Let y = y _ 1 (x) and y = y _ 2 (x) be two distinct solutions of...

Given,

$\dfrac{dy}{dx} = x + y ⇒ \dfrac{dy}{dx} - y = x$

So, $IF = {e}^{∫ - 1dx} = {e}^{- x}$

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

$⇒ y{e}^{- x} = - x{e}^{- x} - {e}^{- x} + c$

$⇒ y = - x - 1 + c{e}^{x}$

Given, ${y}_{1}\left(0\right) = 0 ⇒ c = 1$

$∴ {y}_{1} = - x - 1 + {e}^{x} ⋯ \left(1\right)$

Also given ${y}_{2}\left(0\right) = 1 ⇒ c = 2$

$∴ {y}_{2} = - x - 1 + 2{e}^{x} ⋯ \left(2\right)$

Now from equation \left(1\right) \text{ and } \left(2\right) we get, ${y}_{2} - {y}_{1} = {e}^{x} > 0 ∴ {y}_{2} \ne {y}_{1}$

$∴$ Number of points of intersection of ${y}_{1} and {y}_{2}$ is zero.