Let y = y(x) be the solution of the differential equation x tan ( y/x )dy = ( y tan ( 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/√2 and y = y(x) in the upper half plane is :

Solution

Maths sir · 5 steps

Put y = vx, the equation separates into cos(y/x) = x, and the area is one integration by parts.

1. Substitute y = vx. Divide through to get dy/dx = y/x - cot y/x, which depends only on y/x. Put y = vx, so dy/dx = v + xdv/dx. The v cancels on both sides, leaving xdv/dx = -cot v.
2. Separate and integrate. tan v dv = -dx/x gives -ln|cos v| = -ln|x| + C. So cos y/x = kx for some constant k.
3. Fix the constant. At x = 1/2, y = π/6, so y/x = π/3 and cos π/3 = 1/2 = k/2. So k = 1 and the curve is y = x cos⁻¹ x, which is positive on [0, 1/√2], so it lies in the upper half plane.

   ∫ x cos⁻¹ x dx = x²/2 cos⁻¹ x + 1/4(sin⁻¹ x - x√(1-x²))

4. Evaluate the limits. This comes from integration by parts with u = cos⁻¹ x, then x = sin θ for ∫ x²/√(1-x²) dx. At x = 1/√2: 1/4·π/4 + 1/4(π/4 - 1/2) = π/16 + π/16 - 1/8. At x = 0 the expression is 0.
5. So. The area is π/8 - 1/8 = 1/8(π - 1), which is option (a).

The idea to keep. When the right side of dy/dx depends only on y/x, put y = vx: the v terms cancel and the variables separate. Use the initial condition to fix the constant before you find the area.

Answer from the NTA final answer key. Solution by PrepWiser.