Let y = y 1 x and y = y 2 x be two distinct solutions of the differential equation d y d x = x + y , with y…

Given,

$\frac{dy}{dx}=x+y\Rightarrow \frac{dy}{dx}-y=x$

So, $IF=e^{\int -1dx}=e^{-x}$

Now solution is given by $y{e}^{-x}=\int x{e}^{-x}dx$

$\Rightarrow y{e}^{-x}=-x{e}^{-x}-e^{-x}+c$

$\Rightarrow y=-x-1+c{e}^x$

Given, $y_1\left(0\right)=0\Rightarrow c=1$

$\therefore y_1=-x-1+e^x \cdots \left(1\right)$

Also given $y_2\left(0\right)=1\Rightarrow c=2$

$\therefore y_2=-x-1+2e^x \cdots \left(2\right)$

Now from equation $\left(1\right) \& \left(2\right)$ we get,  $y_2-y_1=e^x>0 \therefore y_2\ne y_1$

$\therefore$  Number of points of intersection of $y_1$ and $y_2$ is zero.