Single Choice

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

A$$-23$$
B$$-24$$
Correct Answer
C$$26$$
D$$-26$$

Solution

$$\displaystyle 2x=y^{\frac{1}{5}}+y^{\frac{-1}{5}}$$

$$\displaystyle 2x=y^{\frac{1}{5}}+\frac{1}{y^{\frac{1}{5}}}$$

$$2x=a+\dfrac{1}{a}$$

$$a^2-2xa+1=0$$

$$a=\dfrac{2x\pm\sqrt{4x^2-4}}{2}$$

$$a=\dfrac{2x\pm 2\sqrt{x^2-1}}{2}$$

$$a=x\pm\sqrt{x^2-1}$$

$$y^{\frac{1}{5}} = x\pm \sqrt{x^2-1}$$

$$y=(x\pm \sqrt{x^2-1})^5$$

$$\dfrac{dy}{dx} = 5\left(x\pm \sqrt{x^2-1}\right)^4 \left(1\pm\dfrac{2x}{2\sqrt{2x-1}} \right)$$

$$=5\left(x\pm\sqrt{x^2-1}\right)^4 \left(\dfrac{\sqrt{x^2-1}\pm x}{\sqrt{x^2-1}}\right)$$

$$\dfrac{dy}{dx} = \dfrac{-5(x\pm\sqrt{x^2-1})^5}{\sqrt{x^2-1}}$$

$$\dfrac{dy}{dx} = \dfrac{-5.y}{\sqrt{x^2-1}}$$ .....(1)

$$\dfrac{d^2y}{dx^2} = \dfrac{\sqrt{x^2-1}\left(-5\dfrac{dy}{dx}\right)-\left(-5y\right)\dfrac{1}{2}\dfrac{2x}{\sqrt{x^2-1}}}{(x^2-1)}$$

$$\therefore (x^2-1)\dfrac{d^2y}{dx^2} = -5\sqrt{x^2-1}\dfrac{dy}{dx}+5y\dfrac{x}{\sqrt{x^2-1}}$$

$$=-5(-5y)+5\frac{1}{-5}\dfrac{dy}{dx}\times x$$

$$(x^2-1)\dfrac{d^2y}{dx^2} = 25y -x\dfrac{dy}{dx}$$

$$(x^2-1)\dfrac{d^2y}{dx^2}+1x\dfrac{dy}{dx}-25y=0$$

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

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