Section A: Pure Mathematics [40 marks] (i) Prove that the substitution $u = x^2 + y^2$ reduces the differential equation [y \frac{dy}{dx} + x = \sqrt{x^2 + y^2}] to [\frac{du}{dx} = 2\sqrt{u}.] Hence, show that the general solution of the differential equation $y\frac{dy}{dx} + x = \sqrt{x^2 + y^2}$ is given by $\sqrt{x^2 + y^2} = x + D$, where $D$ is an arbitrary constant.

1. Differentiate the given substitution $u = x^2 + y^2$ with respect to $x$.

$u = x^2 + y^2$
$\frac{du}{dx} = 2x + 2y\frac{dy}{dx}$
$\frac{1}{2}\frac{du}{dx} = x + y\frac{dy}{dx}$

2. Substitute the expressions for $u$ and $\frac{du}{dx}$ into the given differential equation to deduce the reduced form.

Given the differential equation:
$y \frac{dy}{dx} + x = \sqrt{x^2 + y^2}$
Substituting the LHS with $\frac{1}{2}\frac{du}{dx}$ and the RHS with $\sqrt{u}$:
$\frac{1}{2}\frac{du}{dx} = \sqrt{u}$
$\frac{du}{dx} = 2\sqrt{u} \quad \text{(Shown)}$

3. Solve the reduced differential equation by separating the variables.

$\frac{du}{dx} = 2\sqrt{u}$
$\int \frac{1}{2\sqrt{u}} \, du = \int 1 \, dx$
$\sqrt{u} = x + D$

4. Substitute back $u = x^2 + y^2$ to verify the general solution.

Substituting $u = x^2 + y^2$ into the integrated result:
$\sqrt{x^2 + y^2} = x + D \quad \text{(Shown)}$

1. Substitute the point $(-2,0)$ into the general solution to determine the specific value of the constant $D$.

$\sqrt{(-2)^2 + 0^2} = -2 + D$
$2 = -2 + D \implies D = 4$

The equation of the specific member is $\sqrt{x^2 + y^2} = x + 4$.

2. Rearrange the equation into a standard conics form by squaring both sides.

$\left(\sqrt{x^2 + y^2}\right)^2 = (x + 4)^2$
$x^2 + y^2 = x^2 + 8x + 16$
$y^2 = 8x + 16$
$y^2 = 8(x + 2)$

(Note: The squaring requires $x+4 \ge 0 \implies x \ge -4$. Since $y^2 \ge 0$, $8(x+2) \ge 0 \implies x \ge -2$, which satisfies $x \ge -4$, meaning no domain restriction is needed beyond the parabola's natural domain.)

3. Determine the axial intercepts and sketch the graph.

For $x$-intercepts, let $y = 0$:
$0 = 8(x + 2) \implies x = -2$
So the $x$-intercept is $(-2,0)$, which is also the vertex of the parabola.

For $y$-intercepts, let $x = 0$:
$y^2 = 8(0) + 16 = 16$
$y = \pm 4$
So the $y$-intercepts are $(0, 4)$ and $(0, -4)$.

The graph is a rightward-opening parabola $y^2 = 8(x+2)$ passing through $(-2, 0)$, $(0, 4)$, and $(0, -4)$.

1. Identify the line of symmetry from the curve's equation

From part (ii), substituting $(-2,0)$ into the general solution gives $D = 4$, so the equation of the curve is $\sqrt{x^2+y^2} = x+4$, which simplifies to $y^2 = 8(x+2)$.

Replacing $y$ with $-y$ leaves the equation unchanged, indicating that the graph is a parabola opening to the right, and is symmetric about the $x$-axis.

Therefore, the line of symmetry is $y = 0$.