Let y = y(x) be the solution curve of the differential equation (1 + x2)dy + (y - -1 x)dx = 0, y(0) = 1. Then the value of y(1) is :
- 4e /4 - 2 - 1
- 2e /4 + 4 - 1
- 4e /4 + 2 - 1
- 2e /4 - 4 - 1
Answer
(2) 2e /4 + 4 - 1
Solution
dydx + y1 + x2 = -1 x1 + x2 (linear)
I.F. = e dx1 + x2 = e -1 x
y e -1 x = -1 x1 + x2e -1 xdx
Put -1 x = t: t et dt = (t - 1)et + C
y = -1 x - 1 + C e- -1 x
y(0) = 1 1 = 0 - 1 + C C = 2
y(1) = 4 - 1 + 2e- /4 = 2e /4 + 4 - 1
Answer: (2)