Question

The differential equation of the family of curves x² + y² - 2ay = 0, where a is arbitrary constant, is _______________.

Answer 1

Given: x² + y² − 2ay = 0


$$\begin{gathered} x^2+y^2-2ay=0...\left(1\right) \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}x \\ \Rightarrow \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)-2a\frac{d}{dx}\left(y\right)=0 \\ \Rightarrow 2x+2y\frac{dy}{dx}-2a\frac{dy}{dx}=0 \\ \Rightarrow 2a\frac{dy}{dx}=2x+2y\frac{dy}{dx} \\ \Rightarrow a\frac{dy}{dx}=x+y\frac{dy}{dx} \\ \Rightarrow ay\text{'}=x+yy\text{'},\mathrm{where}\frac{dy}{dx}=y\text{'} \\ \Rightarrow a=\frac{x+yy\text{'}}{y\text{'}} \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}a\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ {x}^2+y^2-2\frac{x+yy\text{'}}{y\text{'}}y=0 \\ \Rightarrow \frac{x^{2}y\text{'}+y^{2}y\text{'}-2\left(x+yy\text{'}\right)y}{y\text{'}}=0 \\ \Rightarrow x^{2}y\text{'}+y^{2}y\text{'}-2xy-2y^{2}y\text{'}=0 \\ \Rightarrow x^{2}y\text{'}-y^{2}y\text{'}-2xy=0 \\ \Rightarrow \left(x^2-y^2\right)\frac{dy}{dx}-2xy=0 \end{gathered}$$

Hence, the differential equation representing the family of curves x² + y² − 2ay = 0, is $\underline{\left(x^2-y^2\right)\frac{dy}{dx}-2xy=0}.$