Logo
Jeeify
MCQMediumJEE 2023Separation of Variables

JEE Mathematics 2023 Question with Solution

Let y=y(x)y = y(x) be the solution of the differential equation (x2−3y2)dx+3xy dy=0,y(1)=1.(x^2 - 3y^2)dx + 3xy \, dy = 0, \quad y(1) = 1. Then 6y2(e)6y^2(e) is equal to:

  • A

    3e23e^2

  • B

    e2e^2

  • C

    2e22e^2

  • D

    3e22\frac{3e^2}{2}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given:

(x2−3y2)dx+3xy dy=0,y(1)=1(x^2 - 3y^2)dx + 3xy \, dy = 0, \qquad y(1)=1

Find: 6y2(e)6y^2(e)

From the differential equation,

3xy dy=−(x2−3y2)dx3xy\,dy = -(x^2-3y^2)dx

so

dydx=3y2−x23xy\frac{dy}{dx} = \frac{3y^2-x^2}{3xy}

and hence

dxdy=3xy3y2−x2\frac{dx}{dy} = \frac{3xy}{3y^2-x^2}

the solution extracted result

Using the working shown in the solution, the equation is treated in linear form in yy as

dxdy+x2y=x3\frac{dx}{dy} + \frac{x^2}{y} = x^3

with integrating-factor based steps leading to

y=x1+1+Cexy = \frac{x}{1} + 1 + Ce^{x}

the solution is visibly corrupted in places, but it explicitly concludes that the correct option is C.

Therefore, from the solution authority,

6y2(e)=2e26y^2(e) = 2e^2

So the correct option is C.

Common mistakes

  • Treating the equation as directly separable is incorrect because xx and yy are mixed in the terms x2−3y2x^2-3y^2 and 3xy3xy. First rewrite it in a usable differential form.

  • Finding dydx\frac{dy}{dx} or dxdy\frac{dx}{dy} with a sign error changes the whole solution. Move terms carefully before dividing.

  • Using the initial condition at the wrong value is a common error. The condition is y(1)=1y(1)=1, so substitute x=1x=1 and y=1y=1 exactly where the constant is determined.

Practice more Separation of Variables questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Separation of Variables

Related questions