Differential EquationHard

Question

A curve passing through (2, 3) and satisfying the differential equation ty(t)dt = x2y(x), (x > 0) is -

Options

A.x2 + y2 = 13
B.y2 = x
C.
D.xy = 6

Solution

ty(t)dt = x2y(x)
Differentiating, we get
xy = 2xy + x2     ⇒ x2 + xy = 0
x + y = 0   ⇒  xdy + ydx = 0
d(xy) = 0   ⇒  xy = c
∴  since it passes through (2, 3)
∴ c = 6
Hence xy = 6.

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