Differential EquationHard
Question
A curve passing through (2, 3) and satisfying the differential equation 
ty(t)dt = x2y(x), (x > 0) is -
ty(t)dt = x2y(x), (x > 0) is -Options
A.x2 + y2 = 13
B.y2 = 
x
xC.
D.xy = 6
Solution
ty(t)dt = x2y(x)Differentiating, we get
xy = 2xy + x2
⇒ x2 
+ xy = 0x
+ y = 0 ⇒ xdy + ydx = 0d(xy) = 0 ⇒ xy = c
∴ since it passes through (2, 3)
∴ c = 6
Hence xy = 6.
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