Let $y=y_{1}(x)$ and $y=y_{2}(x)$ be the solution curves of the differential equation $\frac{d y}{d x}=y+7$ with initial conditions $y_{1}(0)=0$ and $y_{2}(0)=1$ respectively. Then the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ intersect at

<p>The given differential equation is</p> <p>$\frac{d y}{d x} = y + 7$</p> <p>This is a first order linear differential equation and can be solved using an integrating factor. </p> <p>Rearrange the equation to the standard form of a linear differential equation :</p> <p>$\frac{d y}{d x} - y = 7$</p> <p>The integrating factor is $e^{-\int dx} = e^{-x}$.</p> <p>Multiplying each side of the equation by the integrating factor gives :</p> <p>$e^{-x} \frac{d y}{d x} - e^{-x} y = 7e^{-x}$</p> <p>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 :</p> <p>$\frac{d}{d x}(e^{-x}y) = 7e^{-x}$</p> <p>Integrate both sides with respect to $x$ :</p> <p>$e^{-x}y = -7e^{-x} + C$</p> <p>Multiply both sides by $e^{x}$ to isolate $y$ :</p> <p>$y = -7 + Ce^{x}$</p> <p>So, the general solution to the differential equation is $y = -7 + Ce^{x}$.</p> <p>Now, let&#39;s apply the initial conditions to find the particular solutions :</p> <p>For $y_{1}(0)=0$, we substitute into the general solution and solve for $C$ :</p> <p>$0 = -7 + C$</p> <p>So, $C = 7$, and the solution for $y_{1}$ is $y_{1}(x) = 7e^{x} - 7$.</p> <p>For $y_{2}(0)=1$, again substitute into the general solution:</p> <p>$1 = -7 + C$</p> <p>So, $C = 8$, and the solution for $y_{2}$ is $y_{2}(x) = 8e^{x} - 7$.</p> <p>The two curves intersect when $y_{1}(x) = y_{2}(x)$. Setting these equal and solving for $x$ gives :</p> <p>$7e^{x} - 7 = 8e^{x} - 7$</p> <p>$e^{x} = 0$</p> <p>But $e^{x} = 0$ has no solution, because the exponential function never equals zero.</p> <p>So, the curves $y=y_{1}(x)$ and $y=y_{2}(x)$ do not intersect at any point.</p>