Respuesta :
Answer:
[tex]V(t) = b^{(1-e^{-at})}[/tex]
Step-by-step explanation:
We are given the following information in the question:
[tex]\displaystyle\frac{dV}{dt} = a (In b - In V) V[/tex]
where a > 0 and b > 0.
[tex]\displaystyle\frac{dV}{dt} = a (In b-In V) V\\\\\displaystyle\frac{dV}{dt} = -aV(ln\frac{V}{b})\\\\\frac{dV}{V(ln\frac{V}{b})} = (-a)dt\\\\\text{Put } ln\frac{V}{b} = z\\\\\text{Integrating both sides}\\\\\int \frac{dV}{V(ln\frac{V}{b})} = \int (-a)dt\\\\\text{We get}\\\\\int \frac{dz}{z} = \int (-a)dt\\\\\\\text{where C is the constant of integration}[/tex]
[tex]V(0) = 1~ mm^3\\V(t) = b.e^{e^{-at+C}}\\\text{Putting t =0, V(0) = 1}\\\\V(0) = 1 = b.e^{e^{C}}\\\\V(t) = b^{(1-e^{-at})}[/tex]
where v(t) is the required tumor volume as a function of time that has an initial tumor volume of V(0) = 1 cubic mm.
