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MathMedMain 2026 • 8th Apr - Shift 2

Solve First Order Differential Equation with Given Initial Condition

Let y=y(x)y = y(x) be the solution of the differential equation x1−x2dy+(y1−x2−xcos⁡−1x)dx=0,x∈(0,1),lim⁡x→1−y(x)=1x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x)dx = 0, \quad x \in (0, 1), \quad \lim_{x \to 1^-} y(x) = 1. Then y(12)y\left(\frac{1}{2}\right) equals :

Options

A

3 - π3\frac{\pi}{\sqrt{3}}

Correct
B

4 - 3π\sqrt{3} \pi

C

4 - 2π3\frac{2\pi}{\sqrt{3}}

D

3 - π23\frac{\pi}{2\sqrt{3}}

Topics & Concepts

Differential EquationsDifferential EquationsLinear Differential Equations

Step-by-Step Solution

To solve the given differential equation for x∈(0,1)x \in (0, 1):

x1−x2dy+(y1−x2−xcos⁡−1x)dx=0x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x\cos^{-1}x)dx = 0

Rearranging terms to write it in the standard first-order linear form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x)y = Q(x):

x1−x2dydx+y1−x2=xcos⁡−1xx\sqrt{1-x^2} \frac{dy}{dx} + y\sqrt{1-x^2} = x\cos^{-1}x

Dividing both sides by x1−x2x\sqrt{1-x^2}:

dydx+1xy=cos⁡−1x1−x2\frac{dy}{dx} + \frac{1}{x} y = \frac{\cos^{-1}x}{\sqrt{1-x^2}}

Here, P(x)=1xP(x) = \frac{1}{x} and Q(x)=cos⁡−1x1−x2Q(x) = \frac{\cos^{-1}x}{\sqrt{1-x^2}}.

Step 1: Find the Integrating Factor (I.F.)

I.F.=e∫P(x) dx=e∫1x dx=eln⁡x=x\text{I.F.} = e^{\int P(x) \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x

Step 2: Write the General Solution

y⋅(I.F.)=∫Q(x)⋅(I.F.) dx+Cy \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) \, dx + C

y⋅x=∫xcos⁡−1x1−x2 dx+Cy \cdot x = \int \frac{x \cos^{-1}x}{\sqrt{1-x^2}} \, dx + C

Step 3: Evaluate the Integral ∫xcos⁡−1x1−x2 dx\int \frac{x \cos^{-1}x}{\sqrt{1-x^2}} \, dx

Let cos⁡−1x=u  ⟹  x=cos⁡u\cos^{-1}x = u \implies x = \cos u and dx=−sin⁡u dudx = -\sin u \, du. Since x∈(0,1)x \in (0, 1), we have u∈(0,π2)u \in \left(0, \frac{\pi}{2}\right), so 1−x2=sin⁡u\sqrt{1-x^2} = \sin u.

Substituting these into the integral gives:

∫cos⁡u⋅usin⁡u(−sin⁡u du)=−∫ucos⁡u du\int \frac{\cos u \cdot u}{\sin u} (-\sin u \, du) = -\int u \cos u \, du

Using integration by parts:

∫ucos⁡u du=usin⁡u−∫sin⁡u du=usin⁡u+cos⁡u\int u \cos u \, du = u \sin u - \int \sin u \, du = u \sin u + \cos u

Substituting back u=cos⁡−1xu = \cos^{-1}x, sin⁡u=1−x2\sin u = \sqrt{1-x^2}, and cos⁡u=x\cos u = x:

−∫ucos⁡u du=−(1−x2cos⁡−1x+x)-\int u \cos u \, du = -\left( \sqrt{1-x^2} \cos^{-1}x + x \right)

So, the general solution becomes:

y⋅x=−1−x2cos⁡−1x−x+Cy \cdot x = -\sqrt{1-x^2} \cos^{-1}x - x + C

y(x)=−1−1−x2cos⁡−1xx+Cxy(x) = -1 - \frac{\sqrt{1-x^2} \cos^{-1}x}{x} + \frac{C}{x}

Step 4: Determine the Constant CC

Using the given boundary condition lim⁡x→1−y(x)=1\lim_{x \to 1^-} y(x) = 1:

lim⁡x→1−(−1−1−x2cos⁡−1xx+Cx)=1\lim_{x \to 1^-} \left( -1 - \frac{\sqrt{1-x^2} \cos^{-1}x}{x} + \frac{C}{x} \right) = 1

Since lim⁡x→1−1−x2cos⁡−1x=0⋅0=0\lim_{x \to 1^-} \sqrt{1-x^2} \cos^{-1}x = 0 \cdot 0 = 0:

−1−0+C=1  ⟹  C=2-1 - 0 + C = 1 \implies C = 2

Thus, the exact solution function is:

y(x)=−1−1−x2cos⁡−1xx+2xy(x) = -1 - \frac{\sqrt{1-x^2} \cos^{-1}x}{x} + \frac{2}{x}

Step 5: Calculate y(12)y\left(\frac{1}{2}\right)

Substitute x=12x = \frac{1}{2} into y(x)y(x):

y(12)=−1−1−(12)2cos⁡−1(12)12+212y\left(\frac{1}{2}\right) = -1 - \frac{\sqrt{1 - \left(\frac{1}{2}\right)^2} \cos^{-1}\left(\frac{1}{2}\right)}{\frac{1}{2}} + \frac{2}{\frac{1}{2}}

y(12)=−1−32⋅π312+4y\left(\frac{1}{2}\right) = -1 - \frac{\frac{\sqrt{3}}{2} \cdot \frac{\pi}{3}}{\frac{1}{2}} + 4

y(12)=3−3π3=3−π3y\left(\frac{1}{2}\right) = 3 - \frac{\sqrt{3} \pi}{3} = 3 - \frac{\pi}{\sqrt{3}}

Hence, the correct option is A.

Solve First Order Differential Equation with Given Initial Condition | Mathematics PYQ Solution - JEE Challenger