Question

# Make and solve the given equation x dx + y dy=a^{2}frac{x dy - y dx}{x^{2} + y^{2}}

Second order linear equations
Make and solve the given equation $$x\ dx\ +\ y\ dy=a^{2}\frac{x\ dy\ -\ y\ dx}{x^{2}\ +\ y^{2}}$$

2021-02-27
Given:$$x\ dx\ +\ y\ dy=a^{2}\frac{x\ dy\ -\ y\ dx}{x^{2}\ +\ y^{2}}$$
$$x\ dx\ +\ y\ dy=a^{2}\left(\frac{x\ dy\ -\ y\ dx}{x^{2}\ +\ y^{2}}\right)$$
$$=a^{2}\left[\frac{\frac{x\ dy\ -\ y\ dx}{x^{2}}}{\frac{x^{2}\ +\ y^{2}}{x^{2}}}\right]$$
$$=a^{2}\left[\frac{d\left(\frac{y}{x}\right)}{1\ +\ \frac{y^{2}}{x^{2}}}\right]$$
$$=a^{2}\left[\frac{d\left(\frac{y}{x}\right)}{1\ +\ \left(\frac{y}{x}\right)^{2}}\right]$$
$$\Rightarrow\ \int\ xdx\ +\ \int\ ydy=a^{2}\ \int\ \left[\frac{d\left(\frac{y}{x}\right)}{1\ +\ \left(\frac{y}{x}\right)^{2}}\right]\ +\ C$$ where C is a constant
$$\Rightarrow\ \frac{x^{2}}{2}\ +\ \frac{y^{2}}{2}=a^{2}\ \tan^{-1}\left(\frac{y}{x}\right)\ C\ \left[\sin\ ce\ \int\ \frac{dx}{x^{2}\ +\ a^{2}}=\frac{1}{a^{2}}\ \frac{\tan^{-1}x}{a}\ +\ c\right]$$
Hence the solution of the given differential equation is $$x^{2}\ +\ y^{2}=2a^{2}\tan^{-1}\left(\frac{y}{x}\right)$$