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Home Differential Equations Second Order Homogeneous Linear Equations Examples Example 3
 
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Example 3

Find the solution of the initial value problem

y" - 4y' + 4y = 0, y(0) = -3, y'(0) = 1.

Step 1

The characteristic polynomial z2 - 4z + 4 has one real root, z = 2.

Step 2

The general solution is y = Ae2t + Bte2t.

Step 3

Substitute 0 for t and -3 for y.

-3 = Ae0 + B · 0 · e0,

A= -3.

Compute y' for the general solution.

y' = 2Ae2t + 2Bte2t + Be2t.

Substitute 0 for t and 1 for y'.

1 = 2Ae0 + 2B · 0 · e0 + Be0 = 2A + B,

B = 7.

The particular solution, shown in Figure 14.6.2, is

y = -3e2t + 7te2t.

14_differential_equations-195.gif

Figure 14.6.2 Example 3

 
Home Differential Equations Second Order Homogeneous Linear Equations Examples Example 3

Last Update: 2006-11-16