Let $$y = y_1(x)$$ and $$y = y_2(x)$$ be two distinct solutions of the differential equation $$\frac{dy}{dx} = x + y$$, with $$y_1(0) = 0$$ and $$y_2(0) = 1$$ respectively. Then, the number of points of intersection of $$y = y_1(x)$$ and $$y = y_2(x)$$ is

We need to find the negation of $$(p \wedge q) \rightarrow (q \vee r)$$.

Step 1: Recall the negation of implication

The negation of $$P \rightarrow Q$$ is $$P \wedge (\sim Q)$$.

Here, $$P = p \wedge q$$ and $$Q = q \vee r$$.

Step 2: Apply the negation

$$\sim[(p \wedge q) \rightarrow (q \vee r)] = (p \wedge q) \wedge \sim(q \vee r)$$

Step 3: Simplify $$\sim(q \vee r)$$

By De Morgan's law: $$\sim(q \vee r) = (\sim q) \wedge (\sim r)$$

Step 4: Combine

$$(p \wedge q) \wedge (\sim q) \wedge (\sim r) = p \wedge q \wedge (\sim q) \wedge (\sim r)$$

Hence, the correct answer is Option A: $$p \wedge q \wedge (\sim q) \wedge (\sim r)$$.