Question

A model for the populations of two interacting species of animals is

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dy}}\mathbf{=}{\mathbf{k}}_{\mathbf{1}}\mathbf{x}\left(\alpha -x\right) \\ \frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{k}}_{\mathbf{2}}\mathbf{xy} \end{gathered}$$

Solve for $\mathbf{x}$ and $\mathbf{y}$ in terms of $\mathbf{t}$.

Short answer

The equations $\frac{\mathrm{dx}}{\mathrm{dy}}={\mathrm{k}}_1\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)$ and $\frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{k}}_2\mathrm{xy}$ , then$\mathrm{x}$ and$\mathrm{y}$ in terms of$\mathrm{t}$ is$\mathrm{x}\left(\mathrm{t}\right)=\frac{{\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}{1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}$ and$\mathrm{y}\left(\mathrm{t}\right)={\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right)}^{\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}}$

Step-by-step solution

Step 1:Find the equation

We have the population of two interaction species of animals which are described by the differential equations

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dy}}={\mathrm{k}}_1\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right) \\ \frac{\mathrm{dy}}{\mathrm{dt}}={\mathrm{k}}_2\mathrm{xy} \end{gathered}$$

And we have to solve these different equation x and y as the following technique:

First, we have to obtain the solution for the different equation shown in (1) as the following:

Since this differential equation is separable, then we can solve it as

$$\begin{gathered} \int \frac{1}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)}\mathrm{dx}={\mathrm{k}}_1\int \mathrm{dt} \\ \int \frac{1}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)}\mathrm{dx}={\mathrm{k}}_1\mathrm{t}+{\mathrm{c}}_1 \end{gathered}$$

But before we start to find the integration shown above, we have to make partial fraction to the fraction $\frac{1}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)}$as

$$\begin{gathered} \frac{1}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{B}}{\mathrm{\alpha }-\mathrm{x}} \\ =\frac{\mathrm{A}\left(\mathrm{\alpha }-\mathrm{x}\right)+\mathrm{Bx}}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)} \\ =\frac{\left(\mathrm{B}-\mathrm{A}\right)\mathrm{x}+\mathrm{A\alpha }}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)} \end{gathered}$$

Then by comparison, we can have the equations

$$\begin{gathered} \mathrm{B}-\mathrm{A}=0 \\ \mathrm{A\alpha }=1 \end{gathered}$$

Find the solution of x(t)=αek1αt1+ek1αt

Then by solving the two equation (a) and (b), we can obtain the value of constants as

$\mathrm{A}=\frac{1}{\mathrm{\alpha }}\mathrm{and}\mathrm{B}=\frac{1}{\mathrm{\alpha }}$

Then by substituting with A and B, we can have

$\frac{1}{\mathrm{x}\left(\mathrm{\alpha }-\mathrm{x}\right)}=\frac{1}{\mathrm{\alpha }}\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{\alpha }}\frac{1}{\mathrm{\alpha }-\mathrm{x}}$

Then we have

$$\begin{gathered} \frac{1}{\mathrm{\alpha }}\int \frac{1}{\mathrm{x}}\mathrm{dx}-\frac{1}{\mathrm{\alpha }}\int \frac{-1}{\mathrm{\alpha }-\mathrm{x}}\mathrm{dx}={\mathrm{k}}_1\mathrm{t} \\ \frac{1}{\mathrm{\alpha }}\mathrm{In}\left(\mathrm{x}\right)-\frac{1}{\mathrm{\alpha }}\mathrm{In}\left(\mathrm{\alpha }-\mathrm{x}\right)={\mathrm{k}}_1\mathrm{t} \\ \frac{1}{\mathrm{\alpha }}\mathrm{In}\left(\frac{\mathrm{x}}{\mathrm{\alpha }-\mathrm{x}}\right)={\mathrm{k}}_1\mathrm{t} \\ \mathrm{In}\left(\frac{\mathrm{x}}{\mathrm{\alpha }-\mathrm{x}}\right)={\mathrm{k}}_1\mathrm{\alpha t} \\ {}_{\mathrm{e}}\mathrm{In}\left(\frac{\mathrm{x}}{\mathrm{\alpha }-\mathrm{x}}\right)={\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \end{gathered}$$

$$\begin{gathered} {}_{\mathrm{e}}\mathrm{In}\left(\frac{\mathrm{x}}{\mathrm{\alpha }-\mathrm{x}}\right)={\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \\ \frac{\mathrm{x}}{\mathrm{\alpha }-\mathrm{x}}={\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \\ \mathrm{x}={\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}-{\mathrm{xe}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \\ \mathrm{x}+{\mathrm{xe}}^{{\mathrm{k}}_1\mathrm{\alpha t}}={\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \\ \mathrm{x}\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right)={\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}} \end{gathered}$$

Then we have

$\mathrm{x}\left(\mathrm{t}\right)=\frac{{\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}{1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}$ is the solution for x in terms of t.

Step 3:Find the solution of  y(t)=(1+ek1αt)k2k1

Second, we have to obtain the solution for y by substituting with the solution of x shown in (3) into equation (2), then we have

$\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{{\mathrm{k}}_2{\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}{1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}\mathrm{y}$

Since this differential equation is separable, then we can solve it as

$$\begin{gathered} \frac{1}{y}dy=\frac{{\mathrm{k}}_2{\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}{1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}dt \\ \frac{1}{\mathrm{y}}\mathrm{dy}=\frac{{\mathrm{k}}_2\mathrm{\alpha }{\mathrm{k}}_1{\mathrm{\alpha e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}{{\mathrm{k}}_1\mathrm{\alpha }1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}}\mathrm{dt} \\ \mathrm{In}\left(\mathrm{y}\right)=\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\mathrm{In}\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right) \\ \mathrm{In}\left(\mathrm{y}\right)=\mathrm{In}{\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right)}^{\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}} \\ {}_{\mathrm{e}}\mathrm{In}\left(\mathrm{y}\right)={}_{\mathrm{e}}\mathrm{In}{\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right)}^{\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}} \end{gathered}$$

Then we have

$\mathrm{y}\left(\mathrm{t}\right)={\left(1+{\mathrm{e}}^{{\mathrm{k}}_1\mathrm{\alpha t}}\right)}^{\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}}$is the solution for in terms of t.