Linear ODE with an Inverse Cosine Term – JEE Main 2026 PYQ

Quick Summary

Dividing the given differential form by x√(1−x²)dx converts it into the linear ODE dy/dx + y/x = cos⁻¹(x)/√(1−x²), with integrating factor x. The right-hand side integral is handled by the substitution w=cos⁻¹x (so x=cosw, √(1−x²)=sinw), reducing it to the standard by-parts integral ∫w cosw dw. Applying the boundary condition lim_x→1⁻y(x)=1 fixes the constant of integration at C=2, giving the closed form y = [−cos⁻¹(x)√(1−x²) − x + 2]/x. Evaluating at x=1/2 gives y(1/2) = 3 − π/√3.

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Why This Answer

The differential form looks intimidating with the √(1−x²) and cos⁻¹x mixed together, but dividing through by x√(1−x²)dx immediately reveals a completely standard linear ODE with integrating factor x — the √(1−x²) factor was only ever there to make the equation exact in a disguised way, and it cancels cleanly. The genuinely new step is the right-hand-side integral, x·cos⁻¹(x)/√(1−x²), which is not a standard table integral but becomes one after substituting w=cos⁻¹x: it collapses to −∫w cosw dw, solvable by a single integration by parts.

The boundary condition is given as a LIMIT rather than a value at an interior point because y(x) itself is undefined in a simple sense at x=1 (the ODE’s coefficient x√(1−x²) vanishes there) — but the limit exists and is finite, and evaluating it requires recognising that cos⁻¹(x)√(1−x²) → 0 as x→1⁻ (both factors vanish, and their product vanishes even faster, like 2(1−x)).

Core Concept

Reducing to linear form. Many first-order ODEs that look unfamiliar in “differential form” M dx + N dy = 0 become the standard linear dy/dx + P(x)y = Q(x) after dividing by the right combination — always check this before searching for a more exotic technique.

By-parts under a trig substitution. Integrals combining an inverse trig function with an algebraic/trig expression in x are frequently solved by substituting w = (the inverse trig function itself), converting the whole integral into a polynomial-times-trig integral solvable by parts.

Step-by-Step Solution

1. Convert to linear form. Dividing by x√(1−x²)dx: dy/dx + y/x = cos⁻¹(x)/√(1−x²). Integrating factor = e^{∫dx/x} = x.

2. Write the exact-derivative form: d/dx(xy) = x·cos⁻¹(x)/√(1−x²).

3. Integrate the right side via w=cos⁻¹x (x=cosw, √(1−x²)=sinw, dx=−sinw dw):

   ∫x cos⁻¹x/√(1-x²) dx = ∫cosw·w/sinw·(-sinw)dw = -∫w cosw dw
   = -(w sinw + cosw) + C = -[cos⁻¹(x)√(1-x²) + x] + C

4. Write the general solution: xy = −cos⁻¹(x)√(1−x²) − x + C.

5. Apply the boundary condition. As x→1⁻, cos⁻¹(x)→0 and √(1−x²)→0, so their product →0. Then y→(0−1+C)/1=C−1=1 ⇒ C=2.

6. Evaluate at x=1/2: cos⁻¹(1/2)=π/3, √(1−1/4)=√3/2.

   y(1/2) = [-(π/3)(√3/2) – 1/2 + 2] / (1/2)
   = [3/2 – π√3/6] × 2 = 3 – π√3/3 = 3 – π/√3

Final Answer

y(1/2) = 3 − π/√3 exactly. The exam lesson: a messy-looking differential form is worth dividing out immediately to check for hidden linear structure before reaching for a more advanced technique.

Key Facts to Remember

• Always try dividing a differential form by an obvious common factor to reveal a standard dy/dx + Py = Q structure.
• Integrating factor for dy/dx + y/x = Q(x) is x.
• Substituting w = cos⁻¹(x) (giving x=cosw, √(1−x²)=sinw) converts inverse-cosine integrals into polynomial-times-trig integrals.
• ∫w cosw dw = w sinw + cosw + C (by parts).

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Prerequisites

Before practising this question type, make sure the following are in place:

• Linear first-order ODEs and the integrating factor method.
• Substitution w = inverse trig function to simplify mixed algebraic/inverse-trig integrands.
• Integration by parts, especially ∫w cosw dw and ∫w sinw dw.
• Limits involving cos⁻¹(x) and √(1−x²) as x→1⁻.

Revise these from the Differential Equations chapter, then attempt the related questions below.

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