[SOLVED] Differential Equation Help

**Maccaman** · September 20th 2008, 05:47 PM

Hi all,
Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it.

Here it is:
"find the equation of R(t) when

$\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$ ,

$\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$

and

$\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$
where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants.

Now I know that

$S(t) = S_0 e^{-\gamma_{s} t}$

$E(t) = E_0 e^{-\gamma_{b} t}$

and then by substituting these values into $\frac{dR}{dt}$ we obtain

$\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$

this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks

**Prove It** · September 20th 2008, 05:49 PM

Originally Posted by Maccaman

Hi all,
Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it.

Here it is:
"find the equation of R(t) when

$\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$ ,

$\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$

and

$\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$
where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants.

Now I know that

$S(t) = S_0 e^{-\gamma_{s} t}$

$E(t) = E_0 e^{-\gamma_{b} t}$

and then by substituting these values into $\frac{dR}{dt}$ we obtain

$\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$

this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks

Just integrate with respect to t.

**topsquark** · September 20th 2008, 05:50 PM

Originally Posted by Maccaman

Hi all,
Earlier I started a thread with a past exam question that I wanted to understand before I attempted a homework question. It is similar to the question I posted earlier, but Im still having trouble with it.

Here it is:
"find the equation of R(t) when

$\frac{dE}{dt}= -\gamma_{b}E$ where $E(0) = E_0$ and $\gamma_b = \gamma_{b_1} + \gamma_{b_2} + \lambda_p$ ,

$\frac{dS}{dt}= -\gamma_{s}S$ where $S(0) = S_0$ and $\gamma_s = \gamma_{s_1} + \gamma_{s_2} + \lambda_p$

and

$\frac{dR}{dt} = \gamma_{b_1} E + \gamma_{s_1} MS$
where $M, \gamma_{b_1}, \gamma_{b_2}, \gamma_{b}, \gamma_{s_1}, \gamma_{s_2}, \gamma_{s}, \lambda_p$ are all constants.

Now I know that

$S(t) = S_0 e^{-\gamma_{s} t}$

$E(t) = E_0 e^{-\gamma_{b} t}$

and then by substituting these values into $\frac{dR}{dt}$ we obtain

$\frac{dR}{dt} = \gamma_{b_1} E_0 e^{-\gamma_b t} + \gamma_{s_1} M S_0 e^{-\gamma_s t}$

this is where I get lost. How do I continue from here? What type of differential equation is this? Thanks

Well, as the right hand side doesn't depend on R all you need to do is integrate both sides with respect to t.

-Dan

**Maccaman** · September 20th 2008, 05:53 PM

ahhhhhh, of course

I feel so stupid now

thanks

**topsquark** · September 20th 2008, 05:56 PM

Originally Posted by Maccaman

ahhhhhh, of course

I feel so stupid now

thanks

There is nothing more difficult than seeing the obvious after you have done something complicated.

-Dan

**Maccaman** · September 20th 2008, 06:35 PM

Is this answer correct? (my integration rules are a bit dodgy)

$R(t) = - \frac{\gamma_{b_1} E_0 e^{-\gamma_b t}}{\gamma_b} - \frac{\gamma_{s_1} M S_0 e^{-\gamma_s t}}{\gamma_s}$

**topsquark** · September 20th 2008, 06:48 PM

Originally Posted by Maccaman

Is this answer correct? (my integration rules are a bit dodgy)

$R(t) = - \frac{\gamma_{b_1} E_0 e^{-\gamma_b t}}{\gamma_b} - \frac{\gamma_{s_1} M S_0 e^{-\gamma_s t}}{\gamma_s}$

Looks good to me.

-Dan

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