Question

# Find the general solution of : $\dfrac{{dy}}{{dx}} = 1 + x + y + xy$

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Hint: Start by clubbing the terms, the next step is to take the common terms out, and then group the terms such that it is easier to integrate.

$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$
$\dfrac{{dy}}{{dx}} = \left( {1 + x} \right) + y\left( {1 + x} \right)$
$\dfrac{{dy}}{{dx}} = \left( {1 + x} \right)\left( {1 + y} \right)$
$\dfrac{{dy}}{{\left( {1 + y} \right)}} = \left( {1 + x} \right)dx$
$\int {\dfrac{{dy}}{{\left( {1 + y} \right)}}} = \int {\left( {1 + x} \right)dx}$
Answer = ${\log _e}|1 + y| = x + \dfrac{{{x^2}}}{2} + C$