A family of curves is given by the differential equation $x\dfrac{dy}{dx}-y=x(x-y)$.

1. Differentiate the substitution equation with respect to $x$.

Given $y = vx$.
Using the product rule,
$\frac{dy}{dx} = v + x\frac{dv}{dx}$

2. Substitute $y$ and $\frac{dy}{dx}$ into the original differential equation and simplify.

Substitute $y$ and $\frac{dy}{dx}$ into $x\frac{dy}{dx}-y=x(x-y)$:
$x\left(v + x\frac{dv}{dx}\right) - vx = x(x-vx)$
$xv + x^2\frac{dv}{dx} - vx = x^2(1-v)$
$x^2\frac{dv}{dx} = x^2(1-v)$
Assuming $x \neq 0$, dividing by $x^2$ gives:
$\frac{dv}{dx} = 1-v$

3. Separate the variables and integrate.

Separating variables:
$\int \frac{1}{1-v} \, dv = \int 1 \, dx$
$-\ln|1-v| = x + C$

4. Solve for $v$ and introduce the arbitrary constant $A$.

$\ln|1-v| = -x - C$
$|1-v| = e^{-x-C}$
$1-v = \pm e^{-C} e^{-x}$
Let $A = \pm e^{-C}$, which is an arbitrary constant.
$1-v = Ae^{-x}$
$v = 1 - Ae^{-x}$

5. Back-substitute to express $y$ in terms of $x$.

Substitute $v = \frac{y}{x}$:
$\frac{y}{x} = 1 - Ae^{-x}$
$y = x(1 - Ae^{-x}) \text{ (Shown)}$

1. Determine the asymptotic behaviour of the curves.

As $x \to \infty$, $e^{-x} \to 0$, which implies $Axe^{-x} \to 0$.
Therefore, $y = x(1-Ae^{-x}) = x - Axe^{-x} \to x$.
The line $y = x$ is an oblique asymptote for all curves in the family. Since $A > 0$ and $x > 0$ for large $x$, the term $-Axe^{-x}$ is negative, meaning the curves approach the asymptote $y=x$ from below.
As $x \to -\infty$, $e^{-x} \to \infty$, so $y \approx x(-Ae^{-x}) \to \infty$.

2. Identify the axes intercepts for different positive values of $A$.

Substitute $x = 0$ into the equation: $y = 0(1-A) = 0$. Thus, all curves pass through the origin $(0,0)$.
Substitute $y = 0$ to find the $x$-intercepts:
$x(1 - Ae^{-x}) = 0$
$x = 0 \quad \text{or} \quad 1 - Ae^{-x} = 0 \implies e^x = A \implies x = \ln A$
Choosing representative positive values for $A$:

• For $0 < A < 1$, the second intercept $x = \ln A$ is negative.
• For $A = 1$, the only intercept is $x = 0$ (the origin is a turning point).
• For $A > 1$, the second intercept $x = \ln A$ is positive.

3. Sketch the three representative curves on a single diagram, paying attention to their relative positions.

Using a graphing calculator or by deduction, sketch the curves for $A=0.5, A=1, A=2$.
Note that the curves only intersect at the origin since $x(1-A_1e^{-x}) = x(1-A_2e^{-x}) \implies x=0$.
For $x > 0$, a larger $A$ yields a smaller $y$ (curve is lower).
For $x < 0$, a larger $A$ yields a larger $y$ (curve is higher).

• Curve $A=2$: Intersects at $x=0$ and $x=\ln 2 > 0$. Lowest for $x>0$, highest for $x<0$.
• Curve $A=1$: Touches the $x$-axis at the origin (minimum point).
• Curve $A=0.5$: Intersects at $x=\ln 0.5 < 0$ and $x=0$. Highest for $x>0$, lowest for $x<0$.
  Draw all three possessing minimum points and approaching $y=x$ from below.

1. Set $\frac{dy}{dx} = 0$ in the differential equation to find the condition for stationary points.

At a stationary point, $\frac{dy}{dx} = 0$.
Substituting this into the given differential equation $x\frac{dy}{dx}-y=x(x-y)$:
$x(0) - y = x(x-y)$
$-y = x^2 - xy$

2. Rearrange to find the cartesian equation of the locus.

$xy - y = x^2$
$y(x-1) = x^2$
$y = \frac{x^2}{x-1}$

3. Determine the domain restriction based on the condition $A > 0$.

From part (i), $y = x(1-Ae^{-x})$, which can be rearranged to $Ae^{-x} = 1 - \frac{y}{x}$.
Substitute $y = \frac{x^2}{x-1}$ into this expression:
$Ae^{-x} = 1 - \frac{x}{x-1} = \frac{(x - 1) - x}{x-1} = \frac{-1}{x-1} = \frac{1}{1-x}$
Since we are given $A > 0$ and we know $e^{-x} > 0$ for all real $x$, it follows that:
$\frac{1}{1-x} > 0$
$1 - x > 0 \implies x < 1$
Thus, the locus is valid only for $x < 1$.