If a curve y = y(x) passes through the point (1, 2 ) and satisfies the differential equation (7x4 y - ex cosec y ) dxdy = x5, x 1, then at x = 2, the value of y is :
- 2e2 + e64
- 2e2 - e64
- 2e2 - e128
- 2e2 + e128
Answer
(3) 2e2 - e128
Solution
dydx = 7x4 y - ex cosec yx5
Multiplying by y:
y dydx = 7 yx - exx5
Put y = t - y dydx = dtdx
dtdx + 7xt = exx5 (linear in t)
I.F. = e 7xdx = x7
t x7 = x2 ex dx = ex(x2 - 2x + 2) + C
At x = 1, y = 2 t = 0: 0 = e(1 - 2 + 2) + C C = -e
At x = 2: 128 y = e2(4 - 4 + 2) - e = 2e2 - e
y = 2e2 - e128
Answer: (3)