Differential EquationMedium

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 )

Create a free account to view solution

View Solution Free
Topic: Differential Equation·Practice all Differential Equation questions

More Differential Equation Questions

The differential equation representing the family of curves y2 = 2c (x + √c), where c > 0, is a parameter, is of o...The general solution of differential equation (4 + 5 sin x) = cos x is-...At present a firm is manufacturing 2000 items. It is estimated that the rate of change of production P w.r.t. additional...The order of the differential equations whose general solution is given by y = (c1 + c2) cos(x + c3) - c4ex+c5 where c1,...If = e−2y and y = 0 when x = 5, the value of x for y = 3 is-...