Question

If $\frac{dy}{dx}$=1+x+y+xy and y(-1)=0, then function y is

• A. $e^{(1-x{)}^2/2}$
• B. $e^{(1+x{)}^2/2}-1$
• C. $lo{g}_e(1+x)-1$
• D. $1+x$

Answer 1

Answer: (B)

$e^{(1+x{)}^2/2}-1$

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

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

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

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

$\log (1+y)=x+\frac{x^2}{2}+C.....\left(1\right)$

$\because y(-1)=0$

Therefore,

$\log (1+0)=-1+\frac{1}{2}+C$

$\Rightarrow C=\frac{1}{2}$

Substituting the value of $C$ in equation $\left(1\right)$, we get

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

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

$y=e^{\frac{{(x+1)}^2}{2}}-1$