Question

\[\left( x - y \right)\frac{dy}{dx} = x + 2y\]

Answer 1

\[\left( x - y \right) \frac{dy}{dx} = x + 2y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y}\]
This is a homogeneous differential equatiuon .
\[\text{ Putting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}, \text{ we get }\]
\[v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx}\]
\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v - v + v^2}{1 - v}\]
\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + v + v^2}{1 - v}\]
\[ \Rightarrow \frac{1 - v}{1 + v + v^2}dv = \frac{1}{x}dx\]
Integrating both sides, we get 
\[\int\frac{1 - v}{1 + v + v^2}dv = \int\frac{1}{x}dx\]
\[\Rightarrow \int\frac{- (v - 1)}{v^2 + v + 1}dv = \frac{dx}{x}\]
\[ \Rightarrow \int\frac{1}{2} \times \frac{2v - 2}{v^2 + v + 1}dv = \int\frac{- dx}{x}\]
\[ \Rightarrow \int\frac{(2v + 1) - 3}{v^2 + v + 1}dv = - \int\frac{2dx}{x}\]
\[ \Rightarrow \int\frac{(2v + 1)}{v^2 + v + 1}dv - \int\frac{3}{v^2 + v + 1}dv = - \int\frac{2dx}{x}\]
\[\text{ Let }I_1 = \int\frac{(2v + 1)}{v^2 + v + 1}dv\]
\[\text{ and }I_2 = \int\frac{3}{v^2 + v + 1}dv\]
\[I = I_1 + I_2 \]
\[I_1 = \int \frac{2v + 1}{v^2 + v + 1}dv\]
\[\text{ let }v^2 + v + 1 = t \Rightarrow (2 v^2 + 1)dv = dt\]
\[\text{ therefore, }I_1 = \int \frac{2v + 1}{v^2 + v + 1}dv = \int\frac{dt}{t} = \log\left| t \right| = \log\left| v^2 + v + 1 \right|\]
\[\text{ hence, } I_1 = log\left| v^2 + v + 1 \right|\]
\[\text{ Also }, I_2 = \int\frac{3}{v^2 + v + 1}dv = \int\frac{3}{v^2 + 2v\left( \frac{1}{2} \right) + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 + 1}\]
\[ = \int\frac{3}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dv = 3\left( \frac{2}{\sqrt{3}} \right) \tan^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[ I_2 = 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[\text{ Hence, }I = I_1 + I_2 = log\left| v^2 + v + 1 \right| + 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)\]
\[\text{ therefore, }log\left| v^2 + v + 1 \right| + 2\sqrt{3} ta n^{- 1} \left( \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = - 2\log\left| x \right| + C\]
putting the value of v in the above equation we get, 
\[log\left| x^2 + y^2 + xy \right| = 2\sqrt{3} ta n^{- 1} \left( \frac{x + 2y}{x\sqrt{3}} \right) + C\]