Question

The solution of differential equation $(y \log x-1) y d x=x d y$ is

• A. $y\left(\log e^{x}+C x\right)=1$
• B. $\left(\log \frac{x}{e}+C x\right) x=y$
• C. $\left(\log C x^{2}+e x^{2}\right) y=x$
• D. None of these

Answer 1

Answer: (D)

Given differential equation is
$(y \log x-1) y d x=x d y$
$\Rightarrow \frac{d y}{d x}=\frac{(y \log x-1) y}{x}$
$=\frac{y^{2} \log x}{x}-\frac{y}{x}$
$\Rightarrow \frac{d y}{d x}+\frac{y}{x}=\frac{y^{2} \log x}{x}$
$\Rightarrow \frac{1}{y^{2}} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}$ ...(i)
Put $y^{-1}=v$
$\Rightarrow -y^{-2} \frac{d y}{d x}=\frac{d v}{d x}$
From Eq. (i), we have
$-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x}$
$\Rightarrow \frac{d v}{d x}-\frac{v}{x}=-\frac{\log x}{x}$
This is linear differential equation.
Here, $I F=e^{\int\left(-\frac{1}{x}\right) d x}=e^{-\log x}=\frac{1}{x}$
So, solution is $v \cdot I F=\int I F \cdot Q d x+ C$
$v \cdot \frac{1}{x}=\int \frac{1}{x}\left(-\frac{\log x}{x}\right) d x+C$
$\Rightarrow v \cdot \frac{1}{x}=-\int \frac{\log x}{x^{2}} d x+C$
$=-\left[\log x\left(\frac{-1}{x}\right)+\int \frac{1}{x} \cdot \frac{1}{x} d x\right]+C$
$\Rightarrow \frac{1}{x \cdot y}=\frac{\log x}{x}+\frac{1}{x}+C\left[\because v=\frac{1}{y}\right]$
$\Rightarrow 1=y[\log x+1+C x]$
$\Rightarrow 1=y[\log x+\log e+C x][\because 1=\log e]$
$\Rightarrow 1=y[\log e \cdot x+C x]$