Question

The solution of the differential equation $x(x - y)\frac{dy}{dx} = y(x + y)$ is

• A. $\frac{x}{y} + \log(xy) = C$
• B. $\frac{y}{x} + \log(xy) = C$
• C. $\frac{x}{y} + y\log x = C$
• D. $\frac{y}{x} + x\log y = C$

Answer 1

Answer: (A)

$\frac{x}{y} + \log(xy) = C$

Answer 2

Let $\nu x - y \Rightarrow \frac{dy}{dx} = \nu + x\frac{d\nu}{dx} = \frac{\nu x^{2} + \nu^{2}x^{2}}{x^{2} - \nu x^{2}} = \frac{\nu + \nu^{2}}{1 - \nu}$

$\Rightarrow$ $x\frac{d\nu}{dx} = \frac{\nu + \nu^{2} - \nu + \nu^{2}}{1 - \nu} = \frac{2\nu^{2}}{1 - \nu}$

$\therefore$ $\int_{}^{}{\frac{1 - \nu}{2\nu^{2}}d\nu = \int_{}^{}\frac{1}{x}dx}$ [Integrating both sides]

$\Rightarrow$ $\frac{1}{2}\left\lbrack \frac{- 1}{\nu} - \log\nu \right\rbrack = \log x + C$

$\Rightarrow$ $\frac{1}{2}\left\lbrack \frac{- x}{y} - \log\left( \frac{y}{x} \right) - \log(x^{2}) \right\rbrack = C$

$\Rightarrow$ $\frac{x}{y} + \log\left( \frac{y}{x}.x^{2} \right) = C \Rightarrow \frac{x}{y} + {\log*}xy) = C.$