Q94 / 170 · Manipal (MET) 2020 · Mathematics
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$\left(\log C x^2+e x^2\right) y=x$
- Correct answerDNone of these
Solution
Given differential equation is
$(y \log x-1) y d x=x d y$
$\Rightarrow \quad \frac{d y}{d x}=\frac{(y \log x-1) y}{x}$
$=\frac{y^2 \log x}{x}-\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=\frac{y^2 \log x}{x}$
$\Rightarrow \quad \frac{1}{y^2} \frac{d y}{d x}+\frac{y^{-1}}{x}=\frac{\log x}{x}$ ....(i)
Put $\quad y^{-1}=v \Rightarrow-y^{-2} \frac{d y}{d x}=\frac{d v}{d x}$
$\Rightarrow \quad y^{-2} \frac{d y}{d x}=-\frac{d v}{d x}$
From Eq. (i), we have
$-\frac{d v}{d x}+\frac{v}{x}=\frac{\log x}{x}$
$\Rightarrow \quad \frac{d v}{d x}-\frac{v}{x}=-\frac{\log x}{x}$ (ii)
This is linear differential equation.
Here, IF $=e^{\int\left(-\frac{1}{x}\right) d x}=e^{-\log x}=\frac{1}{x}$
So, solution is $v \cdot \mathrm{IF}=\int \mathrm{IF} \cdot Q d x+C$
$\begin{aligned} & v \cdot \frac{1}{x} \\ &=\int \frac{1}{x}\left(-\frac{\log x}{x}\right) d x+C \\ & \Rightarrow v \cdot \frac{1}{x}\end{aligned}=-\int \frac{\log x}{x^2} d x+C$
$=-\left[\log x\left(\frac{-1}{x}\right)+\int \frac{1}{x} \cdot \frac{1}{x} d x\right]+C$
$\Rightarrow \quad \frac{1}{x \cdot y}=\frac{\log x}{x}+\frac{1}{x}+C \quad\left[\because v=\frac{1}{y}\right]$
$\Rightarrow \quad 1=y[\log x+1+C x]$
$\Rightarrow \quad 1=y[\log x+\log \theta+C x] \quad[\because 1=\log e]$
$\Rightarrow \quad 1=y[\log e \cdot x+C x]$
Hence, the correct answer is option: 'D'.