Solve General Equation: ODE Project

[trajan22]

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

Hey guys, I've really been needing some help with this one. I am doing an assignment for Ordinary Differential Equations and I was hoping someone could help me out by looking over my work. I've 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 that's 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?

 

[chaoseverlasting]

You don't 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...

 

[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?