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Solve the Initial Value Problem

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Step 1

-2y = 0 when y = 0. Thus the constant y(t) = 0 is a particular solution.

Step 2

Separate the variables and integrate both sides,

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where C = eB if y > 0, and C = -eB if y < 0.

Step 3

General solution: y(t) = Ce-2t, where C is any constant (C can be 0 from Step 1).

Step 4

Substitute 1 for t and - 5 for y, and solve for C. -5 = Ce-2·1, -5 = Ce-2, C = -5e2. Particular solution: y(t)= -5e2e-2t = -5e(2-2t). The graph of the solution is shown in Figure 14.1.4.

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Figure 14.1.4 Example 1

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Last Update: 2006-11-17