Chapter 7, Problem 3RE

a.

To determine

To sketch : the graph of the solution with initial condition is $y\left(0\right)=1$

Explanation of Solution

Given information : The question is

A direction field for the differential equation $y\text{'}=x^2-y^2$

Graph : The solution curve for the initial condition of $y\left(0.3\right)\approx 0.8$

  

Interpretation :

To draw the solution of the curve. By plotting the initial condition the direction field plot

Starting from this point, the make a curve along the direction field arrows

To find the approximate value of $x=\left(0.3\right)$ . Draw a horizontal

The y-coordinate of point where the horizontal meets the curve is the value of $y\left(0.3\right)\approx 0.8$

b.

To determine

To compare the estimate from part (a)

Answer to Problem 3RE

The estimation is $y\left(0.3\right)\approx 0.75676$

Explanation of Solution

Given information : The question is

Use Euler’s method with step size $0.1$ to estimate $y\left(0.3\right)$ where $y\left(x\right)$ is the solution of the initial problem in part (a)

Calculation :

According to Euler’s method $y\left(x+h\right)\approx y\left(x\right)+h\frac{dy}{dx}\mathrm{......}\left(1\right)$

  $\begin{array}{l}y\left(0\right)=1\left(\text{given}\right)\\ y\left(0.3\right)\left(\text{to find}\right)\\ h=0.1\\ y\left(x+h\right)\approx y\left(x\right)+h\frac{dy}{dx}\\ y\text{'}=x^2-y^2\left(\text{given}\right)\\ \frac{dy}{dx}=x^2-{\left(y\left(x\right)\right)}^2\\ y\left(x+h\right)\approx y\left(x\right)+h\cdot \left[x^2-{\left(y\left(x\right)\right)}^2\right]\\ h=0.1;y\left(0\right)=1;y\left(0+0.1\right)=y\left(0.1\right)\\ y\left(0.1\right)\approx 1+0.1\cdot \left[0-{\left(1\right)}^2\right]\\ y\left(0.1\right)\approx 0.9\\ y\left(x+h\right)\approx y\left(x\right)+h\cdot \left[x^2-{\left(y\left(x\right)\right)}^2\right]\\ h=0.1;y\left(0.1\right)=0.9;y\left(0.1+0.1\right)=y\left(0.2\right)\\ y\left(0.2\right)\approx 0.9+0.1\cdot \left[{\left(0.1\right)}^2-{\left(0.9\right)}^2\right]\\ y\left(0.2\right)\approx 0.082\\ y\left(x+h\right)\approx y\left(x\right)+h\cdot \left[x^2-{\left(y\left(x\right)\right)}^2\right]\\ h=0.1;y\left(0.2\right)=0.82;y\left(0.2+0.1\right)=y\left(0.3\right)\\ y\left(0.3\right)\approx 0.82+0.1\cdot \left[{\left(0.2\right)}^2-{\left(0.82\right)}^2\right]\\ y\left(0.3\right)\approx 0.75676\end{array}$

c.

To determine

at what lines are the center of the horizontal line segment in part (a) located

Explanation of Solution

Given information : The question is

The solution curve crosses these lines

Graph :

The centers of horizontal vectors in the vectors field lie along the lines $y=-x\&y=x$

  

Interpretation :

The solution curve of vector field has the vector of the vector field and its tangent vector

So, if the tangent vector is horizontal that means the curve has zero slope at that point