Question

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

Answer 1

$\frac{dy}{dx}=1+x+y+xy\Rightarrow \frac{dy}{dx}=(1+x)(1+y)$

Separating the variables, $\frac{dy}{1+y}=(1+x)dx$,
Integrating $\log (1+y)=(x+\frac{x^2}{2})+c$

when $x=1,y=0\Rightarrow \log \left(1\right)=\frac{3}{2}+c\Rightarrow 0=\frac{3}{2}+c\Rightarrow c=\frac{-3}{2}$

$\therefore \log (1+y)=(x+\frac{x^2}{2})-\frac{3}{2}$