Solution Curve of First Order Differential Equation

To solve the given differential equation and determine the properties of its solution curve, we proceed as follows:

Step 1: Solving the Differential Equation

The given first-order differential equation is: $(x^2 + xy + 4x + 2y + 4)\frac{dy}{dx} - y^2 = 0, \quad \text{for } x > 0$

By grouping the terms involving $x$, we can factorize the coefficient of $\frac{dy}{dx}$: $(x^2 + 4x + 4 + y(x + 2))\frac{dy}{dx} - y^2 = 0$ $((x + 2)^2 + y(x + 2))\frac{dy}{dx} = y^2$

Let $X = x + 2$. Since $x > 0$, we have $X > 2$. Notice that $\frac{dX}{dx} = 1$, so $\frac{dy}{dx} = \frac{dy}{dX}$. The equation becomes: $(X^2 + X y)\frac{dy}{dX} = y^2$

Rewriting this in terms of $\frac{dX}{dy}$: $\frac{dX}{dy} = \frac{X^2 + XY}{y^2} = \left(\frac{X}{y}\right)^2 + \frac{X}{y}$

Substitute $u = \frac{X}{y}$, which gives $X = u y \implies \frac{dX}{dy} = u + y\frac{du}{dy}$. Substituting into the differential equation yields: $u + y\frac{du}{dy} = u^2 + u$ $y\frac{du}{dy} = u^2$

Separating the variables: $\frac{du}{u^2} = \frac{dy}{y}$

Integrating both sides: $-\frac{1}{u} = \ln y + C$

Substituting $u = \frac{x+2}{y}$ back into the equation: $-\frac{y}{x+2} = \ln y + C$

Step 2: Applying the Initial Condition

The solution curve passes through the point $(1, 3)$. Substituting $x = 1$ and $y = 3$: $-\frac{3}{1+2} = \ln 3 + C \implies -1 = \ln 3 + C \implies C = -(1 + \ln 3)$

Thus, the implicit equation of the solution curve is: $-\frac{y}{x+2} = \ln y - \ln 3 - 1$ $\frac{y}{x+2} = 1 - \ln\left(\frac{y}{3}\right)$

Rearranging for $x + 2$: $x + 2 = \frac{y}{1 - \ln(y/3)}$

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Step 3: Analyzing the Options

Testing Options (A) and (B): Intersection with $y = x + 2$

Substitute $y = x + 2$ into the equation of the solution curve: $x + 2 = \frac{x + 2}{1 - \ln\left(\frac{x+2}{3}\right)}$

Since $x > 0 \implies x + 2 > 2 \neq 0$, we can divide both sides by $x + 2$: $1 = \frac{1}{1 - \ln\left(\frac{x+2}{3}\right)}$ $1 - \ln\left(\frac{x+2}{3}\right) = 1$ $\ln\left(\frac{x+2}{3}\right) = 0 \implies \frac{x+2}{3} = 1 \implies x = 1$

Since $x = 1 > 0$, there is a unique solution $x = 1$, corresponding to the point $(1, 3)$.

• Therefore, the curve intersects $y = x + 2$ at exactly one point.
• (A) is CORRECT, and (B) is INCORRECT.

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Testing Option (C): Intersection with $y = (x + 2)^2$

Substitute $y = (x + 2)^2$ into the equation $\frac{y}{x+2} = 1 - \ln\left(\frac{y}{3}\right)$: $\frac{(x+2)^2}{x+2} = 1 - \ln\left(\frac{(x+2)^2}{3}\right)$ $x + 2 = 1 - 2\ln(x+2) + \ln 3$ $x + 1 + 2\ln(x+2) - \ln 3 = 0$

Define $f(x) = x + 1 + 2\ln(x+2) - \ln 3$ for $x > 0$.

• At $x = 0$: $f(0) = 1 + 2\ln 2 - \ln 3 = 1 + \ln\left(\frac{4}{3}\right) > 0$.
• Derivative: $f'(x) = 1 + \frac{2}{x+2} > 0$ for all $x > 0$.

Since $f(x)$ is strictly increasing for $x > 0$ and $f(0) > 0$, $f(x) > 0$ for all $x > 0$. Hence, $f(x) = 0$ has no solution for $x > 0$.

• Therefore, the solution curve does NOT intersect $y = (x + 2)^2$.
• (C) is INCORRECT.

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Testing Option (D): Intersection with $y = (x + 3)^2$

Substitute $y = (x + 3)^2$ into the equation $\frac{y}{x+2} = 1 - \ln\left(\frac{y}{3}\right)$: $\frac{(x+3)^2}{x+2} = 1 - \ln\left(\frac{(x+3)^2}{3}\right)$ $\frac{(x+3)^2}{x+2} - 1 + 2\ln(x+3) - \ln 3 = 0$

Define $g(x) = \frac{(x+3)^2}{x+2} - 1 + 2\ln(x+3) - \ln 3$ for $x > 0$. Rewriting the first part: $\frac{(x+3)^2}{x+2} - 1 = \frac{x^2 + 6x + 9 - x - 2}{x+2} = \frac{x^2 + 5x + 7}{x+2} = x + 3 + \frac{1}{x+2}$

For $x > 0$:

• $x + 3 + \frac{1}{x+2} > 3 + \frac{1}{2} = 3.5 > 0$
• $x + 3 > 3 \implies 2\ln(x+3) > 2\ln 3 > \ln 3 \implies 2\ln(x+3) - \ln 3 > 0$

Thus, $g(x) > 3.5 > 0$ for all $x > 0$. Hence, $g(x) = 0$ has no solution for $x > 0$.

• Therefore, the solution curve does NOT intersect $y = (x + 3)^2$.
• (D) is CORRECT.

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Conclusion

The correct options are A and D.