Find the differential equation whose general solution is…

Find the differential equation whose general solution is \[ (x-a)^2 + by^2 = 1, \] where $a$ and $b$ are arbitrary constants.

Step 1: Differentiate the general solution

Differentiating $(x-a)^2 + by^2 = 1$ with respect to $x$ gives

\[2(x-a) + 2by\,y' = 0,\]

which simplifies to

\[(x-a) + by\,y' = 0. \qquad (1)\]

Differentiating (1) again: $1 + b(y')^2 + by\,y'' = 0$, so \[1 + b\bigl[(y')^2 + y\,y''\bigr] = 0. \qquad (2)\]

Step 2: Eliminate the arbitrary constants

From (2), \[b = -\frac{1}{(y')^2 + y\,y''}. \qquad (3)\]

From (1), \[x - a = -by\,y' = \frac{y\,y'}{(y')^2 + y\,y''}. \qquad (4)\]

Step 3: Substitute into the original equation

Substitute (3) and (4) into $(x-a)^2 + by^2 = 1$: $\left(\frac{y\,y'}{(y')^2+y\,y''}\right)^2 - \frac{y^2}{(y')^2+y\,y''} = 1.$

Multiply through by $\bigl[(y')^2+y\,y''\bigr]^2$: $(y\,y')^2 - y^2\bigl[(y')^2+y\,y''\bigr] = \bigl[(y')^2+y\,y''\bigr]^2.$

The left-hand side simplifies to $-y^3 y''$, so

\[-y^3 y'' = \bigl[(y')^2+y\,y''\bigr]^2.\]

Step 4: The differential equation

Equivalently, $(y')^2 + y\,y'' = -\sqrt{-y^3 y''}$, or, in implicit form, $\bigl[(y')^2 + y\,y''\bigr]^2 + y^3\,y'' = 0.$

Step 5: Verification

Differentiating $(x-a)^2 + by^2 = 1$ twice recovers (1) and (2), from which $b$ and $(x-a)$ are given by (3) and (4). Substituting back reproduces the boxed equation. For $b>0$ the curves are ellipses centered at $(a,0)$; for $b<0$ they are hyperbolas. When $b>0$, real $y$ requires $|x-a|\le 1$.