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Home Differential Equations First Order Homogenous Linear Equations Example Example 2 | |
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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:
(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. 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.
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Home Differential Equations First Order Homogenous Linear Equations Example Example 2 |