Question

The solution of the differential equation $\frac{dy}{dx}=\frac{ax+g}{by+f}$ represents a circle when
(a) a = b
(b) a = −b
(c) a = −2b
(d) a = 2b

Answer 1

(b) a = −b

We have,
$$\begin{gathered} \frac{dy}{dx}=\frac{ax+g}{by+f} \\ \Rightarrow \left(by+f\right)dy=\left(ax+g\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \left(by+f\right)dy=\int \left(ax+g\right)dx \\ \Rightarrow b\frac{y^2}{2}+fy=a\frac{x^2}{2}+gx+C \\ \Rightarrow b\frac{y^2}{2}+fy-a\frac{x^2}{2}-gx=C \\ \Rightarrow b{y}^2+2fy-a{x}^2-2gx-2C=0 \\ \mathrm{The}\mathrm{above}\mathrm{equation}\mathrm{represents}\mathrm{a}\mathrm{circle}. \\ \mathrm{Therefore},\mathrm{the}\mathrm{coffecients}\mathrm{of}{x}^2\mathrm{and}{y}^2\mathrm{must}\mathrm{be}\mathrm{equal}. \\ \mathrm{i}.\mathrm{e}.-a=b \\ \Rightarrow a=-b \end{gathered}$$