Find the general solution of y''l(1 + y y'r) = y'l(1 +…

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

Step 1: Substitution

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

Step 2: Recognize an arctangent derivative

Divide (1) by $1+p^2$: $\frac{p'}{1+p^2}(1+yp) = p.$

The left factor is a standard derivative, $\dfrac{p'}{1+p^2} = \dfrac{d}{dx}(\arctan p)$, so: \[\frac{d(\arctan p)}{dx} = \frac{p}{1+yp}. \tag{2}\]

Step 3: Change of variable to $\theta = \arctan p$

Since $dy = p\,dx$, equation (2) gives $(1+yp)\,d(\arctan p) = p\,dx = dy$, i.e.

\[ d(\arctan p) = \frac{dy}{1+yp}. \tag{3} \]

Let $\theta = \arctan p$, so $p = \tan\theta$. Then (3) reads $d\theta = \dfrac{dy}{1+y\tan\theta}$, i.e.

\[ \frac{dy}{d\theta} - y\tan\theta = 1. \tag{4} \]

This is a first-order linear ODE in $y(\theta)$.

Step 4: Solve equation (4)

The integrating factor is $\mu = e^{-\int\tan\theta\,d\theta} = \cos\theta$. Multiply (4) by $\cos\theta$: $\frac{d}{d\theta}(y\cos\theta) = \cos\theta.$

Integrate: \[y\cos\theta = \sin\theta + C_1,\qquad y = \tan\theta + C_1\sec\theta. \tag{5}\]

Step 5: Find $x(\theta)$

Since $dx = dy/p = dy/\tan\theta$, and from (5) $dy = (\sec^2\theta+C_1\sec\theta\tan\theta)\,d\theta$: $dx = \left(\frac{\sec^2\theta}{\tan\theta} + C_1\sec\theta\right)d\theta.$

Integrate (first term with $u=\tan\theta$, second is standard): \[x = \ln|\tan\theta| + C_1\ln|\sec\theta+\tan\theta| + C_2. \tag{6}\]

Step 6: General solution

The general solution is parametric with parameter $\theta = \arctan y'$, or equivalently in terms of $p=y'$: where $C_1$ and $C_2$ are arbitrary constants.

Step 7: Verification

From $y'=\tan\theta$, differentiate: $y'' = \sec^2\theta\cdot\dfrac{d\theta}{dx} = \dfrac{\sec^2\theta}{dx/d\theta} = \dfrac{\sec^2\theta\tan\theta}{\sec^2\theta+C_1\sec\theta\tan\theta}$.

Since $1+yy' = 1+(\tan\theta+C_1\sec\theta)\tan\theta = \sec^2\theta+C_1\sec\theta\tan\theta$: \[y''(1+yy') = \sec^2\theta\tan\theta = y'(1+y'^2) = \tan\theta\cdot\sec^2\theta.\ \checkmark\]