Forming an ODE from a Family of Curves

The elimination recipe:

• Count the arbitrary constants — that is the order of the ODE you will get.
• Differentiate the family that many times.
• Eliminate the constants between the original equation and its derivatives; the constant-free relation is the ODE.
• Examples: parabolas $x^2=4ay$ (one constant) → $x\,\dfrac{dy}{dx}=2y$; $y=e^x(a\cos x+b\sin x)$ (two constants) → $y''-2y'+2y=0$.

The step BEFORE the recipe — writing a geometrically described family, with its parameter count (five PYQs turn on this line, not on the differentiation):

• Lines through the origin: $y=mx$ (1) → $y=x\,y'$. All lines: $y=mx+c$ (2) → $y''=0$.
• Circles touching the y-axis at the origin: centre $(a,0)$, radius $a$: $x^2+y^2=2ax$ (1) → $2xy\,y'=y^2-x^2$. Touching the x-axis at the origin: $x^2+y^2=2by$ (1).
• Circles touching both axes in the first quadrant: $(x-r)^2+(y-r)^2=r^2$ (1). Circles with centre on the x-axis: $(x-a)^2+y^2=r^2$ (2) → $1+y'^2+y\,y''=0$. All circles: $x^2+y^2+2gx+2fy+c=0$ (3, so order 3).
• Parabolas with vertex at the origin and axis along $y$: $x^2=4ay$ (1); axis along $x$: $y^2=4ax$ (1) → $y=2x\,y'$.
• Ellipses with axes along the coordinate axes: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (2) → $x\,y\,y''+x\,y'^2-y\,y'=0$. Concentric circles $x^2+y^2=r^2$ (1) → $x+y\,y'=0$.

When a constant enters with a fractional power (e.g. $y=\sqrt{c}\,x+\ldots$, or $y^2=4a(x+a)$): eliminate it as usual, then clear the radical ONLY if the eliminated relation still carries one — squaring raises the degree ($(y^2-2xy\,y')^2=4(y\,y')^3$ has degree 3). Squaring when the relation was already polynomial produces a non-minimal ODE and the wrong degree.

Two directions, one idea:

• Family → ODE: differentiate and eliminate constants (as above), then compare with the given option.
• ODE → family: integrate the separable ODE and read off the curve type.
• A solved family is a circle only when the $x^2$ and $y^2$ coefficients are equal — e.g. $\dfrac{dy}{dx}=\dfrac{ax+h}{by+k}$ integrates to a circle exactly when $a=-b$.