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JEE Mathematics 2025 Question with Solution

Let x=x(y)x = x(y) be the solution of the differential equation y2 dx+(x−1y)dy=0y^2 \, dx + \left( x - \frac{1}{y} \right) dy = 0. If x(1)=1x(1) = 1, then x(12)x\left( \frac{1}{2} \right) is :

  • A

    12+e\frac{1}{2} + e

  • B

    32+e\frac{3}{2} + e

  • C

    3−e3 - e

  • D

    3+e3 + e

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: y2 dx+(x−1y)dy=0y^2 \, dx + \left( x - \frac{1}{y} \right) dy = 0 and x(1)=1x(1)=1.

Find: x(12)x\left(\frac{1}{2}\right).

Rewrite the equation as a differential equation in xx with respect to yy:

dxdy=−x−1yy2=−xy2+1y3\frac{dx}{dy}=-\frac{x-\tfrac{1}{y}}{y^2}=-\frac{x}{y^2}+\frac{1}{y^3}

So,

dxdy+1y2x=1y3\frac{dx}{dy}+\frac{1}{y^2}x=\frac{1}{y^3}

This is a linear differential equation.

The integrating factor is

μ(y)=exp⁡ ⁣(∫1y2 dy)=exp⁡ ⁣(−1y)=e−1/y\mu(y)=\exp\!\left(\int \frac{1}{y^2} \, dy\right)=\exp\!\left(-\frac{1}{y}\right)=e^{-1/y}

Multiplying throughout by the integrating factor,

ddy(xe−1/y)=e−1/yy3\frac{d}{dy}\left(xe^{-1/y}\right)=\frac{e^{-1/y}}{y^3}

Integrating both sides,

xe−1/y=∫e−1/yy3 dy+Cxe^{-1/y}=\int \frac{e^{-1/y}}{y^3} \, dy + C

Detailed Integration and Substitution

Let t=1yt=\frac{1}{y}. Then dt=−1y2 dydt=-\frac{1}{y^2} \, dy, so

∫e−1/yy3 dy\int \frac{e^{-1/y}}{y^3} \, dy

becomes

−∫te−t dt-\int t e^{-t} \, dt

Hence,

−∫te−t dt=(t+1)e−t+C-\int t e^{-t} \, dt=(t+1)e^{-t}+C

Substituting back t=1yt=\frac{1}{y},

xe−1/y=(1y+1)e−1/y+Cxe^{-1/y}=\left(\frac{1}{y}+1\right)e^{-1/y}+C

Therefore,

x=1+1y+Ce1/yx=1+\frac{1}{y}+Ce^{1/y}

Now apply the initial condition x(1)=1x(1)=1:

1=1+1+Ce1=1+1+Ce Ce=−1Ce=-1 C=−1eC=-\frac{1}{e}

So,

x(y)=1+1y−e1/y−1x(y)=1+\frac{1}{y}-e^{1/y-1}

Now evaluate at y=12y=\frac{1}{2}:

x(12)=1+2−e2−1=3−ex\left(\frac{1}{2}\right)=1+2-e^{2-1}=3-e

Therefore, the correct option is C.

The first extracted approach in the solution incorrectly concludes x(12)=e−1/2x\left(\frac{1}{2}\right)=e^{-1/2} and then states the answer as 3−e3-e. The consistent working from the linear differential equation and integrating factor gives x(12)=3−ex\left(\frac{1}{2}\right)=3-e, matching Option 3.

Common mistakes

  • Treating the equation as directly separable in xx and yy is incorrect because dxdy+1y2x=1y3\frac{dx}{dy}+\frac{1}{y^2}x=\frac{1}{y^3} is a linear first-order differential equation. Use the integrating factor method instead.

  • Using the wrong integrating factor is a common error. Since P(y)=1y2P(y)=\frac{1}{y^2}, the integrating factor is e∫y−2dy=e−1/ye^{\int y^{-2}dy}=e^{-1/y}, not e1/ye^{1/y}.

  • After finding the general solution, students may forget to apply the initial condition x(1)=1x(1)=1 correctly. Substitute y=1y=1 carefully to obtain C=−1eC=-\frac{1}{e}.

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