Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, solved using an integrating factor or, in some cases, a suitable substitution.
• Method of Substitution: A technique for simplifying differential equations by introducing a new variable that is a function of the original variables. This often transforms the equation into a separable form.
• Uniqueness Theorem for First-Order ODEs: Given $\frac{dy}{dx} = f(x,y)$, if $f(x,y)$ and $\frac{\partial f}{\partial y}$ are continuous in a region $R$, then for any point $(x_0, y_0)$ in $R$, there exists a unique solution $y(x)$ passing through $(x_0, y_0)$.

Step-by-Step Solution

Step 1: Introduce the Substitution

• Why this step? The given differential equation $\frac{dy}{dx} = x+y$ has the form $\frac{dy}{dx} = f(ax+by+c)$ which can be simplified by substituting $t = x+y$. This will transform the equation into a separable one.
• Let $t = x+y$.
• Differentiate both sides with respect to $x$: $\frac{dt}{dx} = \frac{d}{dx}(x+y) = 1 + \frac{dy}{dx}$
• Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{dt}{dx} - 1$

Step 2: Substitute into the Differential Equation

• Why this step? To rewrite the original differential equation in terms of $t$ and $x$ only, eliminating $y$ and $\frac{dy}{dx}$.
• Substitute $t = x+y$ and $\frac{dy}{dx} = \frac{dt}{dx} - 1$ into the original equation: $\frac{dt}{dx} - 1 = t$
• Rearrange the equation: $\frac{dt}{dx} = t + 1$

Step 3: Separate Variables

• Why this step? To isolate the variables $t$ and $x$ on opposite sides of the equation, making it integrable.
• Assuming $t+1 \neq 0$, divide both sides by $(t+1)$ and multiply by $dx$: $\frac{dt}{t+1} = dx$

Step 4: Integrate Both Sides

• Why this step? To find the general relationship between $t$ and $x$ by performing the inverse operation of differentiation.
• Integrate both sides: $\int \frac{dt}{t+1} = \int dx$
• Evaluate the integrals: $\ln|t+1| = x + C_0$ where $C_0$ is the constant of integration.

Step 5: Solve for $t$ and Back-Substitute

• Why this step? To express the general solution in terms of the original variables $x$ and $y$.
• Exponentiate both sides: $|t+1| = e^{x + C_0} = e^x e^{C_0}$
• Let $K = e^{C_0}$, where $K > 0$. Then: $|t+1| = K e^x$
• This implies $t+1 = \pm K e^x$. Let $C = \pm K$, where $C$ can be any non-zero real number. $t+1 = C e^x, \quad C \neq 0$
• Substitute back $t = x+y$: $x+y+1 = C e^x, \quad C \neq 0$
• Consider the case $t+1=0$: If $t+1=0$, then $x+y+1=0$, or $y=-x-1$. Substituting into the original DE: $-1 = x + (-x-1) \implies -1 = -1$. So $y=-x-1$ is also a solution. This solution corresponds to $C=0$ in the general form $x+y+1 = C e^x$.
• Therefore, the complete general solution is: $x+y+1 = C e^x$ where $C$ is any real number.

Step 6: Apply Initial Conditions to Find Specific Solutions

• Why this step? To determine the particular solutions $y_1(x)$ and $y_2(x)$ that satisfy the given initial conditions.

For $y_1(x)$:

• Given $y_1(0) = 0$. Substitute $x=0$ and $y=0$ into the general solution: $0 + 0 + 1 = C e^0$ $1 = C$
• So, $C_1 = 1$. The specific solution $y_1(x)$ is: $x + y_1(x) + 1 = e^x$ $y_1(x) = e^x - x - 1$

For $y_2(x)$:

• Given $y_2(0) = 1$. Substitute $x=0$ and $y=1$ into the general solution: $0 + 1 + 1 = C e^0$ $2 = C$
• So, $C_2 = 2$. The specific solution $y_2(x)$ is: $x + y_2(x) + 1 = 2 e^x$ $y_2(x) = 2 e^x - x - 1$

Step 7: Find Points of Intersection

• Why this step? To determine if the two solution curves intersect by checking if there are any $x$ values for which $y_1(x) = y_2(x)$.
• Set $y_1(x) = y_2(x)$: $e^x - x - 1 = 2 e^x - x - 1$
• Simplify: $e^x = 2 e^x$ $e^x = 0$
• Analysis: Since $e^x > 0$ for all $x$, the equation $e^x = 0$ has no solution. Therefore, the two curves do not intersect.

Common Mistakes & Tips

• Constant of Integration: Forgetting the constant of integration is a common mistake.
• Checking for $t+1=0$: Always remember to check if the case where the divisor is zero ($t+1=0$ in this case) yields a valid solution.
• Uniqueness Theorem: Recognizing the applicability of the uniqueness theorem can save significant time.

Summary

We solved the given differential equation using substitution to obtain a general solution. Then, we applied the initial conditions to find the particular solutions $y_1(x)$ and $y_2(x)$. Finally, we set $y_1(x) = y_2(x)$ to find the points of intersection and determined that there are no such points. Therefore, the number of points of intersection is 0.

Final Answer

The final answer is \boxed{0}, which corresponds to option (A).