Question

Find a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1

Answer 1

We have,

\[\left( x + y \right)dy + \left( x - y \right)dx = 0\]

\[\frac{dy}{dx} = \frac{y - x}{x + y}\]

Let y = vx

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

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

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

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

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

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

Integrating both sides, we get

\[\int\frac{1 + v}{1 + v^2}dy = - \int\frac{1}{x}dx\]

\[\int\frac{1}{1^2 + v^2}dy + \frac{1}{2}\int\frac{2v}{1 + v^2} = - \int\frac{1}{x}dx\]

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

\[ \Rightarrow 2 \tan^{- 1} v + \log\left( 1 + v^2 \right) + 2\log \left| x \right| = 2C\]

\[ \Rightarrow 2 \tan^{- 1} v + \log\left( 1 + v^2 \right) x^2 = k\text{ where, }k = 2C\]

\[ \Rightarrow 2 \tan^{- 1} \frac{y}{x} + \log\left( 1 + \frac{y^2}{x^2} \right) x^2 = k\]

\[ \Rightarrow 2 \tan^{- 1} \frac{y}{x} + \log \left( x^2 + y^2 \right) = k . . . . . . . . . \left( 1 \right)\]

Now,

When x = 1, y = 1

\[ \therefore 2 \tan^{- 1} 1 + \log \left( 2 \right) = k\]

\[ \Rightarrow k = \frac{\pi}{2} + \log 2\]

Putting the value of `k` in (1), we get

\[2 \tan^{- 1} \frac{y}{x} + \log \left( x^2 + y^2 \right) = \frac{\pi}{2} + \log 2\]