Question

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

Answer 1

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:
${\displaystyle \frac{dy}{dx}}=1+x+y+xy$
Let us club the terms to simplify the process,
${\displaystyle \frac{dy}{dx}}=\left(1+x\right)+y\left(1+x\right)$
Now take the common term out,
${\displaystyle \frac{dy}{dx}}=\left(1+x\right)\left(1+y\right)$
Send the x terms on one side and the y terms to the other,
${\displaystyle \frac{dy}{\left(1+y\right)}}=\left(1+x\right)dx$
On applying integration on both sides, we get,
$\int {\displaystyle \frac{dy}{\left(1+y\right)}}=\int \left(1+x\right)dx$
Answer = ${\log }_e|1+y|=x+{\displaystyle \frac{x^2}{2}}+C$

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