The given differential equation is

$$\frac{d y}{d x} = y + 7$$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$$\frac{d y}{d x} - y = 7$$

The integrating factor is $e^{-\int dx} = e^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

$$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$$

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :

$$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$$

Integrate both sides with respect to $x$ :

$$e^{-x}y = -7e^{-x} + C$$

Multiply both sides by $e^{x}$ to isolate $y$ :

$$y = -7 + Ce^{x}$$

So, the general solution to the differential equation is $y = -7 + Ce^{x}$.

Now, let's apply the initial conditions to find the particular solutions :

For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :

$$0 = -7 + C$$

So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.

For $y_{2}(0)=1$, again substitute into the general solution:

$$1 = -7 + C$$

So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.

The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :

$$7e^{x} - 7 = 8e^{x} - 7$$

$$e^{x} = 0$$

But $e^{x} = 0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.