Given,
dxdy=x+y⇒dxdy−y=x
So, IF=e∫−1dx=e−x
Now solution is given by ye−x=∫xe−xdx
⇒ye−x=−xe−x−e−x+c
⇒y=−x−1+cex
Given, y1(0)=0⇒c=1
∴y1=−x−1+ex⋯(1)
Also given y2(0)=1⇒c=2
∴y2=−x−1+2ex⋯(2)
Now from equation (1)&(2) we get, y2−y1=ex>0∴y2=y1
∴ Number of points of intersection of y1 and y2 is zero.