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JEE Main 2019 · 12 Jan · Shift I

JEE Main 2019 Mathematics question on Differential Equations

Let y=y(x)y = y(x) be the solution of the differential equation, xdydx+y=xlog⁡exx\frac{dy}{dx} + y = x\log_e x, (x>1)(x > 1). If 2y(2)=log⁡e4−12y(2) = \log_e 4 - 1, then y(e)y(e) is equal to
  1. A.
    −e2-\dfrac{e}{2}
  2. B.
    −e22-\dfrac{e^2}{2}
  3. C.
    e4\dfrac{e}{4}
  4. D.
    e24\dfrac{e^2}{4}
Show answer and solution

Correct option: C

Solution

The equation x(dy/dx) + y = x ln x rewrites as dy/dx + y/x = ln x, a linear ODE with integrating factor e^(∫1/x dx) = x.

Multiplying: d(xy)/dx = x ln x. Integrate: xy = ∫x ln x dx = (x²/2)ln x - x²/4 + C, so y = (x/2)ln x - x/4 + C/x.

Apply 2y(2) = ln 4 - 1: 2[(2/2)ln 2 - 2/4 + C/2] = ln 4 - 1 → 2[ln 2 - 1/2 + C/2] = 2ln 2 - 1 → 2ln 2 - 1 + C = 2ln 2 - 1, so C = 0.

Thus y(e) = e/2·(1) - e/4 = e/2 - e/4 = e/4.

[JEE Main 2019 · 12 Jan · Shift I]

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