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JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Differential Equations

If

y^{1/4} + y^{- 1/4} = 2 x

and

(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + α x \frac{d y}{d x} + β y = 0

, then

∣ α - β ∣

is equal to

Answer: 17

Solution

From

y^{1/4} + y^{- 1/4} = 2 x

, put

w = y^{1/4} = x + x^{2} - 1

, for which

x^{2} - 1 w^{'} = w

. Since

y = w^{4}

,

y^{'} = 4 w^{3} w^{'} = \frac{4 w ^{4}}{x ^{2} - 1} = \frac{4 y}{x ^{2} - 1}

, i.e.

x^{2} - 1 y^{'} = 4 y

. Squaring gives

(x^{2} - 1) (y^{'})^{2} = 16 y^{2}

, and differentiating yields

(x^{2} - 1) y^{''} + x y^{'} - 16 y = 0

. Thus

α = 1

,

β = - 16

, and

∣ α - β ∣ = 17.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Integer answer

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