Section: Engineering Mathematics | Type: NAT | Marks: 1
Let π¦(π₯) be the solution of the initial value problem, π₯2π¦β²β² + π₯π¦β² βπ¦= 0, π₯> 0, π¦(1) = 0, π¦β²(1) = 2. Then the value of π¦β² ( 1 2) is equal to (Answer in integer) _______
Answer: 5