Question:
Let \(x^3+3y^2=4\) for all \(x,y\isin\R,\) \(y'=\frac{dy}{dx}\) and \(y''=\frac{d^2y}{dx^2}\). Then
Let \(x^3+3y^2=4\) for all \(x,y\isin\R,\) \(y'=\frac{dy}{dx}\) and \(y''=\frac{d^2y}{dx^2}\). Then
Updated On: Oct 1, 2024
- x2 + yy" + (y')2 = 0
- 2x + y" + 2(y')2 = 0
- x + (y')2 = 0
- x + yy" + (y')2 = 0
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The Correct Option is D
Solution and Explanation
The correct option is (D): x + yy" + (y')2 = 0
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