# Introductory Differential Equations Question

Thank you everyone for being a source of help in previous problems I've posted here. I'm starting an intermediate course in Differential Equations and I'm enjoying it so far, but this one problem on my homework seems to be giving me a problem and I think that I haven't fully grasped the machinery of this sort of mathematics. Help is appreciated!

## Homework Statement

A population $$x$$ has growth as such: $$x'=rx(1-\frac{x}{k})-\lambda$$ where $$r>0, k>0, \lambda\in{\textbf{R}}$$. That is, r is the growth rate, k is the carrying capacity and $$\lambda$$ is a removal rate.

For what value of $$\lambda$$ is the population guaranteed to go extinct?

## The Attempt at a Solution

I've been trying to "solve the equation" so that $$x'=0$$ but then I can only find the trivial solution from this, that is, x=0.

Any ideas?

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SammyS
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Thank you everyone for being a source of help in previous problems I've posted here. I'm starting an intermediate course in Differential Equations and I'm enjoying it so far, but this one problem on my homework seems to be giving me a problem and I think that I haven't fully grasped the machinery of this sort of mathematics. Help is appreciated!

## Homework Statement

A population $$x$$ has growth as such: $$x'=rx(1-\frac{x}{k})-\lambda$$ where $$r>0, k>0, \lambda\in{\textbf{R}}$$. That is, r is the growth rate, k is the carrying capacity and $$\lambda$$ is a removal rate.

For what value of $$\lambda$$ is the population guaranteed to go extinct?

## The Attempt at a Solution

I've been trying to "solve the equation" so that $$x'=0$$ but then I can only find the trivial solution from this, that is, x=0.

Any ideas?
$$\text{If } x'=0\,,\$$ then x is constant. That's not extinction.

Divide both sides of your differential equation by the right hand side - then integrate both sides with respect to time.

$$x'\,dt=dx\,.\$$