Help with integration of first moment from PBE

Question

I'm wondering anyone can help me with the following integration:

$$\frac{d(m_0 V)}{dt} = BV$$ where $B$ is just a constant, $V$ is a variable parameter. Product rule must be applied somehow?

EDIT: m_0 is not constant! it is variable also.

I want to implement it in MATLAB/Simulink. It is basically first moment in the Population Balance Equation, no. 8 from this paper [LINK].

Thanks and Kind Regards.

Answer 1

If $B,m_0$ are constants, then $$m_0\frac{d V}{dt} = BV$$ or $$\frac{d V}{V} = \frac{B}{m_0}dt$$ Integrating both sides as $$\int \frac{d V}{V} =\int \frac{B}{m_0}dt$$ will give you $$\ln V = \frac{B}{m_0}t + K$$ which is equivalent to $$\exp(\ln V) =\exp( \frac{B}{m_0}t + K) =\underbrace{ \exp( K) }_C\exp( \frac{B}{m_0}t) $$ Finally you get $$V = C \exp( \frac{B}{m_0}t) $$

Answer 2

$$\frac{d(m_0V)}{dt}=BV$$ since $m_0$ and $B$ are constants we can say: $$m_0\frac{dV}{dt}=BV$$ we can now rearrange this to say: $$\frac 1V\frac{dV}{dt}=\frac{B}{m_0}$$ Now just integrate both sides: $$\int\frac 1V\frac{dV}{dt}dt=\int\frac{B}{m_0}dt\Rightarrow\ln|V|=\frac{B}{m_0}t+C$$ Now exponentiate both sides: $$V=e^Ce^{\frac{B}{m_0}t}$$ Now notice that when $t=0$,$V=e^C$ so we will let $V_0=e^C$ so: $$V=V_0e^{\frac{B}{m_0}t}$$