Question

• A. x² + yy" + (y')² = 0
• B. 2x + y" + 2(y')² = 0
• C. x + (y')² = 0
• D. x + yy" + (y')² = 0

Answer 1

The Correct Option is D

To solve the problem, we are given the equation \(x^3+3y^2=4\) and need to find a relationship involving higher-order derivatives. We'll begin by differentiating the given equation with respect to \(x\).

1. Differentiate both sides of the equation \(x^3+3y^2=4\) with respect to \(x\):

\[ \frac{d}{dx}(x^3) + \frac{d}{dx}(3y^2) = \frac{d}{dx}(4) \]

1. This yields:

\[ 3x^2 + 6yy' = 0 \]

1. Solving for \(y'\), we get:

\[ y' = -\frac{3x^2}{6y} = -\frac{x^2}{2y} \]

1. Next, differentiate the expression for \(y'\) to find \(y''\):

\[ \frac{dy'}{dx} = y'' = \frac{d}{dx}\left(-\frac{x^2}{2y}\right) \]

1. Using the quotient rule:

\[ y'' = \frac{(2y)(-2x) - (-x^2)(2y')}{(2y)^2} \]

1. Substitute \(y' = -\frac{x^2}{2y}\):

\[ y'' = \frac{-4xy + x^3(2y')}{4y^2} = \frac{-4xy + x^3\left(-\frac{x^2}{2y}\right)}{4y^2} \]

1. Simplifying gives:

\[ y'' = \frac{-4xy - \frac{x^5}{2y}}{4y^2} \]

1. Insert \(y''\) and \((y')^2\) back into the main problem expression:

Check if: \[ x + yy'' + (y')^2 = 0 \]

1. Substitute \(y''\) and \((y')^2 = \left(-\frac{x^2}{2y}\right)^2 = \frac{x^4}{4y^2}\):

Therefore, the correct option is \(x + yy'' + (y')^2 = 0\), matching the derived expression as \(x + yy'' + (y')^2 = 0\).