“To imagine yourself subject to a differential equation, start somewhere. There you are tugged in some direction, so you move that way ... as you move, the tugging forces change, pulling you in a new direction; for your motion to solve the differential equation you must keep drifting with and responding to the ambient forces.”

—From the introduction to Differential Equations, J. H. Hubbard and Beverly West, Springer-Verlag, New York, 1991

EXAMPLE 1 Using Isoclines

Draw the slope field for \[\dot{y}=y-t\] and sketch the integral curves satisfying the initial conditions (a) \(y(0) = 1\) and (b) \(y(1) = -2\).

Solution

A good way to sketch the slope field of \(\dot{y} = F(t, y)\) is to choose several values \(c\) and identify the curve \(F(t,y)=c\), called the isocline of slope \(c\). The isocline is the curve consisting of all points where the slope field has slope \(c\).

In our case, \(F(t,y)=y-t\), so the isocline of fixed slope \(c\) has equation \(y-t=c\), or \(y = t+c\), which is a line. Consider the following values:

• \(c = 0\): This isocline is \(y -t = 0\), or \(y=t\). We draw segments of slope \(c=0\) at points along the line \(y=t\), as in Figure 9.14.
• \(c = 1\): This isocline is \(y -t = 1\), or \(y=t+1\). We draw segments of slope \(1\) at points along \(y=t+1\), as in Figure 9.14.
• \(c = 2\): This isocline is \(y -t = 2\), or \(y=t+2\). We draw segments of slope \(2\) at points along \(y=t+2\), as in Figure 9.14.
• \(c = -1\): This isocline is \(y -t = -1\), or \(y=t-1\) [Figure 9.14.

  A more detailed slope field is shown in Figure 9.14. To sketch the solution satisfying \(y(0) = 1\), begin at the point \((t_0,y_0) = (0,1)\) and draw the integral curve that follows the directions indicated by the slope field. Similarly, the graph of the solution satisfying \(y(1) = -2\) is the integral curve obtained by starting at \((t_0,y_0) = (1,-2)\) and moving along the slope field. Figure 9.14 shows several other solutions (integral curves).

Figure 9.14: Drawing the slope field for \(\dot{y}=y-t\) using isoclines.

GRAPHICAL INSIGHT

Slope fields often let us see the asymptotic behavior of solutions (as \(t\to\infty\)) at a glance. Figure 9.14 suggests that the asymptotic behavior depends on the initial value (the \(y\)-intercept): If \(y(0)>1\), then \(y(t)\) tends to \(\infty\), and if \(y(0)<1\), then \(y(t)\) tends to \(-\infty\). We can check this using the general solution \(y(t) = 1 + t + Ce^t\), where \(y(0)=1+C\). If \(y(0)>1\), then \(C>0\) and \(y(t)\) tends to \(\infty\), but if \(y(0)<1\), then \(C<0\) and \(y(t)\) tends to \(-\infty\). The solution \(y=1+t\) with initial condition \(y(0)=1\) is the straight line shown in Figure 9.14.

EXAMPLE 2 Newton's Law of Cooling Revisited

The temperature \(y(t)\)(\(^\circ\)C) of an object placed in a refrigerator satisfies \(\dot{y} = -0.5(y-4)\)(\(t\) in minutes). Draw the slope field and describe the behavior of the solutions.

Solution

The function \(F(t,y)=-0.5(y-4)\) depends only on \(y\), so slopes of the segments in the slope field do not vary in the \(t\)-direction. The slope \(F(t,y)\) is positive for \(y<4\) and negative for \(y>4\). More precisely, the slope at height \(y\) is \(-0.5(y - 4) = -0.5y + 2\), so the segments grow steeper with positive slope as \(y\to -\infty\), and they grow steeper with negative slope as \(y\to \infty\) (Figure 9.15).

The slope field shows that if the initial temperature satisfies \(y_0>4\), then \(y(t)\) decreases to \(y=4\) as \(t\to\infty\). In other words, the object cools down to \(4^\circ\)C when placed in the refrigerator. If \(y_0<4\), then \(y(t)\) increases to \(y=4\) as \(t\to\infty-\) the object warms up when placed in the refrigerator. If \(y_0 = 4\), then \(y\) remains at \(4^\circ\)C for all time \(t\).

CONCEPTUAL INSIGHT

Most first-order equations arising in applications have a uniqueness property: There is precisely one solution \(y(t)\) satisfying a given initial condition \(y(t_0) = y_0\). Graphically, this means that precisely one integral curve (solution) passes through the point \((t_0,y_0)\). Thus, when uniqueness holds, distinct integral curves never cross or overlap. Figure 9.16 shows the slope field of \(\dot{y}=-\sqrt{|y|}\), where uniqueness fails. We can prove that once an integral curve touches the \(t\)-axis, it either remains on the \(t\)-axis or continues along the \(t\)-axis for a period of time before moving below the \(t\)-axis. Therefore, infinitely many integral curves pass through each point on the \(t\)-axis. However, the slope field does not show this clearly. This highlights again the need to analyze solutions rather than rely on visual impressions alone.

Euler's Method is the simplest method for solving initial value problems numerically, but it is not very efficient. Computer systems use more sophisticated schemes, making it possible to plot and analyze solutions to the complex systems of differential equations arising in areas such as weather prediction, aerodynamic modeling, and economic forecasting.

GRAPHICAL INSIGHT

The values \(y_k\) are defined so that the segment joining \(P_{k-1}\) to \(P_k\) has slope \[ \frac{y_k - y_{k-1}}{t_k-t_{k-1}} = \frac{(y_{k-1}+hF(t_{k-1},y_{k-1})) - y_{k-1}}h = F(t_{k-1},y_{k-1}) \] Thus, in Euler's method we move from \(P_{k-1}\) to \(P_k\) by traveling in the direction specified by the slope field at \(P_{k-1}\) for a time interval of length \(h\)(Figure 9.17).

EXAMPLE 3

Use Euler's Method with time step \(h=0.2\) and \(n=4\) steps to approximate the solution of \(\dot{y}=y-t^2\), \(y(0)=3\).

Solution Our initial value at \(t_0 = 0\) is \(y_0=3\). Since \(h=0.2\), the time values are \(t_1 = 0.2\), \(t_2=0.4\), \(t_3=0.6\), and \(t_4=0.8\). We use Eq. (3) with \(F(t,y) = y - t^2\) to calculate \[ \begin{align*} y_1 &= y_0+hF(t_0,y_0) = 3 + 0.2(3-(0)^2) = 3.6\\ y_2 &= y_1+hF(t_1,y_1) = 3.6 + 0.2(3.6-(0.2)^2) \approx 4.3\\ y_3 &= y_2+hF(t_2,y_2) = 4.3 + 0.2(4.3-(0.4)^2) \approx 5.14 \\ y_4 &= y_3+hF(t_3,y_3) = 5.14 + 0.2(5.14-(0.6)^2) \approx 6.1 \end{align*} \]

Figure 9.18 shows the exact solution \(y(t)=2+2t+t^2+e^t\) together with a plot of the points \((t_k,y_k)\) for \(k=0,1,2,3,4\) connected by line segments.

CONCEPTUAL INSIGHT

Figure 9.18 shows that the time step \(h=0.1\) gives a better approximation than \(h=0.2\). In general, the smaller the time step, the better the approximation. In fact, if we start at a point \((a,y(a))\) and use Euler's Method to approximate \((b,y(b))\) using \(N\) steps with \(h=(b-a)/N\), then the error is roughly proportional to \(1/N\)(provided that \(F(t,y)\) is a well-behaved function). This is similar to the error size in the \(N\) th left- and right-endpoint approximations to an integral. What this means, however, is that Euler's Method is quite inefficient; to cut the error in half, it is necessary to double the number of steps, and to achieve \(n\)-digit accuracy requires roughly \(10^n\) steps. Fortunately, there are several methods that improve on Euler's Method in much the same way as the Midpoint Rule and Simpson's Rule improve on the endpoint approximations (see Exercises 22–27).

EXAMPLE 4

Let \(y(t)\) be the solution of \(\dot{y}=\sin t\;\cos y\), \(y(0)=0\).

1. Use Euler's Method with time step \(h=0.1\) to approximate \(y(0.5)\).
2. Use a computer algebra system to implement Euler's Method with time steps \(h= 0.01\), 0.001, and 0.0001 to approximate \(y(0.5)\).

Solution

1. When \(h=0.1\), \(y_k\) is an approximation to \(y(0+k(0.1)) =y(0.1k)\), so \(y_5\) is an approximation to \(y(0.5)\). It is convenient to organize calculations in the following table. Note that the value \(y_{k+1}\) computed in the last column of each line is used in the next line to continue the process.

   \(t_k\)	\(y_k\)	\(F(t_k,y_k)=\sin t_k\;\cos y_k\)	\(y_{k+1}=y_k+hF(t_k,y_k)\)
   \(t_0=0\)	\(y_0=0\)	\((\sin 0)\cos0=0\)	\(y_1 = 0+0.1(0)=0\)
   \(t_1=0.1\)	\(y_1 = 0\)	\((\sin 0.1)\cos 0\approx 0.1\)	\(y_2 \approx 0+0.1(0.1)=0.01\)
   \(t_2=0.2\)	\(y_2\approx 0.01\)	\((\sin 0.2)\cos(0.01) \approx 0.2\)	\(y_3 \approx 0.01+0.1(0.2)= 0.03\)
   \(t_3=0.3\)	\(y_3\approx 0.03\)	\((\sin 0.3)\cos(0.03) \approx 0.3\)	\(y_4 \approx 0.03+0.1(0.3)= 0.06\)
   \(t_4=0.4\)	\(y_4\approx 0.06\)	\((\sin 0.4) \cos(0.06) \approx 0.4\)	\(y_5 \approx 0.06+0.1(0.4)= 0.10\)

   Thus, Euler's Method yields the approximation \(y(0.5)\approx y_5 \approx 0.1\).
2. When the number of steps is large, the calculations are too lengthy to do by hand, but they are easily carried out using a CAS. Note that for \(h=0.01\), the \(k\) th value \(y_k\) is an approximation to \(y(0+k(0.01)) =y(0.01k)\), and \(y_{50}\) gives an approximation to \(y(0.5)\). Similarly, when \(h = 0.001\), \(y_{500}\) is an approximation to \(y(0.5)\), and when \(h = 0.0001\), \(y_{5{,}000}\) is an approximation to \(y(0.5)\). Here are the results obtained using a CAS:

   Time step \(h = 0.01\)	\(y_{50}\) \(\approx\) \(0.1197\)
   Time step \(h = 0.001\)	\(y_{500}\) \(\approx\) \(0.1219\)
   Time step \(h = 0.0001\)	\(y_{5000}\) \(\approx\) \(0.1221\)

   The values appear to converge and we may assume that \(y(0.5)\approx 0.12\). However, we see here that Euler's Method converges quite slowly.