Question

For the differential equation xy$\frac{dy}{dx}$ = (x + 2) (y + 2). Find the solution curve passing through the point (1, −1).

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ xy\frac{dy}{dx}=\left(x+2\right)\left(y+2\right) \\ \Rightarrow \frac{y}{y+2}dy=\frac{\left(x+2\right)}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{y}{y+2}dy=\int \frac{\left(x+2\right)}{x}dx \\ \Rightarrow \int dy-2\int \frac{1}{y+2}dy=\int dx+2\int \frac{1}{x}dx \\ \Rightarrow y-2\log \left|y+2\right|=x+2\log \left|x\right|+C.....\left(1\right) \\ \mathrm{This}\mathrm{equation}\mathrm{represents}\mathrm{the}\mathrm{family}\mathrm{of}\mathrm{solution}\mathrm{curves}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{differential}\mathrm{equation}. \\ \mathrm{We}\mathrm{have}\mathrm{to}\mathrm{find}\mathrm{a}\mathrm{particular}\mathrm{member}\mathrm{of}\mathrm{the}\mathrm{family},\mathrm{which}\mathrm{passes}\mathrm{through}\mathrm{the}\mathrm{point}\left(1,-1\right). \\ \mathrm{Substituting}x=1\mathrm{and}y=-1\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ -1-2\log \left|1\right|=1+2\log \left|1\right|+C \\ \Rightarrow C=-2 \\ \mathrm{Putting}C=-2\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ y-2\log \left|y+2\right|=x+2\log \left|x\right|-2 \\ \Rightarrow y-x+2=\log \left\{x^2{\left(y+2\right)}^2\right\} \\ \mathrm{Hence},y-x+2=\log \left\{x^2{\left(y+2\right)}^2\right\}\mathrm{is}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{required}\mathrm{curve}. \end{gathered}$$