Question

If $\frac{x+a}{2}=b{\cot }^{-1}\left(b\ln y\right)$,$b>0$, then value of $y{y}^{\prime \prime }+y{y}^{\prime }\ln y$ equals

• A. $y^{\prime }$
• B. $y^{\prime 2}$
• C. $0$
• D. $1$

Answer 1

Answer: (B)

$y^{\prime 2}$

$cot\left(\frac{x+a}{2b}\right)=blny$

$\Rightarrow -cs{c}^2\left(\frac{x+a}{2b}\right).\frac{1}{2b}=b\frac{y^{\prime }}{y}$

we know, $cs{c}^{2}x=1+co{t}^{2}x$

$-2b^2\frac{y^{\prime }}{y}\left(lny\right)=2b^2\left(\frac{y{y}^{\prime \prime }-y^{\prime }{y}^{\prime }}{y^2}\right)$

$\Rightarrow -(1+b^2(lny{)}^2)=-2b^2.\frac{y^{\prime }}{y}$

$\Rightarrow -2b^2\frac{y^{\prime }}{y}\left(lny\right)=2b^2\left(\frac{y{y}^{\prime \prime }-y^{\prime }{y}^{\prime }}{y^2}\right)$

$-y{y}^{\prime }\left(lny\right)=y{y}^{\prime \prime }-y^{\prime }{y}^{\prime }$

So, $y{y}^{\prime \prime }+-y{y}^{\prime }\left(lny\right)=(y^{\prime }{)}^2$