Question

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

• A. 26
• B. -24
• C. -23
• D. -26

Answer 1

Answer: (B)

-24

Given

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

or

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

Differentiate with respect to x on both the sides,

$\therefore \frac{1}{5}.y^{\frac{-4}{5}}\frac{dy}{dx}-\frac{1}{5}.y^{\frac{-6}{5}}.\frac{dy}{dx}=2$

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

$(y^{\frac{1}{5}}-y^{-\frac{1}{5}})\frac{dy}{dx}=10y$

$\left(2\sqrt{x^2-1}\right)\frac{dy}{dx}=10y$$-\left(1\right)$

Again Differentiating above equation with respect to x,$

$2\sqrt{x^2-1}\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}.2.\frac{2x}{2\sqrt{x^2-1}}=10\frac{dy}{dx}$

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

$(x^2-1)\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}=5.5y$ ( from equation (1))

On comparing with the above equation,

we get ,$\lambda =1andk=-25$

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