Key Concepts and Formulas

• Differential Equations: An equation involving derivatives of a function. Solving a differential equation means finding the function that satisfies the equation.
• Exact Differentials: An expression of the form $M(x, y)dx + N(x, y)dy$ is an exact differential if there exists a function $F(x, y)$ such that $dF = M(x, y)dx + N(x, y)dy$. In this problem, we use the fact that $d\left(\frac{u}{v}\right) = \frac{v\,du - u\,dv}{v^2}$, and specifically $-d\left(\frac{x}{y}\right) = \frac{x\,dy - y\,dx}{y^2}$.
• Integration: The process of finding the integral of a function. We use integration to find the general solution of the differential equation.

Step-by-Step Solution

Step 1: Rewriting the Differential Equation

We are given the differential equation: $\frac{dy}{dx} = \frac{xy^2 + y}{x}$ We want to rearrange this equation to a form that's easier to integrate. Multiplying both sides by $x \, dx$, we get: $x \, dy = (xy^2 + y) \, dx$ $x \, dy = xy^2 \, dx + y \, dx$ Now, move the terms involving $dx$ and $dy$ to one side: $x \, dy - y \, dx = xy^2 \, dx$ To create the exact differential form involving $\frac{x}{y}$, we divide the entire equation by $y^2$: $\frac{x \, dy - y \, dx}{y^2} = \frac{xy^2 \, dx}{y^2}$ $\frac{x \, dy - y \, dx}{y^2} = x \, dx$ Recognizing that $\frac{x \, dy - y \, dx}{y^2} = -d\left(\frac{x}{y}\right)$, we rewrite the equation as: $-d\left(\frac{x}{y}\right) = x \, dx$

Step 2: Integrating the Equation

Now, we integrate both sides of the equation: $\int -d\left(\frac{x}{y}\right) = \int x \, dx$ The integral of $-d\left(\frac{x}{y}\right)$ is simply $-\frac{x}{y}$. The integral of $x$ with respect to $x$ is $\frac{x^2}{2} + C$, where $C$ is the constant of integration. Therefore: $-\frac{x}{y} = \frac{x^2}{2} + C$

Step 3: Finding the Constant of Integration (C)

We are given that the curve intersects the line $x + 2y = 4$ at $x = -2$. We need to find the corresponding $y$-coordinate. Substituting $x = -2$ into the line equation: $-2 + 2y = 4$ $2y = 6$ $y = 3$ So, the curve passes through the point $(-2, 3)$. Now, substitute $x = -2$ and $y = 3$ into the general solution to find $C$: $-\frac{-2}{3} = \frac{(-2)^2}{2} + C$ $\frac{2}{3} = \frac{4}{2} + C$ $\frac{2}{3} = 2 + C$ $C = \frac{2}{3} - 2$ $C = \frac{2}{3} - \frac{6}{3}$ $C = -\frac{4}{3}$

Step 4: Determining the Equation of the Curve

Substitute the value of $C$ back into the general solution: $-\frac{x}{y} = \frac{x^2}{2} - \frac{4}{3}$

Step 5: Finding the Value of y for a Given Point

We want to find the value of $y$ when $x = 3$. Substitute $x = 3$ into the equation of the curve: $-\frac{3}{y} = \frac{3^2}{2} - \frac{4}{3}$ $-\frac{3}{y} = \frac{9}{2} - \frac{4}{3}$ $-\frac{3}{y} = \frac{27}{6} - \frac{8}{6}$ $-\frac{3}{y} = \frac{19}{6}$ Now, solve for $y$: $y = -\frac{3 \cdot 6}{19}$ $y = -\frac{18}{19}$

Common Mistakes & Tips

• Sign Errors: Be very careful with signs, especially when dealing with the exact differential $-d(x/y)$.
• Recognizing the Exact Differential: The key to solving this problem efficiently is recognizing the exact differential form. Practice identifying these forms.
• Solving for C: Don't forget to use the given information to solve for the constant of integration, $C$.

Summary

We solved the given differential equation by recognizing and utilizing the exact differential form. We found the general solution, then used the intersection point with the line to determine the constant of integration. Finally, we substituted $x = 3$ into the equation of the curve to find the corresponding value of $y$, which is $-\frac{18}{19}$.

Final Answer

The final answer is \boxed{-\frac{18}{19}}, which corresponds to option (A).