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Example 2

(a)     Find the general solution of the equation ty' + 3y = 0 for t > 0.

(b)    Find the particular solution with the initial value y(1) = 2.

SOLUTION

(a)     We first put the equation into the homogeneous linear form (1) by dividing by t:

y'+ 3t-1y = 0.

Next evaluate the integral,

∫ 3t-1 dt = 3 ln t + B.

The constant of integration B is absorbed into the constant C, and the general solution is

y(t) = Ce-3 ln t = Ct-3.

(b)    The particular solution with initial value y(1) = 2 is

y(t) = 2t-3.

The solution to this example is shown in Figure 14.2.2.

14_differential_equations-39.gif

Figure 14.2.2 Example 2

The coefficient 3t-1 in the equation y' + 3t-1y = 0 of Example 2 is discontinuous at t = 0. However, it is continuous on the interval t > 0 and on the interval t < 0. An initial value at a positive time t0 > 0 will determine a particular solution only for the interval t > 0, while an initial value at a negative time t0 < 0 will determine a particular solution for the interval t < 0. Each interval must be solved separately. The next example is like Example 2 but on the negative time interval.


Last Update: 2006-11-17