Q.The general solution of recurrence relation r r r 1 r 2 a 5a 6a 4,r 2 − − − + = ≥ − + = ≥ − + = ≥ − + = ≥ is:
Correct answer: r r r 1 2 A.2 A.3 8.4 + +, where A1 and A2 are arbitrary constants.
Permanent question page
Question #46,741
Math Chapter Wise question with answer
Read, practise, and share this question using its permanent link.