Question

Reread the discussion following Example 2. Construct a linear first-order differential equation for which all nonconstant solutions approach the horizontal asymptote.

$y=4 as x\to \infty$

Short answer

The equation of linear first-order differential equation of horizontal asymptote$y\left(t\right)=e^{-t}c+4$

Step-by-step solution

Definition of initial-value problem

The unknown function $y\left(x\right)$and its derivatives at a number$x_0$.On some intervalIcontaining$x_0$the problem of solving an nth-order differential equation subject to n side conditions specified at:

Solve:$\frac{d^{n}y}{d{x}^n}=f(x,y,y\text{'},...,y^{(n-1)})$

Subject to: $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$

Where $y_0,y_1,...,y_{n-1}$ are arbitrary constants, is called n-th order Initial Value Problem (IVP). The values of $y\left(x\right)$and its first n-1 derivatives at $x_0$$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$are called Initial Conditions.

Formula used in the Equation

First order linear differential equation

$y\text{'}+P\left(t\right)y=f\left(t\right)$

Determination of differential equation  

First order linear differential equation

$y\text{'}+P\left(x\right)y=f\left(x\right)······\left(1\right)$and

The critical point is$y=4$

The critical point will be solved only if$\frac{dy}{dt}=0$ , substitute$y=4$

$$\begin{gathered} \frac{dy}{dt}=4-y \\ \frac{dy}{dt}+y=4······\left(2\right) \end{gathered}$$

Based on the first order linear differential equation and Eqn (2)

we get

$P\left(t\right)=1 and f\left(t\right)=4$

Now, substituting the above function in formula

$$\begin{gathered} \left(t\right)=e^{-\int P\left(t\right)dt}\left[c+\int f\left(t\right)e^{\int P\left(t\right)dt}dt\right] \\ y\left(t\right)=e^{-\int dt}\left[c+\int 4e^{\int dt}dt\right] \\ y\left(t\right)=e^{-t}\left[c+4\int e^{\int dt}dt\right] \\ y\left(t\right)=e^{-t}\left[c+4e^t\right] \\ y\left(t\right)=e^{-t}c+4 \end{gathered}$$

Graphing

All nonconstant solutions approach the asymptote $y=4$

$$\begin{gathered} y=4+2e^{-x} \\ y=4+e^{-x} \\ y=4 \\ y=4-6e^{-x} \\ y=4-3e^{-x} \end{gathered}$$

Hence, the equation of a linear first-order differential equation of horizontal asymptote$y\left(t\right)=e^{-t}c+4$