Question

The function \[ x+y=\tan^{-1}y \] is the solution of which of the following differential equations?

• A. \(y^2y'+y^2+1=0\)
• B. \(y^2y''-2y'=0\)
• C. \(y^2y'-y^2+1=0\)
• D. \(y^2-2y'+1=0\)

Hint:

While differentiating implicitly, remember that the derivative of \(y\) with respect to \(x\) is always written as \(y'\) or \(\dfrac{dy}{dx}\).

Answer 1

The Correct Option is A

Concept: To determine the differential equation satisfied by a given relation, differentiate the relation implicitly with respect to \(x\), then simplify carefully. Given: \[ x+y=\tan^{-1}y \]

Step 1: Differentiate both sides with respect to \(x\). Differentiating term-by-term: Derivative of \(x\): \[ 1 \] Derivative of \(y\): \[ y' \] Derivative of \(\tan^{-1}y\): \[ \frac{y'}{1+y^2} \] Thus, \[ 1+y'=\frac{y'}{1+y^2} \]

Step 2: Eliminate the denominator. Multiply throughout by \(1+y^2\): \[ (1+y')(1+y^2)=y' \] Expand: \[ 1+y^2+y'+y^2y'=y' \] Subtract \(y'\) from both sides: \[ 1+y^2+y^2y'=0 \] Rearranging, \[ y^2y'+y^2+1=0 \] Hence, \[ \boxed{y^2y'+y^2+1=0} \]