Question

The solution of the equation $x^2\frac{d^{2}y}{d{x}^2}=\log x$ when $x=1,y=0and\frac{dy}{dx}=-1is$

• A. $y=\frac{1}{2}{\left(\log x\right)}^2+\log x$
• B. $y=\frac{1}{2}{\left(\log x\right)}^2-\log x$
• C. $y=-\frac{1}{2}{\left(\log x\right)}^2+\log x$
• D. $y=-\frac{1}{2}{\left(\log x\right)}^2-\log x$

Answer 1

Answer: (D)

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

$x^2\frac{d^{2}y}{d{x}^2}=\log x$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{\log x}{x^2}$

$\Rightarrow \int d\left(\frac{dy}{dx}\right)=\int \frac{\log x}{x^2}dx$

Let $\log x=z\Rightarrow x=e^z$

so, $\frac{1}{x}dx=dz$

so, $intd\left(\frac{dy}{dx}\right)=\int z{e}^{-z}dz$

$\int d\left(\frac{dy}{dx}\right)=z\frac{e^{-z}}{(-1)}+\frac{e^{-z}}{(-1)}+c$

$\int d\left(\frac{dy}{dx}\right)=\frac{dy}{dx}=-e^{-z}(z+1)+c$

Now,

$\Rightarrow z=\log x$

so, $\frac{dy}{dx}=e^{-\log x}(\log x+1)+c$

$\frac{dy}{dx}=\frac{-1}{x}(\log x+1)+c$

now, when $x=1,\frac{dy}{dx}=-1$

so, $-1=\frac{-1}{1}\left(1\right)+c$

so, $c=0$

so, $\frac{dy}{dx}=-(\frac{\log x}{x}+\frac{1}{x})$

$\int dy=\int \left(\frac{-\log x}{x}dx\right)-\int \frac{1}{x}dx$

so, $y=\frac{-1}{2}{\left(logx\right)}^2-\log x+c$

now, at $x=1,y=0$

so, $o=\frac{-1}{2}{\left(0\right)}^2-0+c$

$c=0$

so, $y=\frac{-1}{2}{\log }^{2}x-\log x$

Hence, the answer is $y=\frac{-1}{2}{\log }^{2}x-\log x.$