Suppose that the population of cats in a has a rate of growth proportional to the population itself. Write down a differential equation for the population $P (t)$ at time $t$ ?
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Well, we know that $k*P(t)=dP/dt$. This simply states that $P(t)$ is proportional to $dP/dt$. You may be able to stop here, based on the wording of your question. It simply asks for a differential equation involving P(t) and t, which this classifies as. You can move on if you have to solve for k at some point. First, we rearrange this to be $dP/P = kdt$. If we integrate both sides, $\int dP/P = \int kdt$ This leaves us with $ln(P) = kt + C$ Then, we do the opposite operation of the natural log (putting everything to the power of e) and we are left with: $P = e^{kt+c}$ Then you solve for the initial y and the final form of the equation is: $P=P_{0}e^{kt}$ |
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Hint: how would you express the rate of growth in terms of $P(t)$? |
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To the OP: the equation you stated in a comment to Ross is crucial (and in contradiction with your post, I should say). So I understand that the question is to solve $P'(t)=rP(t)(1-P(t)/K)$, presumably with an initial condition $P(0)$ in $(0,K)$ (and if this is not what you ask for, please say so and I shall delete this answer). Hint: The differential equation $P'(t)=rP(t)(1-P(t)/K)$ can be rewritten as $f(P(t))P'(t)=r$ for a function $f$ you should be able to write down. Let $g$ be any differentiable function and $G(t)=g(P(t))$. Compute $G'(t)$. Choose $g$ such that $G'(t)=P'(t)f(P(t))$. The equation $G'(t)=r$ is equivalent to $G(t)=rt+c$ for a given constant $c$. Use all this to solve the original equation. |
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