Logo
Jeeify
MCQMediumJEE 2024Separation of Variables

JEE Mathematics 2024 Question with Solution

The solution curve of the differential equation 2ydydx+3=5dydx2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}, passing through the point (0,1)(0, 1), is a conic whose vertex lies on the line:

  • A

    2x+3y=92x + 3y = 9

  • B

    2x+3y=−92x + 3y = -9

  • C

    2x+3y=−62x + 3y = -6

  • D

    2x+3y=62x + 3y = 6

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: 2ydydx+3=5dydx2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx} and the curve passes through (0,1)(0,1).

Find: The line on which the vertex of the conic lies.

Rearranging,

2ydydx−5dydx=−32y \frac{dy}{dx} - 5 \frac{dy}{dx} = -3

So,

(2y−5)dydx=−3(2y-5)\frac{dy}{dx} = -3

Hence,

dydx=−32y−5\frac{dy}{dx} = \frac{-3}{2y-5}

Separating variables,

(2y−5) dy=−3 dx(2y-5) \, dy = -3 \, dx

Integrating both sides,

∫(2y−5) dy=−∫3 dx\int (2y-5) \, dy = -\int 3 \, dx

we get

y2−5y=−3x+Cy^2 - 5y = -3x + C

that is,

y2−5y+3x=Cy^2 - 5y + 3x = C

Using the point (0,1)(0,1),

12−5(1)+3(0)=C1^2 - 5(1) + 3(0) = C

so

C=−4C = -4

Therefore the curve is

y2−5y+3x=−4y^2 - 5y + 3x = -4

Completing the square,

y2−5y=−(3x+4)y^2 - 5y = -(3x+4) y2−5y+(52)2=(52)2−(3x+4)y^2 - 5y + \left(\frac{5}{2}\right)^2 = \left(\frac{5}{2}\right)^2 - (3x+4) (y−52)2=94−3x\left(y-\frac{5}{2}\right)^2 = \frac{9}{4} - 3x (y−52)2=−3(x−34)\left(y-\frac{5}{2}\right)^2 = -3\left(x-\frac{3}{4}\right)

So the vertex of the parabola is

(34,52)\left(\frac{3}{4}, \frac{5}{2}\right)

Now check the options at this point:

2(34)+3(52)=32+152=92\left(\frac{3}{4}\right) + 3\left(\frac{5}{2}\right) = \frac{3}{2} + \frac{15}{2} = 9

Therefore, the vertex lies on the line 2x+3y=92x + 3y = 9. The correct option is A.

Direct Vertex Form

Given: 2ydydx+3=5dydx2y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx} with point (0,1)(0,1).

Find: The line containing the vertex.

From

(2y−5)dydx=−3(2y-5)\frac{dy}{dx} = -3

write

(2y−5)dy=−3dx(2y-5)dy = -3dx

Integrating,

y2−5y=−3x+λy^2 - 5y = -3x + \lambda

Using (0,1)(0,1),

λ=−4\lambda = -4

Thus,

y2−5y+3x=−4y^2 - 5y + 3x = -4

Complete the square immediately:

(y−52)2=−3(x−34)\left(y-\frac{5}{2}\right)^2 = -3\left(x-\frac{3}{4}\right)

Hence the vertex is

(34,52)\left(\frac{3}{4}, \frac{5}{2}\right)

and substituting into the required line gives

2x+3y=2⋅34+3⋅52=92x+3y = 2\cdot \frac{3}{4} + 3\cdot \frac{5}{2} = 9

Therefore, the correct option is A.

Common mistakes

  • Students may treat the equation as linear in yy and try to use an integrating factor. That is incorrect because the equation is directly separable after factoring dydx\frac{dy}{dx}. First rewrite it as (2y−5)dydx=−3(2y-5)\frac{dy}{dx}=-3 and then separate variables.

  • A common mistake is using the point (0,1)(0,1) incorrectly while finding the constant. Substituting carefully into y2−5y+3x=Cy^2-5y+3x=C gives 1−5+0=−41-5+0=-4, not any other value. An incorrect constant shifts the entire parabola and changes the vertex.

  • While completing the square, students often write (y−52)2\left(\frac{y-5}{2}\right)^2 instead of (y−52)2\left(y-\frac{5}{2}\right)^2. This is wrong because the square completion for y2−5yy^2-5y requires adding (52)2\left(\frac{5}{2}\right)^2. Use the standard form carefully.

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