Numerical Solutions

We consider the 1st order differential equation $y^{\prime} = f(x,y)$ . We can think of $y'$

as the slope of the tangent line at $(x,y)$

. Then consider the line starting at $(x_{0},y_{0})$ with the slope $f(x_{0},y_{0})$ . To draw this as graph, we first draw a curve $f(x,y) = c$

. This curve is called isocline. Now we plot points on the curve, and at each point we draw a short pointed line with the slope $f(x,y)$

. The collection of these pointed lines are called direction field of the differential equation $y^{\prime} = f(x,y)$ .

Up to this point, we try to find the general solution. Thus we ignore the singular solution. In this section, we treat the singular solution using the direction field. A general solutin with a singular solution is called the complete solution.

Example 1..26 Find the complete solution of $y^{\prime} = \sqrt{1 - y^{2}}$ .

SOLUTION We first find the general solution.

$\displaystyle \frac{dy}{\sqrt{1-y^{2}}} = \int dx$

$\displaystyle \sin^{-1}{y} = x + c$

$\displaystyle y = \sin{(x + c)}$

We note that the derivative of right-hand side becomes negative. Thus the derivative of left-hand side $y'$

becomes negative. But by the give differential equation, $y'$

never becomes negative. Then we draw directional fiels using isocline lines .

**Figure 1.2:** solutio curve
$\begin{figure}\begin{center} \includegraphics[width=10cm]{DFQ/Fig1-2.eps} \end{center}\end{figure}$

From this, $y \equiv 1$ and $y \equiv -1$ are solutions. Also, $y = 1$ と $y = -1$ are connected with $\sin$ curve. Thus the complete solution is $y = 1$ or or

$\displaystyle y = \left\{\begin{array}{ll} -1, & x+c \leq -\frac{\pi}{2} \\ \s... ...2}\\ 1, & x+c \geq \frac{\pi}{2} \end{array}\right.\ensuremath{\ \blacksquare}$

We next consider the way to obtain numerical solution from the graph. The idea is to draw a line with the slope $f(x_0,y_0)$ at the starting point $(x_0,y_0)$ . Then the equation of line is given by

$\displaystyle y = f(x_{0},y_{0})(x - x_{0}) + y_{0} .$

Then take the next point as $x_{1} = x_{0} + h$ and $y_{1} = f(x_{0},y_{0})h + y_{0}$ . Now with the point $(x_{1},y_{1})$ and the slope $f(x_{1},y_{1})$ , we get

$\displaystyle y = f(x_{1},y_{1})(x - x_{1}) + y_{1} .$

Repeat this until we reach the given value of $x$

Example 1..27 Given $y^{\prime} = y + x^{2}, y(0) = 1$ , find the value of $y(3)$

provided

SOLUTION Since $(x_{0},y_{0}) = (0,1)$ , we have $y = 1 + x$ . Now since $(x_{1},y_{1}) = (1,2)$ , we have $y = 3x - 1$ . Finally, for $(x_{2},y_{2}) = (2,5)$ , we have $y = 9x - 13$ . Thus, $y(3) = 14$ . Note that this differential equation is linear. Thus we can find the general solution. $y = 3e^{x} - x^{2} - 2x - 2$ . Then $y(3) = 3e^{3} - 17 \doteq 43.3$ $\ \blacksquare$

As you noticed, the error by the approximation lines is large. To make the error small, we must use smaller $h$ . The smaller the $h$ , the more calsulation for computation. Then we use computer to calculate.

The more useful technique is known as Euler's method.
The approximated solution of

$\displaystyle y^{\prime} = f(x,y), \ y(a) = b$

can be obatinded by the following method.

$\begin{displaymath}\begin{array}{ll} x_{0} = a & y_{0} = b\\ x_{1} = a + h & y_... ...} = a + nh & y_{n} = y_{n-1} + f(x_{n-1},y_{n-1})h \end{array} \end{displaymath}$

Subsections

Exercise