Find the particular solution of the differential equation dy dx =… - Vidyadip

Question

Find the particular solution of the differential equation$\frac{\text{ dy}}{\text{dx}} = 1 +\text{x + y +xy},\text{ given that }\text{y} = 0 \text{ when x } = 1.$

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Answer

$\frac{\text{dy}}{\text{dx}} = 1 + \text{x + y + xy} = (1 + \text{x})( 1 + \text{y})$
$\therefore\int\frac{\text{dy}}{1 + \text{y}} = \int(1 + \text{x})\text{dx}$
$\log|1 + \text{y}| = \text{x} + \frac{\text{x}^{2}}{2} + \text{c}$
$\text{x} = 1 ,\text{y} = 0 \Rightarrow\text{c} = - \frac{3}{2}$
$\therefore\text{ solution is } \log|1 + \text{y} | = \text{x} + \frac{\text{x}^{2}}{2} - \frac{3}{2}.$

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