain rule, so:
$\large \frac{δ}{δt}ln|\mu(t)|=\frac{1}{\mu(t)}\cdot\frac{δ\mu(t)}{δt}$

 

[1MileCrash]

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:

$\frac{d\mu(t)/dt}{\mu(t)} = q$

Integrating both sides

$ln|\mu(t)| = qt$

Then taking the derivative of both sides

$\frac{dln|\mu(t)|}{dt} = q$

Therefore:

$\frac{d\mu(t)/dt}{\mu(t)} = \frac{dln|\mu(t)|}{dt} = q$

Is satisfactory.

 

[vela]

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:

$\frac{d\mu(t)/dt}{\mu(t)} = q$

Integrating both sides

$ln|\mu(t)| = qt$

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.

 

[1MileCrash]

q is constant, I just put it there so I could visualize it thoroughly.

 

[gabbagabbahey]

q is constant, I just put it there so I could visualize it thoroughly.

But you defined $q$ as being equal to $\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}$, so it is only constant if $\frac{ \frac{d\mu(t)}{dt} }{\mu(t)}$ is constant, which it isn't, in general.

 

[1MileCrash]

It is constant, here.

If it weren't, instead of qt it would be an arbitrary integral notation of q. It doesn't matter.

 

[gabbagabbahey]

though I don't know why they dropped the absolute value

Try computing $\frac{d}{dx}\ln(x)$ when $x<0$. There's a reason that calculus books give the antiderivative of $\frac{1}{x}$ as $\ln|x|+C$ instead of $\ln(x)+C$.