Question

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

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

Answer 1

Answer: (D)

Given differential equation is
$(y\log x-1)ydx=xdy$
$\Rightarrow \frac{dy}{dx}=\frac{(y\log x-1)y}{x}$
$=\frac{y^2\log x}{x}-\frac{y}{x}$
$\Rightarrow \frac{dy}{dx}+\frac{y}{x}=\frac{y^2\log x}{x}$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}+\frac{y^{-1}}{x}=\frac{\log x}{x}$ ...(i)
Put $y^{-1}=v$
$\Rightarrow -y^{-2}\frac{dy}{dx}=\frac{dv}{dx}$
From Eq. (i), we have
$-\frac{dv}{dx}+\frac{v}{x}=\frac{\log x}{x}$
$\Rightarrow \frac{dv}{dx}-\frac{v}{x}=-\frac{\log x}{x}$
This is linear differential equation.
Here, $IF=e^{\int \left(-\frac{1}{x}\right)dx}=e^{-\log x}=\frac{1}{x}$
So, solution is $v\cdot IF=\int IF\cdot Qdx+C$
$v\cdot \frac{1}{x}=\int \frac{1}{x}\left(-\frac{\log x}{x}\right)dx+C$
$\Rightarrow v\cdot \frac{1}{x}=-\int \frac{\log x}{x^2}dx+C$
$=-\left[\log x\left(\frac{-1}{x}\right)+\int \frac{1}{x}\cdot \frac{1}{x}dx\right]+C$
$\Rightarrow \frac{1}{x\cdot y}=\frac{\log x}{x}+\frac{1}{x}+C\left[\because v=\frac{1}{y}\right]$
$\Rightarrow 1=y[\log x+1+Cx]$
$\Rightarrow 1=y[\log x+\log e+Cx][\because 1=\log e]$
$\Rightarrow 1=y[\log e\cdot x+Cx]$