1. ## Find solutions

Additional processes due to biotic and abiotic degradation contributing to souce decay can be accounted for by adding a decay term in (3) that is proportional to m(t),
, (4)
where λ is the associated decay rate constant. Assume that α(0) and λ(0) are greater than or equal to zero, then choose one pairs of the vaules of α and λ by yourself and find solutions of Equation (4) using the initial condition m(0)=m0 for the following cases: (i) γ=1, (ii) γ=2, (iii) γ= 0.5. Then , graph the solution for m(t) for each γ

(3)

2. Originally Posted by cowracer3
Additional processes due to biotic and abiotic degradation contributing to souce decay can be accounted for by adding a decay term in (3) that is proportional to m(t),
, (4)
where λ is the associated decay rate constant. Assume that α(0) and λ(0) are greater than or equal to zero, then choose one pairs of the vaules of α and λ by yourself and find solutions of Equation (4) using the initial condition m(0)=m0 for the following cases: (i) γ=1, (ii) γ=2, (iii) γ= 0.5. Then , graph the solution for m(t) for each γ

(3)
I assume that "M" in equation 4 is supposed to be "m" as in equation 3 so that you have just added the term $\displaystyle -\lambda m$.

The problem does not require you to solve that equation in general and I suspect it can't be. In (case i) $\displaystyle \gamma = 1$ you just have $\displaystyle \frac{dm}{dt}= (\alpha + \lambda )m$ which can be separated as $\displaystyle \frac{dm}{m}= (\alpha+ \lambda)dt$ and can be integrated easily.

In (case ii) $\displaystyle \gamma = 2$ you have $\displaystyle \frac{dm}{dt}= \alpha m^2+ \gamma m$ which, again can be separated: $\displaystyle \frac{dm}{\alpha m^2+ \gamma m}= dt$. The denominator on the left can be factored as $\displaystyle m(\alpha m+ \gamma)$ and it can be integrated using "partial fractions".

In (case iii) $\displaystyle \gamma = 2$ you have $\displaystyle \frac{dm}{dt} = \alpha m^{1/2}+ \gamma m$ which can be separated as $\displaystyle \frac{dm}{\alpha m^{1/2}+ \gamma m}= dt$. If you let $\displaystyle u= m^{1/2}$ then $\displaystyle du= (1/2)m^{-1/2}dm$ so $\displaystyle dm= 2m^{1/2}du= 2udu$ and the left side becomes $\displaystyle \frac{2udu}{\alpha u+\gamma u^2}= \frac{2du}{\alpha+ \gamma u}$ and the substitution $\displaystyle v= \alpha+ \gamma u$ reduces it even further. (which means I should have suggested $\displaystyle u= \alpha+ \gamma x^{1/2}$ to begin with!)