Question

Let y = y(x) be the solution of the differential equation `xtan(y/x)dy = (ytan(y/x) - x)dx, -1 ≤ x ≤ 1, y(1/2) = π/6`. Then the area of the region bounded by the curves x = 0, x = `1/sqrt(2)` and y = y(x) in the upper half plane is ______.

Options

• `1/12(π - 3)`

• `1/6(π - 1)`

• `1/8(π - 1)`

• `1/4(π - 1)`

Answer 1

Let y = y(x) be the solution of the differential equation `xtan(y/x)dy = (ytan(y/x) - x)dx, -1 ≤ x ≤ 1, y(1/2) = π/6`. Then the area of the region bounded by the curves x = 0, x = `1/sqrt(2)` and y = y(x) in the upper half plane is `underlinebb(1/8(π - 1))`.

Explanation:

Given differential equation

⇒ `xtan(y/x)dy = ytan(y/x)dx - xdx`

⇒ `tan(y/x)(xdy - ydx) = -xdx`

⇒ `tan(y/x)((xdy - ydx)/x^2) = (-x)/x^2dx`

⇒ `int tan(y/x)(d(y/x)) = int (-1)/xdx`

⇒ In`|sec(y/x)|` = –In x + c

⇒ In`|xsec(y/x)|` = c

Now, apply `y(1/2) = π/6` in above

∴ In`|1/2sec(π/3)|` = c

∴ In`|1/2 xx 2|` = c ⇒ c = In 1 = 0

∴ `sec(y/x) = 1/x`

∴ y = `xsec^-1(1/x)`

So, required bounded area in upper half,

A = `int_0^(1/sqrt(2))xsec^-1(1/x)dx = int_0^(1/sqrt(2)) xcos^-1(x)dx`

Using integration by parts

= `|(x^2/2 cos^-1x)|_0^(1/sqrt(2)) + int_0^(1/sqrt(2)) x^2/(2sqrt(1 - x^2))`

= `(1/4.π/4 - 0) + 1/2 int_0^(1/sqrt(2)) (1 - (1 - x^2))/(2sqrt(1 - x^2))dx`

= `π/16 + 1/2[(sin^-1x)_0^(1/sqrt(2)) - int_0^(1/sqrt(2)) sqrt(1 - x^2)dx]`

= `π/16 + 1/2[π/4 - {1/2xsqrt(1 - x^2) + 1/2sin^-1x}_0^(1/sqrt(2))]`

= `π/16 + 1/2[π/4 - {1/4 + π/8}]`

∴ Area = `(π - 1)/8`