What is the relationship between P and G(t) in Calculus with Exponential?

[Bushy]

For \(\displaystyle P=P_0\times e^{G(t)}\)

and \(\displaystyle G'(t) = a+bt\)

Show \(\displaystyle G(0)=0\)

I get \(\displaystyle G(t) = \int a+bt ~dt = at+\frac{1}{2} b t^2+C\) therefore \(\displaystyle G(0)=C \)

 

[I like Serena]

For \(\displaystyle P=P_0\times e^{G(t)}\)

and \(\displaystyle G'(t) = a+bt\)

Show \(\displaystyle G(0)=0\)

I get \(\displaystyle G(t) = \int a+bt ~dt = at+\frac{1}{2} b t^2+C\) therefore \(\displaystyle G(0)=C \)

Hi Bushy! ;)

I expect that it's intended that $P(0) = P_0$.

Then we have:
$$P_0 e^{G(0)} = P_0 \quad\Rightarrow\quad e^{G(0)} = 1 \quad\Rightarrow\quad G(0)=0
$$