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Last updated date: 27th Nov 2023
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MVSAT Dec 2023

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.

Complete step-by-step answer:
We have been given with a differential equation:
$\dfrac{{dy}}{{dx}} = 1 + x + y + xy$
Let us club the terms to simplify the process,
$\dfrac{{dy}}{{dx}} = \left( {1 + x} \right) + y\left( {1 + x} \right)$
Now take the common term out,
$\dfrac{{dy}}{{dx}} = \left( {1 + x} \right)\left( {1 + y} \right)$
Send the x terms on one side and the y terms to the other,
$\dfrac{{dy}}{{\left( {1 + y} \right)}} = \left( {1 + x} \right)dx$
On applying integration on both sides, we get,
$\int {\dfrac{{dy}}{{\left( {1 + y} \right)}}} = \int {\left( {1 + x} \right)dx} $
Answer = ${\log _e}|1 + y| = x + \dfrac{{{x^2}}}{2} + C$

Note: We started by arranging the terms such that it is easier to integrate and then integrated by using the formulas.