Solve the differential equation y=1+(y')^2

Solve the differential equation $y=1+(y')^2$.

Step 1: Solve for $y'$ and separate

\[(y')^2=y-1\ \Longrightarrow\ \frac{dy}{\sqrt{y-1}}=\pm dx,\qquad y\ge1.\]

Step 2: Integrate

\[\int\frac{dy}{\sqrt{y-1}}=\pm\int dx \quad\Longrightarrow\quad 2\sqrt{y-1}=\pm x+C.\]

Squaring and solving for $y$ (relabelling the constant as $c$):

\[\boxed{\,y=1+\frac{(x-c)^2}{4}.}\]

Step 3: Singular solution

Setting $y'=0$ in the original equation forces $y=1$, and indeed $1+(0)^2=1$ holds, so $y\equiv1$ also solves the equation. This is the envelope of the family of upward parabolas with vertices $(c,1)$, tangent to $y=1$ at each vertex.