Question

(a) The autonomous first-order differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{(}\mathbf{y}\mathbf{-}\mathbf{3}\mathbf{)}$has no critical points. Nevertheless, place 3 on the phase line and obtain a phase portrait of the equation. Computerole="math" localid="1667825621291" ${\mathbf{d}}^2\mathbf{y}\mathbf{/}{\mathbf{dx}}^2$to determine where solution curves are concave up and where they are concave down (see Problems 35 and 36 in Exercises 2.1). Use the phase portrait and concavity to sketch, by hand, some typical solution curves.

(b) Find explicit solutions${\mathbf{y}}_1\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{y}}_2\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{,}{\mathbf{y}}_3\mathbf{\left(}\mathbf{x}\mathbf{\right)}$, and${\mathbf{y}}_4\mathbf{\left(}\mathbf{x}\mathbf{\right)}$of the differential equation in part (a) that satisfy, in turn, the initial conditions${\mathbf{y}}_1\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}\mathbf{,}{\mathbf{y}}_2\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{y}}_3\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}$, and${\mathbf{y}}_4\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{4}$. Graph each solution and compare with your sketches in part (a). Give the exact interval of definition for each solution.

Short answer

1. Concave up when$\frac{d^{2}y}{d{x}^2}>0$ for $y<3$, and concave down when $\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}<0$for $\mathrm{y}>3$.
2. For $y_1$interval is $x\in \left[-\frac{1}{2},\infty \right]$, $y_2$interval is $x\in \left[-\frac{1}{2},\infty \right]$, $y_3$interval is$x\in \left(\frac{1}{2},\infty \right)$ and $y_4$for interval is $x\in \left[\frac{-3}{2},\infty \right]$.

Step-by-step solution

Phase diagram(a)

The phase portrait is shown below:

Compute d2ydx2:

Now,$\frac{d^{2}y}{d{x}^2}=\frac{-\left(\frac{dy}{dx}\right)}{{(y-3)}^2}$

$-\frac{-\left(\frac{1}{y}-3\right)}{{(y-3)}^2}=\frac{-1}{{(y-3)}^3}$

The solution curve is concave up when$\frac{d^{2}y}{d{x}^2}>0$ for $y<3$, and concave down when $\frac{d^{2}y}{d{x}^2}<0$for $y>3$.

Graph

Solve by separating variable(b)

By separating variables and integrating we obtain, (y-3) dy=dx on integration, we get

$\begin{array}{c}\frac{y^2}{2}-3y=x+c_1\\ y^2-6y=2x+c_2\\ y^2-6y+9=2x+c\\ {\left(y-3\right)}^2=2x+c\\ \therefore y=3\pm \sqrt{2x+c}\end{array}$

We will use initial conditions to determine plus or minus sign.

$\begin{array}{c}{y}_1\left(0\right)=4\\ 4=3\pm c\\ 1=c\end{array}$

$\therefore y_1\left(x\right)=3+\sqrt{2x+1}$, for $x\ge -\frac{1}{2}$, i.e.,$x\in [-\frac{1}{2},\infty )$

Graph

The graph of $y_1\left(x\right)=3+2\sqrt{2x+1}$is shown below:

Find value of constant

The condition$y_2\left(0\right)=2$ implies$2=3-c$

$\Rightarrow c=1$

${\mathrm{y}}_2\left(\mathrm{x}\right)=3-\sqrt{2\mathrm{x}+1}$, for $x\ge -\frac{1}{2}$, i.e., $\mathrm{x}\in [-\frac{1}{2},\mathrm{\infty })$.

Graph

The graph of $y_2\left(x\right)=3-2\sqrt{2x+1}$is shown below:

Find value of constant

$\begin{array}{c}{y}_3\left(1\right)=2\text{implies}2=3-\sqrt{2+c}\\ -1=-\sqrt{2+c}\\ 1=2+c\\ c=-1\end{array}$

Therefore$y_3\left(x\right)=3-\sqrt{2x-1},\text{for}x\ge \frac{1}{2},\text{i.e.,}x\in [\frac{1}{2},\infty )$

Graph

The graph of $y_3\left(x\right)=3-\sqrt{2x-1}$is shown below

Find value of constant

$\begin{array}{c}{y}_4(-1)=4\text{implies}4=3+\sqrt{-2+c}\\ 1=\sqrt{-2+c}\\ 1=-2+c\\ c=3\end{array}$

Therefore,$y_4\left(x\right)=3+\sqrt{2x+3}$ for,$x\ge \frac{-3}{2}$ i.e.,$x\in [\frac{-3}{2},\infty )$ .

Graph

The graph of $y_4\left(x\right)=3+\sqrt{2x+3}$is shown below: