$${{dy} \over {dx}} = x + y$$

Let $$x + y = t$$

$$1 + {{dy} \over {dx}} = {{dt} \over {dx}}$$

$${{dt} \over {dx}} - 1 = t \Rightarrow \int {{{dt} \over {t + 1}} = \int {dx} } $$

$$\ln |t + 1| = x + C'$$

$$|t + 1| = C{e^x}$$

$$|x + y + 1| = C{e^x}$$

For $${y_1}(x),\,{y_1}(0) = 0 \Rightarrow C = 1$$

For $${y_2}(x),\,{y_2}(0) = 1 \Rightarrow C = 2$$

$${y_1}(x)$$ is given by $$|x + y + 1| = {e^x}$$

$${y_2}(x)$$ is given by $$|x + y + 1| = 2{e^x}$$

At point of intersection

$${e^x} = 2{e^x}$$

No solution

So, there is no point of intersection of $${y_1}(x)$$ and $${y_2}(x)$$.