Let Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y-y=Y (x)(X-x) and the co-ordinate axes, where (x, y) is any point on the curve, is always -y22 Y (x)+1, Y (x) 0. If Y(1)=1, then 12 Y(2) equals .
Answer
20
Solution
[figure]
Tangent at (x, y) meets the axes at (x- yY', 0 ) and (0, y-xY')
Area = 12 (x- yY' )(y-xY')=xy- x2Y'2- y22Y'
=- y22Y'+1 2xy-x2Y'=2
dydx- 2xy=- 2x2 (linear), I.F. =e-2 x= 1x2
yx2= - 2x4dx= 23x3+c
Y(1)=1 c= 13 y= 23x+ x23
Y(2)= 13+ 43= 53 12Y(2)=20
Answer: 20