Section: Engineering Mathematics | Type: NAT | Marks: 2
Let π’(π₯, π‘) be the solution of the initial boundary value problem ππ’ ππ‘β π2π’ ππ₯2 = 0, π₯β(0, 2), π‘> 0 π’(π₯, 0) = sin(ππ₯), π₯β(0, 2) π’(0, π‘) = π’(2, π‘) = 0. Then the value of ππ2 (π’( 1 2 , 1) βπ’( 3 2 , 1)) is __________ (in integer).
Answer: 2