Formation of Differential Equations from Geometrical Problems

When a family of curves is given by an equation containing $n$ arbitrary constants, the differential equation the whole family satisfies is found by differentiating $n$ times and eliminating the constants from the resulting $(n+1)$ equations (§10.4). Two symmetry checks help along the way: eliminating one arbitrary constant always yields a first-order equation; eliminating two always yields a second-order equation; and so on.

Example 10.2 (one constant → order 1). The family of all straight lines through the origin is $y=mx$, with $m$ the only arbitrary constant (Fig. 10.1 sketches four members of this family: $y=x,\ y=2x,\ y=-x,\ y=-2x$). Differentiating once gives $\dfrac{dy}{dx}=m$; substituting back into $y=mx$ eliminates $m$, leaving the first-order equation

Example 10.3 (two constants → order 2). For $y=\mathrm A\cos x+\mathrm B\sin x$ (two constants $\mathrm A,\mathrm B$), differentiating twice gives $y'=-\mathrm A\sin x+\mathrm B\cos x$ and $y''=-\mathrm A\cos x-\mathrm B\sin x=-y$. So directly, $\dfrac{d^2y}{dx^2}+y=0$ — no further substitution was even needed, since the second derivative reproduces $-y$ exactly.

Example 10.4 (circles through two fixed points). For the family of circles through the fixed points $(a,0)$ and $(-a,0)$, the centre must lie on the $y$-axis: writing the centre as $(0,b)$ with $b$ arbitrary gives $x^2+(y-b)^2=a^2+b^2$. Differentiating once: $2x+2(y-b)y'=0\Rightarrow b=y+\dfrac{x}{y'}$. Substituting back and simplifying yields the first-order equation

Example 10.5 (parabola family). For $y^2=4ax$ (one constant $a$), differentiating gives $2y y'=4a\Rightarrow a=\dfrac{yy'}{2}$; substituting back and simplifying gives the first-order equation

$\dfrac{dy}{dx}=\dfrac{2y}{x}.$ …