Let y = y(x) be the solution curve of the differential equation x (x2 + ex )dy + (ex(x - 2)y - x3 )dx = 0, x > 0, passing through the point (1, 0). Then y(2) is equal to :
- 22 + e2
- 44 - e2
- 44 + e2
- 22 - e2
Answer
(3) 44 + e2
Solution
Dividing the equation by x3 gives
(1 + exx2 )dy + y ex(x - 2)x3 dx - dx = 0
ddx ( exx2 ) = x2ex - 2xexx4 = ex(x - 2)x3, this can be written as
dy + [ exx2dy + y d ( exx2 ) ] = dx
d (y + yexx2 ) = dx
y (1 + exx2 ) = x + C
At (1, 0), 0 = 1 + C, so C = −1 and
y(2) (1 + e24 ) = 2 - 1 = 1
y(2) = 44 + e2
Answer: (3)