Chapter 7.3, Problem 4QQ

a.

To determine

To sketch: the solution curve that passes through the point (0, 1) and sketch the solution curve that goes through the point (0, -1).

Answer to Problem 4QQ

Explanation of Solution

Given information: Consider the differential equation $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2\mathrm{y}-4\mathrm{x}$

Calculation:

Draw the solution curve passing through (0, 1) (shown in red) and (0, -1) (shown in green) using the slope of the line segments in the slope field to approximate the slope of the curves at different values of x .

  

b.

To determine

To find: the value of b for which $\mathrm{y}=2\mathrm{x}+\mathrm{b}$ is a solution to the differential equation.

Answer to Problem 4QQ

  $\mathrm{b}=1$ .

Explanation of Solution

Given information:

Calculation:

The slope of $\mathrm{y}=2\mathrm{x}+\mathrm{b}$ is 2 for all x so $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2$ . Substitute $\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2$ and $\mathrm{y}=2\mathrm{x}+\mathrm{b}$ into the differential equation.

  $\begin{array}{c}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2\mathrm{y}-4\mathrm{x}\\ 2=2(2\mathrm{x}+\mathrm{b})-4\mathrm{x}\\ 2=4\mathrm{x}+2\mathrm{b}-4\mathrm{x}\\ 2=2\mathrm{b}\\ \mathrm{b}=1.\end{array}$

c.

To determine

To prove: analytically that g has a local maximum at the point (0, 0) using the differential equation.

Answer to Problem 4QQ

(0, 0) is a local maximum.

Explanation of Solution

Given information: Let g be the function that satisfies the given differential equation

with the initial condition g (0) = 0.

Calculation:

  $\begin{array}{l}\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2\mathrm{y}-4\mathrm{x}.\\ {\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)}_{0.0}=2(0)-4(0)=0\\ \frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}=2\mathrm{y}-4\mathrm{x})\\ =2\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}-4\\ {\left(\frac{{\mathrm{d}}^2\mathrm{y}}{\mathrm{d}{\mathrm{x}}^2}\right)}_{0,0}=2{\left(\frac{\mathrm{d}\mathrm{y}}{\mathrm{d}\mathrm{x}}\right)}_{0,0}-4=2(0)-4=-4\Rightarrow (0,0)\text{is}\text{a}\text{local}\text{maximum}\text{.}\end{array}$

By second derivative lest, any critical point, in this case (0, 0) is a local maximum.