Let x=x(y) be the solution of the differential equation 2y^2 dx/dy - 2xy + x^2 = 0, y > 1, x(e) = e. Then x(e^2) is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q20 – JEE-MAIN 2026

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Home Maths Differential EquationsQuestion #173

Let

x = x (y)

be the solution of the differential equation

2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0

y > 1

x (e) = e

. Then

x (e^{2})

is equal to :

Clue 1 of 3: Recognize the type of differential equation

The given equation $2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0$ can be transformed into a standard form.

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● Video Walkthrough

Step 1: Transforming the Differential Equation✦ Active

The given differential equation is $2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0$ . This is a Bernoulli-type equation if we consider $x$ as the dependent variable. Divide the equation by $x^{2}$ to get:

2 y^{2} \frac{1}{x^{2}} \frac{d x}{d y} - 2 y \frac{1}{x} + 1 = 0

Let $v = \frac{1}{x}$ . Then $\frac{d v}{d y} = - \frac{1}{x^{2}} \frac{d x}{d y}$ . Substituting these into the equation yields:

- 2 y^{2} \frac{d v}{d y} - 2 y v + 1 = 0

Rearranging it into a standard linear first-order differential equation form $\frac{d v}{d y} + P (y) v = Q (y)$ :

\frac{d v}{d y} + \frac{1}{y} v = \frac{1}{2 y^{2}}

Step 2: Solving the Linear ODE and Applying Initial Condition○ Expand

The integrating factor (IF) for this linear ODE is $e^{\int \frac{1}{y} d y} = e^{\ln y} = y$ (since $y > 1$ ). Multiplying the equation by the IF:

y \frac{d v}{d y} + v = \frac{1}{2 y}

The left side is the derivative of $(v y)$ with respect to $y$ . So, $\frac{d}{d y} (v y) = \frac{1}{2 y}$ . Integrating both sides:

v y = \int \frac{1}{2 y} d y = \frac{1}{2} \ln y + C

Substitute back $v = \frac{1}{x}$ :

\frac{y}{x} = \frac{1}{2} \ln y + C

Using the initial condition $x (e) = e$ (i.e., when $y = e, x = e$ ):

\frac{e}{e} = \frac{1}{2} \ln e + C ⟹ 1 = \frac{1}{2} (1) + C ⟹ C = \frac{1}{2}

Thus, the particular solution is:

\frac{y}{x} = \frac{1}{2} \ln y + \frac{1}{2} = \frac{1}{2} (\ln y + 1)

Step 3: Calculating

x (e^{2})

○ Expand

To find $x (e^{2})$ , substitute $y = e^{2}$ into the particular solution:

\frac{e^{2}}{x (e^{2})} = \frac{1}{2} (\ln (e^{2}) + 1)

Since $\ln (e^{2}) = 2$ :

\frac{e^{2}}{x (e^{2})} = \frac{1}{2} (2 + 1) = \frac{3}{2}

Solving for $x (e^{2})$ :

x (e^{2}) = \frac{2 e^{2}}{3}

This matches option 2.

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Maths Question 20 – JEE-MAIN 2026

Clue 1 of 3: Recognize the type of differential equation

🧠 Full Solution Path