Question

Solve the following differential equation:- \[\left( x - y \right)\frac{dy}{dx} = x + 2y\]

Answer 1

We have,

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

\[ \Rightarrow \frac{dy}{dx} = \frac{x + 2y}{x - y} . . . . . \left( 1 \right)\]

Clearly this is a homogeneous equation,

Putting y = vx

\[ \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}\]

\[\text{Substituting }y = vx\text{ and }\frac{dy}{dx} = v + x\frac{dv}{dx}\text{ (1) becomes,} \]

\[v + x\frac{dv}{dx} = \frac{x + 2vx}{x - vx}\]

\[ \Rightarrow v + x\frac{dv}{dx} = \frac{1 + 2v}{1 - v}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v}{1 - v} - v\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{1 + 2v - v + v^2}{1 - v}\]

\[ \Rightarrow x\frac{dv}{dx} = \frac{v^2 + v + 1}{1 - v}\]

\[ \Rightarrow \frac{1 - v}{v^2 + v + 1}dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ \frac{- v}{v^2 + v + 1} + \frac{1}{v^2 + v + 1} \right]dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ - \frac{1}{2} \times \frac{2v + 1 - 1}{v^2 + v + 1} + \frac{1}{v^2 + v + 1} \right]dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ - \frac{1}{2} \times \frac{2v + 1}{v^2 + v + 1} + \frac{1}{2} \times \frac{1}{v^2 + v + 1} + \frac{1}{v^2 + v + 1} \right]dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ - \frac{1}{2} \times \frac{2v + 1}{v^2 + v + 1} + \frac{3}{2} \times \frac{1}{v^2 + v + 1} \right]dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ - \frac{1}{2} \times \frac{2v + 1}{v^2 + v + 1} + \frac{3}{2} \times \frac{1}{v^2 + v + \frac{1}{4} + \frac{3}{4}} \right]dv = \frac{1}{x}dx\]

\[ \Rightarrow \left[ - \frac{1}{2} \times \frac{2v + 1}{v^2 + v + 1} + \frac{3}{2} \times \frac{1}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right]dv = \frac{1}{x}dx\]

Integrating both sides, we get

\[ \Rightarrow \int\left[ - \frac{1}{2} \times \frac{2v + 1}{v^2 + v + 1} + \frac{3}{2} \times \frac{1}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right]dv = \int\frac{1}{x}dx\]

\[ \Rightarrow - \frac{1}{2}\int\frac{2v + 1}{v^2 + v + 1}dv + \frac{3}{2}\int\frac{1}{\left( v + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dv = \int\frac{1}{x}dx\]

\[ \Rightarrow - \frac{1}{2}\log \left| v^2 + v + 1 \right| + \frac{3}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \frac{v + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \log \left| x \right| + C\]

\[ \Rightarrow - \frac{1}{2}\log \left| \left( \frac{y}{x} \right)^2 + \frac{y}{x} + 1 \right| + \frac{3}{2} \times \frac{1}{\frac{\sqrt{3}}{2}} \tan^{- 1} \frac{\frac{y}{x} + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \log \left| x \right| + C\]

\[ \Rightarrow - \frac{1}{2}\log \left| \frac{y^2 + xy + x^2}{x^2} \right| + \sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log \left| x \right| + C\]

\[ \Rightarrow - \frac{1}{2}\log \left| y^2 + xy + x^2 \right| + \frac{1}{2}\log \left| x^2 \right| + \sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log \left| x \right| + C\]

\[ \Rightarrow - \frac{1}{2}\log \left| y^2 + xy + x^2 \right| + \log \left| x \right| + \sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) = \log \left| x \right| + C\]

\[ \Rightarrow - \frac{1}{2}\log \left| y^2 + xy + x^2 \right| + \sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) = C\]

\[ \Rightarrow \log \left| y^2 + xy + x^2 \right| - 2\sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) = - 2C\]

\[ \Rightarrow \log \left| y^2 + xy + x^2 \right| = 2\sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) - 2C\]

\[ \Rightarrow \log \left| y^2 + xy + x^2 \right| = 2\sqrt{3} \tan^{- 1} \left( \frac{2y + x}{\sqrt{3}x} \right) + k\text{ Where, }k = - 2C\]