If $$2x = y^{\frac{1}{5}} + y^{-\frac{1}{5}}$$ and $$(x^2 - 1)\frac{d^2y}{dx^2} + \lambda x \frac{dy}{dx} + ky = 0$$, then $$\lambda + k$$ is equal to

The relation between $$x$$ and $$y$$ is given as
$$2x = y^{\frac15}+y^{-\frac15}$$.

Introduce the substitution $$t = y^{\frac15}\; \Longrightarrow\; y = t^{5}$$.
With this, the relation becomes

$$t + \frac1t = 2x \; \Longrightarrow\; x = \frac{t+\frac1t}{2}$$ $$-(1)$$

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Step 1 : First derivatives

Differentiating $$2x = t + \dfrac1t$$ with respect to $$t$$:

$$2\frac{dx}{dt} = 1 - \frac1{t^{2}} \; \Longrightarrow\; \frac{dx}{dt} = \frac{1-\frac1{t^{2}}}{2} \;\Longrightarrow\; \frac{dt}{dx} = \frac{2}{1-\frac1{t^{2}}}= \frac{2t^{2}}{t^{2}-1}$$ $$-(2)$$

Since $$y = t^{5}$$, we have $$\dfrac{dy}{dt}=5t^{4}$$.
Using $$(2)$$,

$$\frac{dy}{dx}= \frac{dy}{dt}\,\frac{dt}{dx}=5t^{4}\cdot\frac{2t^{2}}{t^{2}-1} =\frac{10t^{6}}{\,t^{2}-1\,}$$ $$-(3)$$

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Step 2 : Second derivative

Let $$M(t)=\dfrac{10t^{6}}{t^{2}-1}$$. Differentiate with respect to $$t$$:

$$\frac{dM}{dt} = 10\!\left[\frac{6t^{5}}{t^{2}-1} -\frac{t^{6}\cdot2t}{(t^{2}-1)^{2}}\right] = \frac{60t^{5}}{t^{2}-1}-\frac{20t^{7}}{(t^{2}-1)^{2}}$$

Using $$(2)$$ again,

$$\frac{d^{2}y}{dx^{2}} = \frac{dM}{dt}\,\frac{dt}{dx} =\left(\frac{60t^{5}}{t^{2}-1}-\frac{20t^{7}}{(t^{2}-1)^{2}}\right) \frac{2t^{2}}{t^{2}-1}$$
$$\Longrightarrow\; \frac{d^{2}y}{dx^{2}} =\frac{120t^{7}}{(t^{2}-1)^{2}} -\frac{40t^{9}}{(t^{2}-1)^{3}} =\frac{40t^{7}(2t^{2}-3)}{(t^{2}-1)^{3}}$$ $$-(4)$$

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Step 3 : Express $$(x^{2}-1)$$ in terms of $$t$$

From $$(1)$$,
$$x^{2}-1 =\left(\frac{t+\frac1t}{2}\right)^{2}-1 =\frac{(t-\frac1t)^{2}}{4} =\frac{(t^{2}-1)^{2}}{4t^{2}}$$ $$-(5)$$

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Step 4 : Substitute in the differential equation

The given ODE is $$(x^{2}-1)\,y'' + \lambda x\,y' + ky = 0$$.
Insert $$(3)$$, $$(4)$$ and $$(5)$$:

Term by term,

$$(x^{2}-1)y'' = \frac{(t^{2}-1)^{2}}{4t^{2}}\; \cdot\frac{40t^{7}(2t^{2}-3)}{(t^{2}-1)^{3}} =\frac{10t^{5}(2t^{2}-3)}{t^{2}-1}$$

$$\lambda x\,y' = \lambda\!\left(\frac{t+\frac1t}{2}\right) \cdot\frac{10t^{6}}{t^{2}-1} =\frac{5\lambda\,(t^{7}+t^{5})}{t^{2}-1}$$

$$ky = kt^{5}$$

Multiply the whole equation by $$(t^{2}-1)$$ to clear the denominator:

$$10t^{5}(2t^{2}-3) +5\lambda(t^{7}+t^{5}) +k\,t^{5}(t^{2}-1)=0$$

Simplify and collect like powers:

$$\bigl(20+5\lambda+k\bigr)\,t^{7} +\bigl(-30+5\lambda-k\bigr)\,t^{5}=0$$

Because the equation must hold for every $$t$$, the coefficients of $$t^{7}$$ and $$t^{5}$$ are zero:

$$20+5\lambda+k=0\;\; -(i)$$
$$-30+5\lambda-k=0\;\; -(ii)$$

Add $$(i)$$ and $$(ii)$$:
$$-10+10\lambda=0 \;\Longrightarrow\; \lambda = 1$$

Substitute $$\lambda=1$$ in $$(i)$$:
$$20+5(1)+k=0 \;\Longrightarrow\; k=-25$$

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Step 5 : Required sum

$$\lambda + k = 1 + (-25) = -24$$

Hence, $$\lambda + k = -24$$, which corresponds to Option B.