y^(')=(x+1)e^(-x^(2)-2x)

\

y(x)=\\\\int (x+1)e^(-x^(2)-2x)dx

. Introduce

u=x^(2)+2x

then

du=2(x+1)dx

and

y(x)=

\

(1)/(2)\\\\int e^(-u)du=-(1)/(2)e^(-u)+C=-(1)/(2)e^(-x^(2)-2x)+C

.\ Hence,

y(0)=-(1)/(2)+C=1

and

C=(3)/(2)

. what are the steps

y=(x+1)ex22x y(x)=(x+1)ex22xdx. Introduce u=x2+2x then du=2(x+1)dx and y(x)= 21eudu=21eu+C=21ex22x+C. Hence, y(0)=21+C=1 and C=3/2