Question
Solve the following differential equation:
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
✓
Answer
We have,
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
$\Rightarrow\text{x dx + y dx}-\text{x dy}-\text{y dy}=\text{dx + dy}$
$\Rightarrow(\text{x + y}-1)\text{dx}=(\text{x + y}+1)\text{dy}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{x}+\text{y}-1}{\text{x}+\text{y}+1}$
Let $\text{ x} + \text{y} = \text{v}$
$\therefore 1+ \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
$\therefore\frac{\text{dv}}{\text{dx}}-1 = \frac{\text{v}-1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1}{\text{v}+1}+1$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1+\text{v}+1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{2\text{v}}{\text{v}+1}$
$\Rightarrow \frac{\text{v}+1}{2\text{v}}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}+1}{2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\int\text{dv}+\frac{1}{2}\int\frac{1}{\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\text{v}+\frac{1}{2}\log|\text{v}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{x}+\text{y})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{y}-\text{x})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{C}$
$(\text{x + y})(\text{dx}-\text{dy})=\text{dx + dy}$
$\Rightarrow\text{x dx + y dx}-\text{x dy}-\text{y dy}=\text{dx + dy}$
$\Rightarrow(\text{x + y}-1)\text{dx}=(\text{x + y}+1)\text{dy}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{x}+\text{y}-1}{\text{x}+\text{y}+1}$
Let $\text{ x} + \text{y} = \text{v}$
$\therefore 1+ \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\Rightarrow \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
$\therefore\frac{\text{dv}}{\text{dx}}-1 = \frac{\text{v}-1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1}{\text{v}+1}+1$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{\text{v}-1+\text{v}+1}{\text{v}+1}$
$\Rightarrow \frac{\text{dv}}{\text{dx}} = \frac{2\text{v}}{\text{v}+1}$
$\Rightarrow \frac{\text{v}+1}{2\text{v}}\text{dv} = \text{dx}$
Integrating both sides, we get
$\int \frac{\text{v}+1}{2\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\int\text{dv}+\frac{1}{2}\int\frac{1}{\text{v}}\text{dv} = \int\text{dx}$
$\Rightarrow \frac{1}{2}\text{v}+\frac{1}{2}\log|\text{v}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{x}+\text{y})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{x}+\text{C}$
$\Rightarrow \frac{1}{2}(\text{y}-\text{x})+\frac{1}{2}\log|\text{x}+\text{y}| = \text{C}$
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