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

[Bushy]

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

and $$G'(t) = a+bt$$

Show $$G(0)=0$$

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

 

[I like Serena]

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

and $$G'(t) = a+bt$$

Show $$G(0)=0$$

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

Hi Bushy! ;)

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

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