Engineering Board Exam Reviewer · Second-Order Linear Homogeneous Equations · medium

Solve y''-y'-6y=0 with y(0)=0 and y'(0)=5, and find y( 2).

  1. 8.25
  2. 7.75
  3. 7
  4. 15.5

Answer

B. 7.75

Solution

r2-r-6=0 r=3,-2, so y=C1e3x+C2e-2x. From C1+C2=0 and 3C1-2C2=5: C1=1, C2=-1. Then y( 2)=23-2-2=8- 14=7.75.

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