Let y = y(x) be the solution of the differential equation (1 + x2 ) dydx + y = e -1x, y(1) = 0. Then y(0) is
- 14 (e /2 - 1 )
- 12 (1 - e /2 )
- 14 (1 - e /2 )
- 12 (e /2 - 1 )
Answer
(2) 12 (1 - e /2 )
Solution
Linear D.E.: dydx + y1 + x2 = e -1x1 + x2
I.F. = e dx1 + x2 = e -1x
y e -1x = e2 -1x1 + x2 dx
= 12e2 -1x + C (put -1x = t)
y(1) = 0: 0 = 12e /2 + C C = - 12e /2
At x = 0: y(0) 1 = 12 - 12e /2 = 12 (1 - e /2 )
Answer: (2)