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JEE Mathematics 2025 Question with Solution

If x=f(y)x = f(y) is the solution of the differential equation

(1+y2)+(x−2etan⁡−1y)dydx=0,y∈(−π2,π2),(1 + y^2) + (x - 2e^{\tan^{-1}y}) \frac{dy}{dx} = 0, \quad y \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right),

with f(0)=1f(0) = 1, then f(13)f\left( \frac{1}{\sqrt{3}} \right) is equal to:

  • A

    eπ3e^{\frac{\pi}{3}}

  • B

    eπ12e^{\frac{\pi}{12}}

  • C

    eπ6e^{\frac{\pi}{6}}

  • D

    eπ4e^{\frac{\pi}{4}}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

(1+y2)+(x−2etan⁡−1y)dydx=0,(1 + y^2) + (x - 2e^{\tan^{-1}y}) \frac{dy}{dx} = 0,

with x=f(y)x = f(y) and f(0)=1f(0) = 1.

Find: f(13)f\left( \frac{1}{\sqrt{3}} \right).

Rewrite the differential equation in terms of dxdy\frac{dx}{dy}:

dydx=−1+y2x−2etan⁡−1y\frac{dy}{dx} = -\frac{1+y^2}{x-2e^{\tan^{-1}y}}

So,

dxdy=−x−2etan⁡−1y1+y2\frac{dx}{dy} = -\frac{x-2e^{\tan^{-1}y}}{1+y^2}

which gives

dxdy+x1+y2=2etan⁡−1y1+y2.\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{2e^{\tan^{-1}y}}{1+y^2}.

Linear Differential Equation Approach

This is a linear first-order differential equation in xx as a function of yy, with

P(y)=11+y2,Q(y)=2etan⁡−1y1+y2.P(y) = \frac{1}{1+y^2}, \qquad Q(y) = \frac{2e^{\tan^{-1}y}}{1+y^2}.

The integrating factor is

I(y)=e∫11+y2 dy=etan⁡−1y.I(y) = e^{\int \frac{1}{1+y^2} \, dy} = e^{\tan^{-1}y}.

Multiplying the equation by the integrating factor,

etan⁡−1ydxdy+etan⁡−1yx1+y2=2e2tan⁡−1y1+y2.e^{\tan^{-1}y}\frac{dx}{dy} + \frac{e^{\tan^{-1}y}x}{1+y^2} = \frac{2e^{2\tan^{-1}y}}{1+y^2}.

The left-hand side becomes

ddy(xetan⁡−1y)=2e2tan⁡−1y1+y2.\frac{d}{dy}\left(xe^{\tan^{-1}y}\right) = \frac{2e^{2\tan^{-1}y}}{1+y^2}.

Integrate both sides:

xetan⁡−1y=∫2e2tan⁡−1y1+y2 dy=e2tan⁡−1y+C.xe^{\tan^{-1}y} = \int \frac{2e^{2\tan^{-1}y}}{1+y^2} \, dy = e^{2\tan^{-1}y} + C.

Hence,

x=etan⁡−1y+Ce−tan⁡−1y.x = e^{\tan^{-1}y} + Ce^{-\tan^{-1}y}.

Now use the initial condition f(0)=1f(0)=1:

1=e0+Ce01 = e^0 + Ce^0

So,

C=0.C = 0.

Therefore,

x=etan⁡−1y.x = e^{\tan^{-1}y}.

Now evaluate at y=13y = \frac{1}{\sqrt{3}}:

f(13)=etan⁡−1(13).f\left( \frac{1}{\sqrt{3}} \right) = e^{\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)}.

Since

tan⁡−1(13)=π6,\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6},

we get

f(13)=eπ6.f\left( \frac{1}{\sqrt{3}} \right) = e^{\frac{\pi}{6}}.

Therefore, the correct option is C.

The second provided approach contains inconsistent substitutions and even contradicts the initial condition, so the valid conclusion is taken from the correct linear differential equation method above.

Common mistakes

  • Treating the equation as separable directly in xx and yy is incorrect here. It is linear in xx as a function of yy. First rewrite in terms of dxdy\frac{dx}{dy} and then use the integrating factor method.

  • Using the initial condition incorrectly by substituting into the original differential equation instead of the solved general solution leads to errors. Apply f(0)=1f(0)=1 only after obtaining x=etan⁡−1y+Ce−tan⁡−1yx = e^{\tan^{-1}y} + Ce^{-\tan^{-1}y}.

  • Confusing tan⁡−1(13)\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) with other standard angles gives the wrong final option. Use the standard value tan⁡π6=13\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}, so the angle is π6\frac{\pi}{6}.

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