Find the general integral of yy'' = y'^2 + y'√(y^2+y'^2)

Find the general integral of $yy'' = y'^2 + y'\sqrt{y^2+y'^2}$.

Step 1: The key insight: an angle-ratio substitution

The presence of $\sqrt{y^2+y'^2}$ alongside $y$ and $y'$ suggests treating $t=y'/y$ as the new unknown, i.e.\ writing $p=y'=yt$. Then

\[ y'' = y't + yt' = yt^2 + yt', \qquad \sqrt{y^2+y'^2} = y\sqrt{1+t^2}\ \ (y>0). \]

Step 2: Substitute and simplify

Substituting into $yy'' = y'^2 + y'\sqrt{y^2+y'^2}$: $y^2(t^2+t') = y^2t^2 + y^2t\sqrt{1+t^2}.$

Dividing by $y^2$, the $t^2$ terms cancel, leaving a first-order equation purely in $t$: $t' = t\sqrt{1+t^2}.$

Step 3: Separate and integrate

Separating variables: $\frac{dt}{t\sqrt{1+t^2}} = dx.$

Substitute $u=\sqrt{1+t^2}$, so $u^2=1+t^2$ and $t\,dt = u\,du$: \[\int\frac{dt}{t\sqrt{1+t^2}} = \int\frac{du}{u^2-1} = \frac12\ln\left|\frac{u-1}{u+1}\right|.\] So $\frac12\ln\left|\frac{u-1}{u+1}\right| = x + C_1.$

Step 4: Solve for $u$, then for $t$

Writing $C_1=\ln A$, $\dfrac{u-1}{u+1}=A^2e^{2x}$, so

\[ u = \frac{1+A^2e^{2x}}{1-A^2e^{2x}}. \]

Then $t^2=u^2-1$, and using a difference of squares on the numerator, \[t^2 = \frac{4A^2e^{2x}}{(1-A^2e^{2x})^2} \implies t = \frac{2Ae^x}{1-A^2e^{2x}}\]

(positive branch).

Step 5: Return to $y$

Recall $t=y'/y$, so

\[ \frac{y'}{y} = \frac{2Ae^x}{1-A^2e^{2x}}. \]

Substitute $\xi=Ae^x$, so $d\xi = Ae^x\,dx$: \[\int \frac{2Ae^x\,dx}{1-A^2e^{2x}} = \int \frac{2\,d\xi}{1-\xi^2} = \ln\left|\frac{1+\xi}{1-\xi}\right|.\]

Integrating $y'/y$ on the left gives $\ln y$, so

\[ \ln y = \ln\left|\frac{1+Ae^x}{1-Ae^x}\right| + \ln B. \] $y = B\cdot\frac{1+Ae^x$.

where $A$ and $B$ are arbitrary constants.

Step 6: Verification

Write $\xi=Ae^x$, so $y=B(1+\xi)/(1-\xi)$ with $\xi'=\xi$. Differentiating, $y' = \dfrac{2B\xi}{(1-\xi)^2}$ and $y'' = \dfrac{2B\xi(1+\xi)}{(1-\xi)^3}$. Substituting into $yy''-y'^2-y'\sqrt{y^2+y'^2}$, every term carries a common factor of $\dfrac{4B^2\xi}{(1-\xi)^4}$, and what remains reduces to the algebraic identity

\[ (1+\xi)^2 - 2\xi - (1+\xi^2) = 0, \]

a clean perfect-square-type cancellation, confirming the boxed solution satisfies the original equation identically. $\checkmark$