Question
Solve the differential equation:
$y + x \frac{dy}{dx} = x - y \frac{dy}{dx}$
$y + x \frac{dy}{dx} = x - y \frac{dy}{dx}$
✓
Answer
The differential equation can be re-written as:
$\frac{\text{dy}}{\text{dx}} = \frac{\text{x -y}}{\text{x + y}}, \text{put y = vx,} \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$\Rightarrow\text{v + x}\frac{\text{dv}}{\text{dx}} = \frac{1 - \text{v}}{1 + \text{v}}\Rightarrow\frac{\text{1 + v}}{\text{1 - 2v - v}^{2}}\text{dv} = \frac{\text{1}}{\text{x}} \text{dx}$
integrating we get
$\Rightarrow\frac{1}{2}\int\frac{\text{2V + 2}}{\text{V}^{2} + \text{2V - 1}}\text{dv} = -\int\frac{1}{\text{x}} \text{dx}=\frac{1}{2}\log|\text{V}^{2} + \text{2V} - 1| = -\log\text{ x }+ \log \text{ C}$
$\therefore $ Solution of the differential equation is:
$\frac{1}{2}\log\bigg|\frac{\text{y}^{2}}{\text{x}^{2}} + \frac{\text{2y}}{\text{x}} - 1\bigg| = \log\text{C} - \log\text{x or,}\text{ y}^{2} + \text{2xy - x}^{2} = \text{C}^{2}$
$\frac{\text{dy}}{\text{dx}} = \frac{\text{x -y}}{\text{x + y}}, \text{put y = vx,} \frac{\text{dy}}{\text{dx}} = \text{v + x} \frac{\text{dv}}{\text{dx}}$
$\Rightarrow\text{v + x}\frac{\text{dv}}{\text{dx}} = \frac{1 - \text{v}}{1 + \text{v}}\Rightarrow\frac{\text{1 + v}}{\text{1 - 2v - v}^{2}}\text{dv} = \frac{\text{1}}{\text{x}} \text{dx}$
integrating we get
$\Rightarrow\frac{1}{2}\int\frac{\text{2V + 2}}{\text{V}^{2} + \text{2V - 1}}\text{dv} = -\int\frac{1}{\text{x}} \text{dx}=\frac{1}{2}\log|\text{V}^{2} + \text{2V} - 1| = -\log\text{ x }+ \log \text{ C}$
$\therefore $ Solution of the differential equation is:
$\frac{1}{2}\log\bigg|\frac{\text{y}^{2}}{\text{x}^{2}} + \frac{\text{2y}}{\text{x}} - 1\bigg| = \log\text{C} - \log\text{x or,}\text{ y}^{2} + \text{2xy - x}^{2} = \text{C}^{2}$
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