Solve the following differential equation: (x+y)^2 dy dx = 1 - Vidyadip

Question

Solve the following differential equation:
$(\text{x}+\text{y})^2\frac{\text{dy}}{\text{dx}} = 1$

✓

Answer

$(\text{x}+\text{y})^2\frac{\text{dy}}{\text{dx}} = 1$
Let $\text{x}+\text{y} = \text{v}$
$1 + \frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}}$
$\frac{\text{dy}}{\text{dx}} = \frac{\text{dv}}{\text{dx}} - 1$
So,
$\text{v}^2\Big(\frac{\text{dv}}{\text{dx}}-1\Big) = 1$
$\frac{\text{dv}}{\text{dx}} = \frac{1}{\text{v}^2}+1$
$\frac{\text{dv}}{\text{dx}} = \frac{\text{v}^2+1}{\text{v}^2}$
$\frac{\text{v}^2}{\text{v}^2+1}\text{dv} = \text{dx}$
$\int\frac{\text{v}^2+1-1}{\text{v}^2+1}\text{dv} = \int \text{dx}$
$\int\Big(1-\frac{1}{\text{v}^2+1}\Big)\text{dv} = \int\text{dx}$
$\text{v}-\tan^{-1}(\text{v}) = \text{x} + \text{C}$
$\text{x}+\text{y}-\tan^{-1}(\text{x}+\text{y}) = \text{x}+\text{C}$
$\text{y}-\tan^{-1}(\text{x}+\text{y}) = \text{C}$

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