1. We are given the differential equation

$\frac{dy}{dx}=x+y.$

We need two solutions:

• $y_1(x)$ with $y_1(0)=0$
• $y_2(x)$ with $y_2(0)=1$

Then we must find how many points of intersection these two curves have.

2. Solve the differential equation.

Rewrite it as

$\frac{dy}{dx}-y=x.$

This is a linear first-order differential equation of the form

$\frac{dy}{dx}+P(x)y=Q(x),$

with

$P(x)=-1, \qquad Q(x)=x.$

The integrating factor is

$\text{I.F.}=e^{\int -1\,dx}=e^{-x}.$

Multiplying the equation by $e^{-x}$,

$e^{-x}\frac{dy}{dx}-e^{-x}y=xe^{-x}.$

The left side becomes

$\frac{d}{dx}(ye^{-x})=xe^{-x}.$

So,

$ye^{-x}=\int xe^{-x}\,dx + C.$

Now,

$\int xe^{-x}\,dx=-(x+1)e^{-x}+C.$

Hence,

$ye^{-x}=-(x+1)e^{-x}+C.$

Multiplying by $e^x$,

$y=-(x+1)+Ce^x.$

Thus the general solution is

$y=Ce^x-x-1.$

3. Find $y_1(x)$ using $y_1(0)=0$.

Substitute $x=0$ and $y=0$:

$0=C\cdot e^0-0-1=C-1.$

So,

$C=1.$

Therefore,

$y_1(x)=e^x-x-1.$

4. Find $y_2(x)$ using $y_2(0)=1$.

Substitute $x=0$ and $y=1$:

$1=C\cdot e^0-0-1=C-1.$

So,

$C=2.$

Therefore,

$y_2(x)=2e^x-x-1.$

5. Find points of intersection.

At intersection points,

$y_1(x)=y_2(x).$

So,

$e^x-x-1=2e^x-x-1.$

Canceling common terms,

$e^x=2e^x.$

This gives

$e^x=0.$

But $e^x>0$ for all real $x$, so this is impossible.

Hence, the two curves never intersect.

6. Therefore, the number of points of intersection is

$0.$

So the correct option is A.