# Differential Equation Problem

• Aug 29th 2007, 05:24 PM
Jrb599
Differential Equation Problem
I can't remember Diff EQ to save my life.

but I have a'(t)=a(t)*a'(0) where a'(0) is constant

my book does

a'(t)/a(t) = d/dt *ln[a(t)] = a'(0)

then they integrate both sides from T to 0 and get

ln[a(t)] = t*a'(0)

we know that if t = 1 ten a(1) = 1 + i, and i some interest rate

then we know ln (1+i) = a'(0)

so basically you get a(t)=(1+i)^t

but I thought from Diff Eq any function of the forum
y'=c*y would give you y =Ce^xt

so how come this one doesn't?
• Aug 29th 2007, 09:17 PM
CaptainBlack
Quote:

Originally Posted by Jrb599
I can't remember Diff EQ to save my life.

but I have a'(t)=a(t)*a'(0) where a'(0) is constant

my book does

a'(t)/a(t) = d/dt *ln[a(t)] = a'(0)

then they integrate both sides from T to 0 and get

ln[a(t)] = t*a'(0)

we know that if t = 1 ten a(1) = 1 + i, and i some interest rate

then we know ln (1+i) = a'(0)

so basically you get a(t)=(1+i)^t

but I thought from Diff Eq any function of the forum
y'=c*y would give you y =Ce^xt

so how come this one doesn't?

You have:

$
a(t)=(1+i)^t=e^{[\ln(1+i)]t}
$

which is of the form:

$
a(t)=C e^{\lambda t}
$

which is the general form of the solution to

$
y'=\lambda y
$

RonL
• Aug 30th 2007, 02:59 AM
Jrb599
Are you able to show the steps, or is there not enough information?
• Aug 30th 2007, 04:56 AM
CaptainBlack
Quote:

Originally Posted by Jrb599
Are you able to show the steps, or is there not enough information?

You have:

Because for any posive number $k$ we have $k=e^{\ln(k)}$, so if we put $k=(1+i)$ we may rewrite your solution:

$
a(t)=(1+i)^t=e^{[\ln(1+i)]t}
$

Now put $C=1$, and $x= \ln(1+i)$, then:

$
a(t)=Ce^{x t}
$

RonL