Find the general solution of (1+x^2) y'' + 1 + (y')^2 = 0

Find the general solution of $(1+x^2)\,y'' + 1 + (y')^2 = 0$.

Step 1: Substitution

Let $p = y'$, so $p' = y''$. The equation becomes: \[(1+x^2)\,p' + 1 + p^2 = 0. \tag{1}\]

This is a first-order separable ODE in $p(x)$.

Step 2: Separate variables

Rearrange (1): $(1+x^2)\,p' = -(1+p^2),\qquad \frac{dp}{1+p^2} = -\frac{dx}{1+x^2}.$

Step 3: Integrate both sides

\[ \int \frac{dp}{1+p^2} = -\int \frac{dx}{1+x^2},\qquad \arctan p = -\arctan x + C_1. \]

Step 4: Solve for $p$

\[ p = \tan(C_1 - \arctan x). \]

Using the tangent subtraction formula with $\tan C_1 = c$ (renaming the constant): $p = \tan\!\left(\arctan c - \arctan x\right) = \frac{c - x}{1 + cx}.$

Thus: \[y' = \frac{C_1 - x}{1 + C_1 x}. \tag{2}\]

Step 5: Second integration

Write $C_1 - x = -\dfrac{1}{C_1}\bigl(C_1 x + 1\bigr) + \dfrac{1+C_1^2}{C_1}$, so: \[\frac{C_1 - x}{1+C_1 x} = -\frac{1}{C_1} + \frac{(1+C_1^2)/C_1}{1+C_1 x}.\]

Integrate term by term: $y = -\frac{x}{C_1} + \frac{1+C_1^2}{C_1^2}\ln|1+C_1 x| + C_2.$

For the special case $C_1 = 0$: $y' = -x$, giving $y = -\dfrac{x^2}{2} + C_2$.

Step 6: General solution

For $C_1 \neq 0$ the solution is recorded in the result below. For $C_1 = 0$:

\[ y = -\frac{x^2}{2} + C_2, \] where $C_1$ and $C_2$ are arbitrary constants.

Step 7: Verification

Differentiate $y' = \dfrac{C_1 - x}{1 + C_1 x}$: $y'' = \frac{-(1+C_1 x) - C_1(C_1-x)}{(1+C_1 x)^2} = \frac{-(1+C_1^2)}{(1+C_1 x)^2}.$

Check $(1+x^2)y'' + 1 + (y')^2$: \[(1+x^2)\cdot\frac{-(1+C_1^2)}{(1+C_1 x)^2} + 1 + \frac{(C_1-x)^2}{(1+C_1 x)^2} = \frac{-(1+x^2)(1+C_1^2) + (C_1-x)^2}{(1+C_1 x)^2} + 1.\]

Expand the numerator: $-(1+C_1^2+x^2+C_1^2 x^2) + (C_1^2 - 2C_1 x + x^2) = -1 - 2C_1 x - C_1^2 x^2 = -(1+C_1 x)^2.$

Thus: $\frac{-(1+C_1 x)^2}{(1+C_1 x)^2} + 1 = -1 + 1 = 0.$ $\checkmark$