What is the solution method for this differential equation?

Divide by x√(1−x²) to get linear ODE dy/dx+(1/x)y=cos⁻¹x/√(1−x²). IF=x. Integrate RHS using u=cos⁻¹x to get xy=−cos⁻¹x·√(1−x²)−x+C.

3−π/√3. Applying BC: C=2. At x=1/2: y/2=3/2−π√3/6, so y=3−π/√3.

ODE x√(1−x²)dy+(y√(1−x²)−xcos⁻¹x)dx=0, lim y=1 as x→1⁻, find y(1/2)

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MathDiff Equations

QMCQDifferential Equations

Let $y=y(x)$ be the solution of the differential equation $x\sqrt{1-x^2}\,dy+\bigl(y\sqrt{1-x^2}-x\cos^{-1}x\bigr)dx=0$, $x\in(0,1)$, $\displaystyle\lim_{x\to1^-}y(x)=1$. Then $y\!\left(\dfrac{1}{2}\right)$ equals:

A) $3-\dfrac{\pi}{\sqrt{3}}$ B) $4-\sqrt{3}\,\pi$ C) $4-\dfrac{2\pi}{\sqrt{3}}$ D) $3-\dfrac{\pi}{2\sqrt{3}}$

✅ Correct Answer

A) 3−π/√3

✓

Solution

Convert to standard linear ODE form

Divide through by $x\sqrt{1-x^2}$:

$\dfrac{dy}{dx}+\dfrac{y}{x}=\dfrac{\cos^{-1}x}{\sqrt{1-x^2}}$

This is linear: $P(x)=\dfrac{1}{x}$, $Q(x)=\dfrac{\cos^{-1}x}{\sqrt{1-x^2}}$.

Find Integrating Factor and general solution

$\text{IF}=e^{\int(1/x)dx}=e^{\ln x}=x$

$\dfrac{d(xy)}{dx}=x\cdot\dfrac{\cos^{-1}x}{\sqrt{1-x^2}}$

$xy=\displaystyle\int\frac{x\cos^{-1}x}{\sqrt{1-x^2}}\,dx+C$

Evaluate the integral using substitution u = cos⁻¹x

Let $u=\cos^{-1}x$, so $x=\cos u$, $dx=-\sin u\,du$, $\sqrt{1-x^2}=\sin u$:

$\displaystyle\int\frac{\cos u\cdot u}{\sin u}\cdot(-\sin u)\,du=-\int u\cos u\,du$

$=-\bigl[u\sin u+\cos u\bigr]+C=-(u\sin u+\cos u)+C$

Back-substitute: $=\ -\cos^{-1}x\cdot\sqrt{1-x^2}-x+C$

$\therefore\quad xy=-\cos^{-1}x\cdot\sqrt{1-x^2}-x+C$

Apply BC and find y(1/2)

As $x\to1^-$: $y\to1$, so $1\cdot1=-\cos^{-1}(1)\cdot0-1+C=0-1+C$. Hence $C=2$.

$xy=-\cos^{-1}x\cdot\sqrt{1-x^2}-x+2$

At $x=\dfrac{1}{2}$: $\cos^{-1}\!\dfrac{1}{2}=\dfrac{\pi}{3}$, $\sqrt{1-\dfrac{1}{4}}=\dfrac{\sqrt{3}}{2}$

$\dfrac{1}{2}\cdot y=-\dfrac{\pi}{3}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{1}{2}+2=-\dfrac{\pi\sqrt{3}}{6}+\dfrac{3}{2}$

$y=2\!\left(\dfrac{3}{2}-\dfrac{\pi\sqrt{3}}{6}\right)=3-\dfrac{\pi\sqrt{3}}{3}=3-\dfrac{\pi}{\sqrt{3}}$

$$y\!\left(\tfrac{1}{2}\right)=\boxed{3-\dfrac{\pi}{\sqrt{3}}}$$

📘 Linear First-Order ODE — Standard Method

Standard form: $dy/dx+P(x)y=Q(x)$. Integrating factor: $\mu=e^{\int P\,dx}$. Solution: $y\cdot\mu=\int Q\cdot\mu\,dx+C$. Here IF $=x$, and the RHS integral $\int x\cos^{-1}x/\sqrt{1-x^2}\,dx$ is solved by substitution $u=\cos^{-1}x$, reducing to $-\int u\cos u\,du$.

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Important Concepts

Standard Linear ODEAfter dividing by $x\sqrt{1-x^2}$: $dy/dx+y/x=\cos^{-1}x/\sqrt{1-x^2}$. The IF is $e^{\int 1/x\,dx}=x$, giving $d(xy)/dx=x\cos^{-1}x/\sqrt{1-x^2}$.

Integration by Substitution u=cos⁻¹xWith $u=\cos^{-1}x$: $x=\cos u$, $dx=-\sin u\,du$, $\sqrt{1-x^2}=\sin u$. Integral $=\int\cos u\cdot u\cdot(-du)=-\int u\cos u\,du=-(u\sin u+\cos u)+C$.

Back Substitution$u=\cos^{-1}x$, $\sin u=\sqrt{1-x^2}$, $\cos u=x$. So $-(u\sin u+\cos u)=-\cos^{-1}x\cdot\sqrt{1-x^2}-x+C$.

Boundary Condition ApplicationAs $x\to1^-$: $y\to1$. The RHS $\to -0\cdot0-1+C=C-1$. LHS $=1\cdot1=1$. So $C=2$. Then at $x=1/2$: $y/2=-\pi\sqrt{3}/6+3/2$, giving $y=3-\pi/\sqrt{3}$.

FAQs

How to identify P(x) and Q(x) here?

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Divide the ODE by $x\sqrt{1-x^2}$ to get $dy/dx+(1/x)y=\cos^{-1}x/\sqrt{1-x^2}$. So $P=1/x$ and $Q=\cos^{-1}x/\sqrt{1-x^2}$.

Why does the substitution u=cos⁻¹x work?

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It converts $\cos^{-1}x$ (the awkward part) to the simple variable $u$, and $1/\sqrt{1-x^2}$ becomes $1/\sin u$ which cancels with the $\sin u$ from $dx=-\sin u\,du$, leaving just $-\int u\cos u\,du$.

How to integrate ∫u·cos(u)du?

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Integration by parts: $\int u\cos u\,du=u\sin u-\int\sin u\,du=u\sin u+\cos u+C$.

What is cos⁻¹(1/2)?

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$\cos^{-1}(1/2)=\pi/3$. And $\sqrt{1-(1/2)^2}=\sqrt{3}/2$.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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