1. ## 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?

2. 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:

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

which is of the form:

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

which is the general form of the solution to

$\displaystyle y'=\lambda y$

RonL

3. Are you able to show the steps, or is there not enough information?

4. Originally Posted by Jrb599
Are you able to show the steps, or is there not enough information?
You have:

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

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

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

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

RonL