Let y = y(x) be the solution of the differential equation (1 + y2)e xdx + 2 x (1 + e2 x )dy = 0, y(0) = 1. Then y ( 4 ) is equal to
- 1e2
- 2e2
- 2e
- 1e
Answer
(4) 1e
Solution
Separate: dy1 + y2 = - e x 2 x dx1 + (e x )2
Put e x = t e x 2 x dx = dt
-1y = - -1 (e x ) + C
y(0) = 1: 4 = - 4 + C C = 2
x = 4: -1y = 2 - -1e = -1e = -1 1e
y ( 4 ) = 1e
Answer: (4)