Question

The solution of $ydx-xdy+\log x$ $dx=0$ is:

• A. $y-\log x-1=cx$
• B. $x+\log x+1=cx$
• C. $y+\log x+1=cx$
• D. $y+\log x-1=cx$

Answer 1

The correct option is A $y-\log x-1=cx$

Given,

$ydx-xdy+\log xdx=0$

$\Rightarrow y-x\frac{dy}{dx}+\log x=0$

$\Rightarrow y-\log x=x\frac{dy}{dx}$

$\Rightarrow \frac{y}{x}-\frac{\log x}{x}=\frac{dy}{dx}$

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

Here, the above equation is a linear differential equation of the form, $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

where, $P\left(x\right)=\frac{-1}{x}$ and $Q\left(x\right)=-\frac{\log x}{x}$

Thus, Integrating factor $=e^{\int P\left(x\right)dx}=e^{\int \frac{-1}{x}dx}=e^{-\log x}=x^{-1}$

Hence, the general solution is $y\times I.F.=\int Q\left(x\right)\times I.F.dx$

$\Rightarrow y\times \left(x^{-1}\right)=\int (-\frac{\log x}{x})\times x^{-1}dx$

$\Rightarrow \frac{y}{x}=-\int \frac{\log x}{x^2}dx=-\left[\right(\log x)\left(\frac{-1}{x}\right)-\int \frac{1}{x}\times \frac{-1}{x}dx]$

$\Rightarrow \frac{y}{x}=\frac{\log x}{x}+\frac{1}{x}+c$

$\Rightarrow y=\log x+1+cx$

$\Rightarrow y-\log x-1=cx$