Re: differential equations,

You are producing A at rate k and it is decaying at a rate $\displaystyle \alpha{A}$, so $\displaystyle \frac{dA}{dt} = k - \alpha{A}$. Your equation $\displaystyle A=kte^{-\alpha{t}}$ is incorrect since some of the A produced has decayed for a long time and some for a short time.

To solve this equation, substitute $\displaystyle A=Y+\frac{k}{\alpha}$. Everything else is pretty straightforward. Let us know if you have any trouble.

- Hollywood

Re: differential equations,

Thanks,

This is my solution:

$\displaystyle \frac{dA}{dt} = k-A\alpha$

$\displaystyle \frac{dB}{dt} = A\alpha$

$\displaystyle A(t) = \frac{k}{\alpha} $

I got the general solution using the integrating factor method

for part v)

$\displaystyle A(t) = \frac{k}{\alpha} - \frac{k}{\alpha}{e}^{-\alpha t}$

$\displaystyle A(0) > C$ C is some constant greater than k/a

$\displaystyle A(t) = \frac{k}{\alpha} + (C - \frac{k}{\alpha}{e}^{- \alpha t} $

for vi I subbed in the given B(t) into the differential equation I wrote for dB/dt

$\displaystyle {c}_{1} = k$ and $\displaystyle {c}_{2} = -{A}_{0}$ (not sure about this)

then for B(t) I get

$\displaystyle B(t) = kt - {A}_{0}{e}^{- \alpha t} + {A}_{0}$