Question

If the slope of the tangent at (x, y) to a curve passing through `(1, π/4)` is given by `y/x - cos^2(y/x)`, then the equation of the curve is ______.

विकल्प

• y = `tan^-1 log(e/x)`

• y = `e^(1+cot (y/x))`

• y = `x tan^-1 log(e/x)`

• y = `e^(1 + tan(y/x)`

Answer 1

If the slope of the tangent at (x, y) to a curve passing through `(1, π/4)` is given by `y/x - cos^2(y/x)`, then the equation of the curve is `underlinebb(y = x  tan^-1 log(e/x))`.

Explanation:

Given, `dy/dx = y/x - cos^2(y/x)`

Putting y = vx so that `dy/dx = v + x (dv)/dx`

We get, `v + x (dv)/dx` = v – cos² v

`\implies (dv)/(cos^2v) = -(dv)/x`

`\implies` sec² v dv = `-dx/x`

Integrating, we get, tan v = – ln x + ln c

`tan(y/x)` = – ln x + ln c

This passes through `(1, π/4)` `\implies` ln c = 1

∴ y = `xtan^-1(log  e/x)`