Question

The solution of the differential equation $ydx+(x+x^{2}y)dy=0$is

Answer 1

Step 1 Compute the solution of the differential equation

The given equation can be written as

$\begin{array}{rl}& \mathrm{ydx}=-({\mathrm{x}}^2\mathrm{y}+\mathrm{x})\mathrm{dy}\\ & =>\mathrm{ydx}+\mathrm{xdy}=-{\mathrm{x}}^2\mathrm{ydy}\\ & \Rightarrow \frac{\mathrm{ydx}+\mathrm{xdy}}{(\mathrm{xy}{)}^2}=\frac{-\mathrm{dy}}{\mathrm{y}}\\ & \Rightarrow \frac{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{xy}\right)}{(\mathrm{xy}{)}^2}=-\frac{\mathrm{dy}}{\mathrm{y}}\\ & -\frac{1}{\mathrm{xy}}=-\log \mathrm{y}+\mathrm{k}\\ & \Rightarrow -\frac{1}{\mathrm{xy}}+\log \mathrm{y}=\mathrm{c}\end{array}$

The solution of the differential equation is $\mathbf{-}\frac{\mathbf{1}}{\mathbf{x}\mathbf{y}}\mathbf{+}\mathit{l}\mathit{o}\mathit{g}\mathbf{}\mathbf{y}\mathbf{=}\mathbf{c}$