If y=y(x) is the solution of the differential equation, √(4-x²) (d y)/(d x)=((sin ⁻¹(x/2))²-y) sin ⁻¹(x/2),-2 ≤ x ≤ 2, y(2)=(π²-8)/4, then y²(0) is equal to ___________.

Solution

Maths sir · 4 steps

Put t = sin⁻¹(x/2) and this messy equation becomes a clean linear one in t.

1. Change the variable. Let t = sin⁻¹(x/2). Then dt/dx = 1/√(4-x²), so the left side √(4-x²) dy/dx is exactly dy/dt. The equation becomes dy/dt = (t² - y) t, that is, dy/dt + t y = t³.
2. Integrating factor. The integrating factor is e^(∫ t dt) = e^(t²/2). For the right side, put u = t²/2, so t dt = du and t² = 2u. Then ∫ t³ e^(t²/2) dt = ∫ 2u eᵘ du = 2(u-1)eᵘ = (t²-2)e^(t²/2).

   y e^(t²/2) = (t²-2) e^(t²/2) + C ⇒ y = t² - 2 + C e^(-t²/2)

3. Use the condition. At x = 2, t = π/2, so y = π²/4 - 2 + C e^(-π²/8). This must equal (π²-8)/4 = π²/4 - 2, so C = 0 and y = t² - 2.
4. So. At x = 0, t = 0, so y(0) = -2 and y²(0) = **4**.

The common slip: -2. Stopping at y(0) gives -2. The question asks for y²(0), which is 4.

The idea to keep. When the same awkward expression and its derivative both sit in the equation, substitute for it. Here 1/√(4-x²) is the derivative of sin⁻¹(x/2), so the equation turns linear in t.

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