Differential EquationHard

Question

A curve passes through the point & its slope at any point is given by - cos2 . Then the curve has the equation -

Options

A.y = x tan-1
B.y = x tan-1(1n + 2)
C.y = tan-1
D.none

Solution

- cos2
Put y = vx then = v + x
v + x = v - cos2v

⇒  tan v = - ln x + c   ⇒ tan = - ln x + c
As it passes through the point (1, )
so   c = 1
tan = -ln x + 1   ⇒ tan = ln
∴  y = x tan-1 (ln )

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