1. Oct 26, 2007

### trajan22

Im sorry, I posted this in the wrong section, feel free to move it to the homework section.

Hey guys, Ive really been needing some help with this one. Im doing an assignment for Ordinary Differential Equations and I was hoping someone could help me out by looking over my work. Ive been working on this assignment for a couple of weeks now and finally decided that I just needed help.
There is a second part of this problem, but Ill post that later depending on whether or not this part is correct.
Anyway, I know this is a really long post but I will truly appreciate the help of anyone thats willing to go through it all.
Basically I just need to solve the general equation below for different values of gamma and lambda.
Thanks

$$General Equation$$
$$\frac {dm}{dt}=-\alpha m^{\gamma}-\lambda m$$

$$\lambda = constant$$

$$Case (1)$$

$$\gamma=1$$

1a.) $$\frac{dm}{dt}=-\alpha(m)-\lambda(m)$$

2a.) $$\frac{dm}{dt}=-m(\alpha+\lambda)$$

3a.) $$\int \frac{1}{m}dm=\int (\alpha+\lambda)dt$$

4a.) $$\ln(m)=-(\alpha+\lambda)t+C$$

5a.) $$m=e^{-(\alpha+\lambda)t+C}$$

$$Case (2)$$

constraints

$$\gamma cannot =0$$ $$\lambda=0$$

Simplifying the general equation I get

1b.) $$\frac{dm}{dt}= -\alpha (m)^\gamma$$

2b.) $$\int \frac{1}{m^\gamma}dm=-\int \alpha dt$$

3b.) $$\frac{m^{(-\gamma + 1)}}{-\gamma+1}=-\alpha (t)+C$$

4b.) $$m=[(-\alpha (t)+C)(-\gamma+1)]^{1/(-\gamma+1)}$$

$$Case (3)$$

Constraints

$$\gamma not=1 \lambda not =0$$

1c.) $$\frac{dm}{dt}=-\alpha (m^{\gamma)}-\lambda (m)$$

I then solved this using the Bernoulli method

2c.) $$v\equiv m^{(1-\gamma)} for \gamma cannot =1$$

3c.) $$\frac{dv}{dt}=(1-\gamma)m^{(-\gamma)} \frac{dm}{dt}$$

4c.) $$\frac{dm}{dt}+\lambda (m) =-\alpha (m^{\gamma)}$$

5c.) $$m^{(-\gamma)}\frac{dm}{dt}=-\alpha-\lambda m^{(1-\gamma)}$$

6c.) $$m^{(-\gamma)} \frac{dm}{dt}=-\alpha-\lambda (v)$$

Substituting dv/dt into the equation I get

7c.) $$\frac{dv}{dt}=(1-\gamma) m^{-\gamma} m^{\gamma}(-\alpha-\lambda v)$$

8c.) $$\frac{dv}{dt}=(1-\gamma)(-\alpha-\lambda v)$$

9c.) $$\frac{dv}{dt}+(1-\gamma)(\lambda)(v)= -(1-\gamma)\alpha$$

This is a linear first order homogeneous equation and can be solved by making

10c.) $$\mu=e^{\int(1-\gamma)\lambda dt} =e^{(1-\gamma)\lambda(t)}$$

using this as my integrating factor I get this

11c.) $$v=\frac{1}{e^{(1-\gamma(t)}}\int e^{(1-\gamma)t} (1-\gamma) (-\alpha)dt$$

this yields

12c.) $$v=-\alpha +\frac{C}{e^{((1-\gamma)t)}}$$

substituting for v we get

13c.) $$m=(-\alpha+ \frac{C}{e^{((1-\gamma)t))}}^{(\frac{1}{1-\gamma})}$$

Am I doing anything wrong?

Last edited: Oct 26, 2007
2. Oct 27, 2007

### chaoseverlasting

You dont seem to be doing anything wrong. Another approach to the last case could be to take $$\alpha m^{\gamma -1}+\lambda =u$$
This gives $$\alpha (\gamma -1) \frac{m^{\gamma -1}}{m}dm=du$$

$$\frac{dm}{m}=\frac{dt}{\alpha m^{\gamma -1}{\gamma -1}}$$

Substituting this in equation 1c. you get $$\frac{du}{(\gamma -1)u-\lambda}=-dt$$

Integrating this, you get $$ln(\alpha m^{\gamma -1})=-(\gamma -1)t +C$$.
Isolating m, I suspect you would get the same answer, but this method is a lot simpler...

3. Oct 27, 2007

### trajan22

Thanks for looking it over. Your right that is a much easier method.
But in the second part of the question it states "show that when $$\gamma \geq 1$$ then the source must have an infinite lifetime.

The way I took this is that if it is to have an infinite life then the $$\lim _{t \rightarrow \infty}m=\infty$$
However when I actually do this I get that the source approaches zero.
Am I simply misunderstanding the question?