Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ can be solved using an integrating factor.
• Integrating Factor: The integrating factor for a first-order linear differential equation is given by $IF = e^{\int P(x) dx}$.
• Uniqueness Theorem for ODEs: Under certain continuity conditions, a first-order initial value problem has a unique solution.

Step-by-Step Solution

• Step 1: Identify the type of differential equation and rewrite it in standard form.

  The given differential equation is $\frac{dy}{dx} = y + 7$. This is a first-order linear differential equation. To rewrite it in standard form, we subtract $y$ from both sides:

  $\frac{dy}{dx} - y = 7$

  This is now in the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -1$ and $Q(x) = 7$.

• Step 2: Calculate the Integrating Factor.

  The integrating factor (IF) is given by $IF = e^{\int P(x) dx}$. In this case, $P(x) = -1$, so:

  $IF = e^{\int -1 dx} = e^{-x}$

  We omit the constant of integration when calculating the integrating factor.

• Step 3: Multiply the differential equation by the Integrating Factor.

  Multiply both sides of the equation $\frac{dy}{dx} - y = 7$ by $e^{-x}$:

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

  The left side is now the derivative of $ye^{-x}$ with respect to $x$:

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

• Step 4: Integrate both sides with respect to $x$.

  Integrate both sides of the equation $\frac{d}{dx}(ye^{-x}) = 7e^{-x}$ with respect to $x$:

  $\int \frac{d}{dx}(ye^{-x}) dx = \int 7e^{-x} dx$

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

  Remember to include the constant of integration, $C$.

• Step 5: Solve for $y$ to find the general solution.

  Multiply both sides of the equation $ye^{-x} = -7e^{-x} + C$ by $e^x$:

  $y = -7 + Ce^x$

  This is the general solution to the differential equation.

• Step 6: Find the particular solution $y_1(x)$ using the initial condition $y_1(0) = 0$.

  Substitute $x = 0$ and $y = 0$ into the general solution $y = -7 + Ce^x$:

  $0 = -7 + Ce^0$

  $0 = -7 + C$

  $C = 7$

  So, $y_1(x) = -7 + 7e^x = 7e^x - 7$.

• Step 7: Find the particular solution $y_2(x)$ using the initial condition $y_2(0) = 1$.

  Substitute $x = 0$ and $y = 1$ into the general solution $y = -7 + Ce^x$:

  $1 = -7 + Ce^0$

  $1 = -7 + C$

  $C = 8$

  So, $y_2(x) = -7 + 8e^x = 8e^x - 7$.

• Step 8: Determine if the curves intersect by setting $y_1(x) = y_2(x)$.

  Set the two particular solutions equal to each other:

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

• Step 9: Solve for $x$.

  Add 7 to both sides:

  $7e^x = 8e^x$

  Subtract $7e^x$ from both sides:

  $0 = e^x$

• Step 10: Analyze the result.

  The equation $e^x = 0$ has no real solutions, since $e^x > 0$ for all real $x$.

• Step 11: Conclude whether the curves intersect.

  Since there is no real value of $x$ for which $y_1(x) = y_2(x)$, the curves do not intersect. This is consistent with the uniqueness theorem.

Common Mistakes & Tips

• Remember to include the constant of integration when solving differential equations.
• The exponential function $e^x$ is always positive and never equal to zero for real values of $x$.
• Understanding the uniqueness theorem for ODEs helps in interpreting the results.

Summary

We solved the given first-order linear differential equation and found two particular solutions based on the provided initial conditions. By setting these solutions equal to each other, we found that there is no real value of $x$ for which they intersect, meaning the two curves do not intersect. This is consistent with the uniqueness theorem for ODEs.

The final answer is \boxed{no point}, which corresponds to option (A).