Question

Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.

Answer 1

According to the question,
$$\begin{gathered} \frac{dy}{dx}=x+xy \\ \Rightarrow \frac{dy}{dx}=x\left(1+y\right) \\ \Rightarrow \frac{1}{y+1}dy=xdx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y+1}dy=\int xdx \\ \Rightarrow \log \left|y+1\right|=\frac{x^2}{2}+\log C \\ \Rightarrow \log \left|\frac{y+1}{C}\right|=\frac{x^2}{2} \\ \Rightarrow y+1=C{e}^{\frac{x^2}{2}} \\ \mathrm{Since},\mathrm{the}\mathrm{curve}\mathrm{passes}\mathrm{through}\left(0,1\right). \\ \mathrm{It}\mathrm{satisfies}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}. \\ \therefore 1+1=C{e}^0 \\ \Rightarrow C=2 \\ \mathrm{Puting}\mathrm{the}\mathrm{value}\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{curve}.\mathrm{We}\mathrm{get} \\ y+1=2e^{\frac{x^2}{2}} \\ \Rightarrow y=-1+2e^{\frac{x^2}{2}} \end{gathered}$$