Question

Solution of the differential equation $(y\ln x-1)ydx=xdy,x>0$ is:
(where $C$ is integration constant)

• A. $y\left[\ln \right(ex)+Cx]=1$
• B. $y\left[2\ln \right(ex)+C{x}^2]=1$
• C. $y\left[\ln \right(ex)+C{x}^2]=1$
• D. $y^2\left[\ln \right(ex)+Cx]=1$

Answer 1

The correct option is A $y\left[\ln \right(ex)+Cx]=1$

The given differential equation can be written as:
$x\frac{dy}{dx}+y=y^2\ln x\cdots \left(i\right)$
Dividing by $x{y}^2,$ we get
$\frac{1}{y^2}\cdot \frac{dy}{dx}+\frac{1}{xy}=\frac{1}{x}\ln x$
Let $\frac{1}{y}=v\Rightarrow -\frac{1}{y^2}\cdot \frac{dy}{dx}=\frac{dv}{dx}$
So, equation $\left(i\right)$ can be written as $\frac{dv}{dx}-\frac{1}{x}v=-\frac{1}{x}\ln x$
which is the standard linear differential equations, with
$P=-\frac{1}{x},Q=-\frac{1}{x}\ln x$
I.F.$=e^{\int \frac{-1}{x}dx}=e^{-\ln |x|}=e^{\ln |x^{-1}|}=\frac{1}{x}$
The solution is given by:
$v\cdot \frac{1}{x}=\int \frac{1}{x}(-\frac{1}{x}\ln x)dx\Rightarrow v\cdot \frac{1}{x}=-\int \frac{\ln x}{x^2}dx$
Integrating by using by parts, we have:
$v\cdot \frac{1}{x}=\frac{\ln x}{x}-\int \frac{1}{x}\cdot \frac{1}{x}dx=\frac{\ln x}{x}+\frac{1}{x}+C$
$\Rightarrow v=1+\ln x+Cx=\ln \left(ex\right)+Cx$
$\Rightarrow \frac{1}{y}=\ln \left(ex\right)+Cx$
$\Rightarrow y\left[\ln \right(ex)+Cx]=1$