Eliminate the arbitrary constants and obtain the differential equation satisfied by it: $y = (\frac{a}{x^{2}}) + b x$

x^{2} y^{''} + 2 x y^{'} - 2 y = 0

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x^{2} y^{''} + 2 x^{2} y^{'} - 2 y = 0

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x^{2} y^{''} + 2 x y^{'} + 2 x y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} y^{''} - 2 x y^{'} + 2 x y = 0

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Solution

The correct option is A $x^{2} y^{''} + 2 x y^{'} - 2 y = 0$
Given, $y = \frac{a}{x^{2}} + b x$ ...(1)
$\Rightarrow a = x^{2} y - b x^{3}$
Differentiating (1) w.r.t x, we get
$y^{'} = - 2 a x^{- 3} + b$ ....(2)

Substituting value of $a$ from (2)
$y^{'} = - 2 (x^{2} y - b x^{3}) x^{- 3} + b = - 2 x^{- 1} y + 3 b$

Differentiating it again w.r.t x, we get
$y^{''} = 2 x^{- 2} - 2 x^{- 1} y \Rightarrow x^{2} y^{''} + 2 x y^{'} - 2 y = 0$

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Eliminate the arbitrary constants and obtain the differential equation satisfied by it: y=(ax2)+bx

Eliminate the arbitrary constants and obtain the differential equation satisfied by it: $y = (\frac{a}{x^{2}}) + b x$