Question

The solution of the differential equation $ydx-xdy+\log xdx=0$ is

• A. $y=\log x+cx$
• B. $y=1+\log x+c$
• C. $y+cx=\log \frac{1}{x}$
• D. None of these

Answer 1

Answer: (D)

None of these

Given differential equation

$ydx-xdy+\log xdx=0$

Here, $M=y+\log x$ , $N=-x$

$\frac{\delta M}{\delta y}=1$

$\frac{\delta N}{\delta x}=-1$

$\Rightarrow \frac{\delta M}{\delta y}\ne \frac{\delta N}{\delta x}$
Given differential eqn is not exact.

To make it exact, multiply given differential eqn by $\frac{1}{x^2}$

$\frac{ydx-xdy}{x^2}+\frac{\log x}{x^2}dx=0$

$\Rightarrow \frac{xdy-ydx}{x^2}-\frac{\log x}{x^2}dx=0$

$\Rightarrow d\left(\frac{y}{x}\right)-\frac{\log x}{x^2}dx=0$

Integrating , we get

$\Rightarrow \frac{y}{x}-\int \frac{\log x}{x^2}dx=0$

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

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

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