Let y=y(x) be the solution of the differential equation x√(1-x^2) dy/dx + (y√(1-x^2)-xcos^-1x) = 0, x ∈ (0,1), lim x→1- y(x) = 1. Then y(1/2) equals:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2026

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Home Maths Differential EquationsQuestion #180

Let

y = y (x)

be the solution of the differential equation

x \sqrt{1 - x^{2}} \frac{d y}{d x} + (y \sqrt{1 - x^{2}} - x \cos^{- 1} x) = 0

x \in (0, 1)

lim_{x \to 1^{-}} y (x) = 1

. Then

y (\frac{1}{2})

equals:

Clue 1 of 3: Recognize the type of differential equation

The given equation can be rearranged into a standard first-order linear differential equation.

🧠 Full Solution Path

Step 1: Rearrange the differential equation into standard linear form.✦ Active

The given differential equation is $x \sqrt{1 - x^{2}} \frac{d y}{d x} + (y \sqrt{1 - x^{2}} - x \cos^{- 1} x) = 0$ . Divide the entire equation by $x \sqrt{1 - x^{2}}$ (since $x \in (0, 1)$ , $x \neq 0$ and $\sqrt{1 - x^{2}} \neq 0$ ) to obtain the standard linear form:

\frac{d y}{d x} + \frac{1}{x} y = \frac{\cos^{- 1} x}{\sqrt{1 - x^{2}}}

This is a first-order linear differential equation of the form $\frac{d y}{d x} + P (x) y = Q (x)$ , where $P (x) = \frac{1}{x}$ and $Q (x) = \frac{\cos^{- 1} x}{\sqrt{1 - x^{2}}}$ .

Step 2: Find the integrating factor and solve the differential equation.○ Expand

The integrating factor (IF) is calculated as $e^{\int P (x) d x}$ . For $P (x) = \frac{1}{x}$ :

I F = e^{\int \frac{1}{x} d x} = e^{\ln | x |} = | x |

Since $x \in (0, 1)$ , $x$ is positive, so $I F = x$ . The general solution is given by $y \cdot I F = \int Q (x) \cdot I F d x + C$ . Substituting the values:

x y = \int \frac{\cos^{- 1} x}{\sqrt{1 - x^{2}}} \cdot x d x + C

To evaluate the integral $\int \frac{x \cos^{- 1} x}{\sqrt{1 - x^{2}}} d x$ , we use substitution and integration by parts. Let $u = \cos^{- 1} x$ , so $d u = - \frac{1}{\sqrt{1 - x^{2}}} d x$ and $x = \cos u$ . The integral becomes $- \int u \cos u d u$ . Applying integration by parts ( $\int v d w = v w - \int w d v$ with $v = u, d w = \cos u d u$ ):

- \int u \cos u d u = - [u \sin u - \int \sin u d u] = - [u \sin u + \cos u]

Substitute back $u = \cos^{- 1} x$ and $\sin (\cos^{- 1} x) = \sqrt{1 - x^{2}}$ :

x y = - (\cos^{- 1} x) \sqrt{1 - x^{2}} - x + C

Dividing by $x$ , the general solution is:

y (x) = - \frac{\cos^{- 1} x \sqrt{1 - x^{2}}}{x} - 1 + \frac{C}{x}

💡 Teacher's Secret Hint

Be careful with the signs during integration by parts and when substituting back.

Step 3: Apply the initial condition and calculate

y (\frac{1}{2})

.○ Expand

We are given the initial condition $lim_{x \to 1^{-}} y (x) = 1$ . Let's evaluate the limit of our general solution:

lim_{x \to 1^{-}} (- \frac{\cos^{- 1} x \sqrt{1 - x^{2}}}{x} - 1 + \frac{C}{x})

As $x \to 1^{-}$ , $\cos^{- 1} x \to 0$ and $\sqrt{1 - x^{2}} \to 0$ . So, the first term approaches $0$ . The second term is $- 1$ , and the third term approaches $C$ . Thus:

0 - 1 + C = C - 1

Given that the limit is $1$ , we have $C - 1 = 1 ⟹ C = 2$ . The particular solution is:

y (x) = - \frac{\cos^{- 1} x \sqrt{1 - x^{2}}}{x} - 1 + \frac{2}{x}

Finally, we need to find $y (\frac{1}{2})$ . Substitute $x = \frac{1}{2}$ into the particular solution:

y (\frac{1}{2}) = - \frac{\cos^{- 1} (\frac{1}{2}) \sqrt{1 - {(\frac{1}{2})}^{2}}}{\frac{1}{2}} - 1 + \frac{2}{\frac{1}{2}}

We know $\cos^{- 1} (\frac{1}{2}) = \frac{π}{3}$ and $\sqrt{1 - {(\frac{1}{2})}^{2}} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$ . Substitute these values:

y (\frac{1}{2}) = - \frac{\frac{π}{3} \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} - 1 + 4

y (\frac{1}{2}) = - \frac{π \sqrt{3}}{6} \cdot 2 + 3

y (\frac{1}{2}) = - \frac{π \sqrt{3}}{3} + 3 = 3 - \frac{π}{\sqrt{3}}

This matches option 1.

💡 Teacher's Secret Hint

Ensure correct evaluation of inverse trigonometric functions and simplification of fractions.

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Recognize the type of differential equation

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