Question

The solution of the differential equation y dx + (x + xy) dy = 0 is ________________.

Answer 1

Given: y dx + (x + xy) dy = 0


$$\begin{gathered} ydx+\left(x+xy\right)dy=0 \\ \Rightarrow ydx=-\left(x+xy\right)dy \\ \Rightarrow ydx=-x\left(1+y\right)dy \\ \Rightarrow \frac{dx}{-x}=\frac{\left(1+y\right)}{y}dy \\ \Rightarrow \frac{\left(1+y\right)}{y}dy=-\frac{dx}{x} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \Rightarrow \int \frac{\left(1+y\right)}{y}dy=-\int \frac{dx}{x} \\ \Rightarrow \int \left(\frac{1}{y}+1\right)dy=-\int \frac{dx}{x} \\ \Rightarrow \log \left|y\right|+y=-\log \left|x\right|+\log \left|C\right|,\mathrm{where}\log \left|C\right|\mathrm{is}\mathrm{arbitrary}\mathrm{constant} \\ \Rightarrow y=\log \left|\frac{C}{x}\right|-\log \left|y\right| \\ \Rightarrow y=\log \left|\frac{C}{xy}\right| \\ \Rightarrow \left|\frac{C}{xy}\right|=e^y \\ \Rightarrow e^y\left|xy\right|=\left|C\right| \\ \Rightarrow e^{y}xy=\pm C \\ \Rightarrow e^{y}xy=A,\mathrm{where}A=\pm C \end{gathered}$$


Hence, the solution of the differential equation y dx + (x + xy) dy = 0 is $\underline{e^{y}xy=A}$.