If x^2 + y^2 = 2 and 2y" + ay^-3 = 0, then value of is equal to

JEE MainMathematicsModerateDifferentiation

If x^2 + y^2 = 2 and 2y" + ay^-3 = 0, then value of
$\frac{5a}{2}$ is equal to

Show answer and solution

x^2 + y^2 = 2
Differentiating both sides, we get
2x + 2yy' = 0

\Rightarrow

y^{'} = \frac{- x}{y}

.....(i)
Differentiating (i) w.r.t x

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

2y^{''} + ay^{- 3} = 0

\Rightarrow

\frac{- 4}{y^{3}} + \frac{a}{y^{3}} = 0

\Rightarrow

a = 4

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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