Question

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

Answer 1

$\frac{dy}{dx}=1+x+y+xy$ Given y=0 when x=1

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

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

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

on intagrating both sides

$\int \frac{dy}{1+y}=\int (1+x)dx\Rightarrow log(1+y)=x+\frac{x^2}{2}+c$

when x=1 y=0

log 1 $=1+\frac{1}{2}+c$

$[c=\frac{-3}{2}]$ $[\because log1=0]$

then log $(1+y)=x+\frac{x^2}{2}-\frac{-3}{2}$ is solution of given

different equation.