Find the general and singular solutions (if any) of…

Find the general and singular solutions (if any) of $y^2\big((y')^2+1\big)=1$.

Step 1: Separate variables

$(y')^2=\dfrac{1-y^2}{y^2}\Rightarrow y'=\pm\dfrac{\sqrt{1-y^2}}{y}\Rightarrow\dfrac{y\,dy}{\sqrt{1-y^2}}=\pm\,dx$. Integrating (using $u=1-y^2$) gives $\mp\sqrt{1-y^2}=x+\text{const}$, i.e. $\sqrt{1-y^2}=\pm(C-x)$.

Step 2: Square to get the implicit general solution

$1-y^2=(x-C)^2$, i.e.

\[(x-C)^2+y^2=1,\]

a family of unit circles centered at $(C,0)$.

Step 3: $p$-discriminant test

With $F(x,y,p)=y^2(p^2+1)-1$: $F_p=2y^2p=0$ gives $y=0$ (rejected: substituting into $F=0$ gives $-1=0$, impossible) or $p=0$. With $p=0$: $y^2=1$, i.e. $y=\pm1$.

Step 4: Confirm $y=\pm1$ are genuine envelopes

Both satisfy the ODE directly ($y'=0$, $y^2(0+1)=y^2=1$). Eliminating $C$ from $G=(x-C)^2+y^2-1=0,\ G_C=-2(x-C)=0$ gives $C=x$, hence $y^2=1$, matching the $p$-discriminant result exactly. Since $y=1$ (or $y=-1$) meets each circle at only a single point rather than coinciding with it, neither line is a member of the circle family for any fixed $C$.