Question

The solution of differential equation $(1+x)ydx+(1-y)xdy=0$ is

• A. ${\log }_e\left(xy\right)+x–y=C$
• B. ${\log }_e\left(\frac{x}{y}\right)+x+y=C$
• C. ${\log }_e\left(\frac{x}{y}\right)-x+y=C$
• D. ${\log }_e\left(xy\right)-x+y=C$

Answer 1

The correct option is A

${\log }_e\left(xy\right)+x–y=C$

Explanation for the correct option:

Find the solution of the given differential equation.

Given: $(1+x)ydx+(1-y)xdy=0$

We can also write as,

$$\begin{gathered} \Rightarrow (1+x)ydx=-(1-y)xdy \\ \Rightarrow \left(\frac{1+x}{x}\right)dx=-\left(\frac{1-y}{y}\right)dy \\ \Rightarrow \left(\frac{1}{x}+1\right)dx=\left(\frac{-1}{y}+1\right)dy \end{gathered}$$

On integrating both sides, we get

$$\begin{gathered} \int \left(\frac{1}{x}+1\right)dx=\int \left(\frac{-1}{y}+1\right)dy \\ \Rightarrow \int \frac{1}{x}dx+\int 1dx=-\int \frac{1}{y}dy+\int 1dy \\ \Rightarrow {\log }_{e}x+x=-{\log }_{e}y+y+C \\ \Rightarrow {\log }_{e}x+{\log }_{e}y=y-x+C \\ \Rightarrow {\log }_e\left(xy\right)+x-y=C \end{gathered}$$

Hence, option (A) is the correct answer.