Two mice (say $A$ and $B$) eat a piece of food. The rate of consumption for $A$ is proportional to amount of food left(Proportionality constant = $2$) and rate of consumption for $B$ is proportional to amount of food eaten by $A$(Proportionality constant = $1/2$ ).
Find the time it takes the food to be finished.
Also find the time after which the rate of consumption of both mice is same.
My Attempt
Let $F_A$ and $F_B$ be the amount of food eaten by A and B after time t.
$\frac{dF_A}{dt}=-2\{N-(F_A+F_B)\}$
$\frac{dF_B}{dt}=\frac{1}{2}F_A\Rightarrow \frac{d^2F_B}{dt^2}=\frac{1}{2}\frac{dF_A}{dt}$
So eliminating $F_A$ and $\frac{dF_A}{dt}$ we get
$\frac{d^2F_B}{dt^2}-2\frac{dF_B}{dt}-F_B+N=0$
Are my deductions correct. How to proceed further now
Answer given is $1$ time unit and $0.66$ time units
Best Answer
As per what Gonçalo said I followed his advice.
Let $N$ be the amount of food in the beginning.
$$\frac{dF_A}{dt}=2\{N-(F_A+F_B)\}$$
$$\frac{dF_B}{dt}=\frac{1}{2}F_A\Rightarrow \frac{d^2F_B}{dt^2}=\frac{1}{2}\frac{dF_A}{dt}$$
So eliminating $F_A$ and $\frac{dF_A}{dt}$ we get
$$\frac{d^2F_B}{dt^2}+2\frac{dF_B}{dt}+F_B=N$$
$$e^t\frac{d^2F_B}{dt^2}+2e^t\frac{dF_B}{dt}+e^tF_B=Ne^t$$
$$\frac{d^2(e^tF_B)}{dt^2}=Ne^t$$
Integrating twice
$$e^tF_B=Ne^t+Ct+D$$
$$F_B=N+(Ct+D)e^{-t}$$
Clearly $F_B(0)=0$ and $F'_B(0)=0$
so we have $$F_B=N-(1+t)Ne^{-t}$$
and $$F_A=2Nte^{-t}$$
When the food is finished we have $$F_A+F_B=N$$
Substituting for $F_A$ and $F_B$ in above equation we get $t=1$
To find the time after which the rates of consumption are equal we obviously put
$$F'_A=F'_B$$ to obtain $$t=\frac{2}{3}$$