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AP EAMCET Engineering 2013 Solved Paper

Section: Mathematics

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Question : 148 of 160

Marks: +1, -0

If

\cos^{-1}\left(\frac{y}{b}\right)=2\log\left(\frac{x}{2}\right)

x>0

x^{2}\frac{d^{2} y}{d x^{2}}+x\frac{d y}{d x}

is equal to

$4y$
$-4y$
0
$-8y$

Solution:

\cos^{-1}\left(\frac{y}{b}\right)=2\log\left(\frac{x}{2}\right)

,\;x>0

On differentiating w.r.t.

x

we get

-\frac{1}{\sqrt{1-\frac{y^{2}}{b^{2}}}}\cdot\frac{1}{b}\frac{d y}{d x}=2\cdot\frac{1}{\left(\frac{x}{2}\right)}\cdot\frac{1}{2}

\Rightarrow -\frac{1}{\sqrt{b^{2}-y^{2}}}\cdot\frac{d y}{d x}=\frac{2}{x}

\Rightarrow x\frac{d y}{d x}=-2\sqrt{b^{2}-y^{2}}

......(i)

Again, differentiating w.r.t.

x

we get,

x\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-2\cdot\frac{1}{2}(b^{2}-y^{2})^{-1/2}

(-2y)\frac{d y}{d x}

\Rightarrow x\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{2y}{\sqrt{b^{2}-y^{2}}}\cdot\frac{d y}{d x}

\Rightarrow x\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=\frac{2y}{-\frac{x}{2}\cdot\frac{d y}{d x}}\cdot\frac{d y}{d x}

[from Eq. (i).

\Rightarrow x^{2}\frac{d^{2} y}{d x^{2}}+x\frac{d y}{d x}=-4y

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