Logo
Jeeify
MCQMediumJEE 2024Linear Differential Equations

JEE Mathematics 2024 Question with Solution

Let y=y(x)y = y(x) be the solution of the differential equation sec⁡(x) dy+{2(1−x)tan⁡(x)+x(2−x)} dx=0\sec(x) \, dy + \{2(1 - x) \tan(x) + x(2 - x)\} \, dx = 0 such that y(0)=2y(0) = 2. Then y(2)y(2) is equal to:

  • A

    22

  • B

    2{1−sin⁡(2)}2\{1 - \sin(2)\}

  • C

    2{sin⁡(2)+1}2\{\sin(2) + 1\}

  • D

    11

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: sec⁡x dy+{2(1−x)tan⁡x+x(2−x)} dx=0\sec x \, dy + \{ 2(1 - x) \tan x + x(2 - x) \} \, dx = 0 and y(0)=2y(0) = 2.

Find: y(2)y(2).

From the differential equation,

dydx=−{2(1−x)sin⁡x+x(2−x)cos⁡x}\frac{dy}{dx} = -\{ 2(1 - x) \sin x + x(2 - x) \cos x \}

Integrate both sides:

y(x)=−∫{2(1−x)sin⁡x+x(2−x)cos⁡x} dx+Cy(x) = - \int \{ 2(1 - x) \sin x + x(2 - x) \cos x \} \, dx + C

Separating the integrals,

y(x)=−∫2(1−x)sin⁡x dx−∫x(2−x)cos⁡x dx+Cy(x) = - \int 2(1 - x) \sin x \, dx - \int x(2 - x) \cos x \, dx + C

Using the working from the solution, this simplifies to

y(x)=(x2−2x)sin⁡x+Cy(x) = (x^2 - 2x) \sin x + C

Apply the initial condition y(0)=2y(0) = 2:

y(0)=0+C=2y(0) = 0 + C = 2

So,

C=2C = 2

Hence,

y(x)=(x2−2x)sin⁡x+2y(x) = (x^2 - 2x) \sin x + 2

Now substitute x=2x = 2:

y(2)=(22−2×2)sin⁡2+2=2y(2) = (2^2 - 2 \times 2) \sin 2 + 2 = 2

Therefore, the correct option is A.

Alternative Approach from the solution

Given: sec⁡(x)dydx+[2(1−x)tan⁡(x)+x(2−x)]=0\sec(x) \frac{dy}{dx} + [2(1 - x) \tan(x) + x(2 - x)] = 0 and y(0)=2y(0) = 2.

Find: y(2)y(2).

The solution presents a second approach, but its intermediate algebra is inconsistent with the original equation. However, it still concludes that y(2)=2y(2) = 2.

Since the first approach correctly derives

y(x)=(x2−2x)sin⁡x+2y(x) = (x^2 - 2x) \sin x + 2

we use that validated expression to evaluate at x=2x = 2.

Therefore,

y(2)=2y(2) = 2

So the answer is 22, which corresponds to A.

Common mistakes

  • Dividing by sec⁡x\sec x incorrectly. Since tan⁡x/sec⁡x=sin⁡x\tan x / \sec x = \sin x and 1/sec⁡x=cos⁡x1 / \sec x = \cos x, the equation must be converted carefully before integrating.

  • Treating the equation as separable in the form used in the second approach. The correct step is to rewrite it as an explicit expression for dydx\frac{dy}{dx} and then integrate with respect to xx.

  • Forgetting to apply the initial condition y(0)=2y(0) = 2 after integration. Without determining the constant CC, the final value of y(2)y(2) cannot be found.

Practice more Linear Differential Equations questions

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

Start free trialMore on Linear Differential Equations

Related questions