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Home Differential Equations Equations with Seperable Variables Examples Find the general solution
 
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Find the general solution

After solving the initial value problem find the particular solution with j;(0) = ½.

Step 1

y2 = 0 when y = 0. Thus y(t) = 0 is a constant solution.

Step 2

y-2 dy = sin t dt, -y-1 = -cos t + C, y = (cos t - C)-1.

Step 3

General solution: y(t) = 0 and y(t) = (cos t - C)-1.

Step 4

y(0) = ½ = (cos 0 - C)-1 = (1 - C)-1, 2 = 1 - C, C= -1.

Particular solution:

y(t) = (cos t + 1)-1.

The particular solution to Example 2 is illustrated in Figure 14.1.5. It is defined only for -π ≤ t < π and approaches ∞ as t approaches π. It is said to have an explosion at t = π.

14_differential_equations-17.gif

Figure 14.1.5 Example 2

To avoid errors, the general solution can be checked by differentiating. The solution for Example 2 is checked as follows.

14_differential_equations-18.gif

as required.
Home Differential Equations Equations with Seperable Variables Examples Find the general solution

Last Update: 2006-11-17