=∫dx ln|t+1|=x+C′ |t+1|=Cex |x+y+1|=Cex For y1(x),y1(0)=0⇒C=1 For y2(x),y2(0)=1⇒C=2 y1(x) is given by |x+y+1|=ex y2(x) is given by |x+y+1|=2ex At point of intersection ex=2ex No solution So, there is no point of intersection of y1(x) and y2(x).