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The equation of the curve passing throug...

The equation of the curve passing through the point `(1,pi/4)` and having a slope of tangent at any point (x,y) as `y/x - cos^2(y/x)` is

Text Solution

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Here, slope is given as `y/x-cos^2(y/x)`
`:. dy/dx = y/x-cos^2(y/x)->(1)`
Let, `y/x = v => y = vx => dy/dx = v+x(dv)/dx`
so, (1) becomes,
`=>v+x(dv)/dx = v- cos^2v`
`=>(dv)/cos^2v = -dx/x`
Integrating both sides,
`int (dv)/cos^2v = - int dx/x`
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