Differential EquationsmediumFree

Transforming a Differential Equation by x=tan⁻¹t | JEE

JEE Maths question with a full step-by-step solution.

Question

If the substitution

x = \tan^{-1}(t)

transforms the differential equation

\dfrac{d^2y}{dx^2} + xy\dfrac{dy}{dx} + \sec^2 x = 0

into

(1 + t^2)\dfrac{d^2y}{dt^2} + \big(2t + y\tan^{-1}(t)\big)\dfrac{dy}{dt} = k

, then the value of

k

is:

1

-1

correct

2

-2

Solution

Step 1: From

x = \tan^{-1}t

\dfrac{dx}{dt} = \dfrac{1}{1 + t^2}

, so

\frac{dy}{dx} = (1 + t^2)\frac{dy}{dt}.

Step 2: Differentiate again:

\frac{d^2y}{dx^2} = \frac{d}{dt}\left[(1 + t^2)\frac{dy}{dt}\right]\cdot\frac{dt}{dx} = (1 + t^2)\left[2t\frac{dy}{dt} + (1 + t^2)\frac{d^2y}{dt^2}\right].

Step 3: Also

\sec^2 x = 1 + \tan^2 x = 1 + t^2

and

x = \tan^{-1}t

. Substitute into the equation:

(1 + t^2)\left[2t\frac{dy}{dt} + (1 + t^2)\frac{d^2y}{dt^2}\right] + (\tan^{-1}t)\,y\,(1 + t^2)\frac{dy}{dt} + (1 + t^2) = 0.

Step 4: Cancel

(1 + t^2)

throughout:

(1 + t^2)\frac{d^2y}{dt^2} + \big(2t + y\tan^{-1}t\big)\frac{dy}{dt} = -1 \Rightarrow k = -1.

Correct answer: (2)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.