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if $\small 2x= y^{\frac{1}{5}}+y^{-\frac{1}{5}}$ and $\small \left ( x^{2}-1 \right )\frac{d^{2}y}{dx^{2}}+\lambda x\frac{dy}{dx} +ky=0$

then $\small \lambda +k$ is equal to :

Option 1)
-23

Option 2)
-24

Option 3)
26

Option 4)
-26

Answers (1)

best_answer

As we learnt in

Differential Equations -

An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable
$\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )$

- wherein

eg:

$\frac{d^{2}y}{dx^{2}}- 3\frac{dy}{dx}+5x=0$

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

$=y^{\frac{1}{5}}+\frac{1}{y^{\frac{1}{5}}}$

$\therefore 2x=\frac{\left ( y^\frac{1}{5} \right )^{2}+1}{y^{\frac{1}{5}}}$

$\therefore \left ( y^{\frac{1}{5}} \right )^{2}-2.x.y^{\frac{1}{5}}+x^{2}=x^{2}-1$

$\therefore \left ( x-y^{\frac{1}{5}} \right )^{2}=x^{2}-1$

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

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

$\therefore y=\left ( x-\sqrt{x^{2}-1} \right )^{5}$

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

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

$\frac{-5.y}{\sqrt{x^{2}-1}}$ ......................(i)

$=>\sqrt{x^{2}-1}\:\frac{dy}{dx}=-5y$

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

$=>\left ( x^{2}-1 \right )\frac{d^{2}y}{dx^{2}}+x.\frac{dy}{dx}$ $=-5\sqrt{x^{2}-1}.\frac{dy}{dx}$ $=-5\times \left ( -5y \right )$ $=25y$ $=25y$ from (i)

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

$\therefore \lambda =1, k=-25$

$\therefore \lambda +k=-24$

Option 1)

-23

Incorrect option

Option 2)

-24

Correct option

Option 3)

26

Incorrect option

Option 4)

-26

Incorrect option

Posted by

divya.saini

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