Question

If e^(rt) is given by e^(lambda*t)cosut+ie^(lambda*t)sinut. Show that d/dt (e^(rt)) =re^(rt) for any complex number, r If e^(rt) is given by e^(lambda*t)cosut+ie^(lambda*t)sinut. Show that d/dt (e^(rt)) =re^(rt) for any complex number, r @Mathematics

Answer 1

e^(lambda*t)cosut+ie^(lambda*t)sinut
\[ = e^{\lambda t + ui} \]
\[ = e^{rt} \]
Now, it's actually a slightly strange question this one because we should expect by general properties of the exponential function that
\[ (d/dt) \ e^{rt} = r . e^{rt} \]

But it looks like the question wants you to prove that this is the case here.

So we have
\[ e^{rt} = e^{\lambda t} (\cos ut + i \sin ut) \]
Now differentiate and use the product rule.

Answer 2

\[ (d/dt) (function \ above) \]
\[ = \lambda e^{\lambda t} (\cos ut + i \sin ut) + e^{\lambda t} (-u \sin ut + ...) \]
\[ = ... \]
You take it from here.

Answer 3

Why didn't you multiply e^rt into the brackets and just take the derivative?

Answer 4

I got for the last bit e^λt (-usin(ut) + uicost (ut)) I still can't solve for it?

Answer 5

Now you know that you want the entire expression to be equal to
r e^(rt)
= ( λ + iu) e^(rt)
= ( λ + iu) e^λt (cos ut + i sin ut)

So just manipulate the expression until you make it equal to that. It works, try it.

Answer 6

How did you get

= ( λ + iu) e^(rt)
= ( λ + iu) e^λt (cos ut + i sin ut).

You mean manipulate the expression above where I used the product rule right

Answer 7

By definition
e^rt = e^λt ( cos ut + i sin ut)

Now you should also know that

cos ut + i sin ut = e^iut

hence

e^rt = e^λt . e^iut = e^(λ+iu)t

Thus r = (λ+iu)

What you asked to prove using the property of real-valued functions and their derivatives, is that

d/dt e^rt = r . e^rt

Now, as we know what r is and e^rt is in terms of λ and u, we have that

r . e^rt = (λ+iu) e^(λt (cos ut + i sin ut) ------- (*)

Hence you need to show that

d/dt e^rt = (*)

So that's what we're doing.

e^rt = e^λt ( cos ut +i sin ut)

Differentiating that expression

d/dt e^rt = .... stuff .... -------- (**)

You want to show now that

(**) = (*)

Answer 8

Ohh I didn't realize cos ut + i sin ut = e^iut yeah I got it now. Thanks a lot for your help!

Answer 9

Reviving this thread in case anyone else comes across it.

I had this same problem. The key is to note that
(-u sin(ut) + ui cos(ut)) = - i u (-i sin(ut) - cos(ut))

Answer 10

That is, we can get away with multiplying that expression by 1 (obviously), and it just so happens that (-1)( i )( i ) = 1.