JEE Main 2017 — Differential Equations Question with Solution

It is given that

$$2x=y^{1/5}+y^{-1/5}$$

$$\Rightarrow 2x=y^{1/5}+1/y^{1/5}$$

Therefore, $$2x=a+\frac{1}{a}\Rightarrow a^2-2ax+1=0$$

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

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

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

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

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

Therefore,

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

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

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

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

Therefore, $$(x^2-1)\frac{d^{2}y}{d{x}^2}=-5\sqrt{x^2-1}\frac{dy}{dx}+5y\frac{x}{\sqrt{x^2-1}}$$

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

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

Therefore, $$\lambda$$ = 1, k = $$-$$25; hence,

$$\lambda +k=-24$$