Let $\mathrm{y}={\mathrm{y}}_1\left(\mathrm{x}\right)\text{ and }\mathrm{y}={\mathrm{y}}_2\left(\mathrm{x}\right)$ be two distinct solutions of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}+\mathrm{y},$ with ${\mathrm{y}}_1\left(0\right)=0\text{ and }{\mathrm{y}}_2\left(0\right)=1$ respectively. Then, the number of points of intersection of $\mathrm{y}={\mathrm{y}}_1\left(\mathrm{x}\right)\text{ and }\mathrm{y}={\mathrm{y}}_2\left(\mathrm{x}\right)$ is

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}+\mathrm{y}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=\mathrm{x}\\ \mathrm{I}.\mathrm{F}={\mathrm{e}}^{-\mathrm{x}}\\ \therefore \text{ solution is }{\mathrm{ye}}^{-\mathrm{x}}=\int {\mathrm{xe}}^{-\mathrm{x}}\mathrm{dx}\\ \Rightarrow {\mathrm{ye}}^{-\mathrm{x}}=-{\mathrm{xe}}^{-\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}+\mathrm{c}\\ \Rightarrow \mathrm{y}=-\mathrm{x}-1+{\mathrm{ce}}^{\mathrm{x}}\\ {\mathrm{y}}_1\left(0\right)=0\Rightarrow \mathrm{c}=1\\ \therefore {\mathrm{y}}_1=-\mathrm{x}-1+{\mathrm{e}}^{\mathrm{x}} ....\left(1\right)\\ {\mathrm{y}}_2\left(0\right)=1\Rightarrow \mathrm{c}=2\\ \therefore {\mathrm{y}}_2=-\mathrm{x}-1+2{\mathrm{e}}^{\mathrm{x}} ....\left(2\right)\end{array}$

Now  ${\mathrm{y}}_2-{\mathrm{y}}_1={\mathrm{e}}^{\mathrm{x}}>0 \therefore {\mathrm{y}}_2\ne {\mathrm{y}}_1$

$\therefore$ Number of points of intersection of y₁ & y₂ is zero.