If xdy = y(dx + y dy), y(1) = 1 and y(x) > 0, then y(- 3) is equal to
3
2
1
0
A.
Given,xdy = y (dx + y dy), y > 0⇒ xdy - ydxy2 = dy⇒ dxy = - dyOn integrating both sides, we getxy = - y + c ...(i)As y(1) = 1 ⇒ x = l, y = 1∴ c = 2∴ Eq. (i) becomes, xy + y = 2Again for x = - 3⇒ - 3 + y2 = 2y ⇒ y2 - 2y - 3 = 0⇒ y + 1y - 3 = 0Also, y > 0 ⇒ y = 3, ∵ neglecting y = -1
If ∫abx3dx = 0 and ∫abx2dx = 23, then the values of a and b are respectively
1, - 1
- 1, 1
1, 1
- 1, - 1
The differential equation representing the family of curves y = 2c (x + c3), where c is a positive parameter, is of
order 1, degree 1
order 1, degree 2
order 1, degree 3
order 1, degree 4
The differential equation representing the family of curves y = xecx (c is a constant) is
dydx = yx1 - logyx
dydx = yxlogyx + 1
dydx = yx1 + logyx
dydx + 1 = yxlogyx
The solution of dydx = 1 + y + y2 + x + xy + xy2 is
tan-12y + 13 = x + xy + xy2
4tan-12y + 13 = 322x + x2
3tan-13y + 13 = 41 + x + x2 + c
tan-12y + 13 = 32x + x2 + c
The integrating factor of the differential equation
dydx + y1 - xx = 1 - x is
1 - x1 + x
1 + x1 - x
x1 - x