Solve the equation (*): y y'' + (y')^2 - (y y')/(√(1+x^2)) =…

Solve the equation $(*)\!: \qquad y\,y'' + (y')^2 - \frac{y\,y'}{\sqrt{1+x^2}} = 0$: 1° by finding a multiplier that turns the left side into an exact derivative; 2° by the substitution $y'=yz(x)$. Show the two routes agree.

Step 1: 1° — Recognizing an exact derivative

The key observation is the product-rule identity

\[ \frac{d}{dx}(y\,y') = y\,y'' + (y')^2, \]

which is exactly the first two terms of $(*)$. So $(*)$ can be rewritten, with $v:=y\,y'$, as the first-order linear equation

\[ v' - \frac{v}{\sqrt{1+x^2}} = 0. \]

Step 2: Integrating factor for $v$

The integrating factor is

\[ \mu(x) = \exp\!\left(-\int\frac{dx}{\sqrt{1+x^2}}\right) = \exp\bigl(-\ln(x+\sqrt{1+x^2})\bigr) = \frac{1}{x+\sqrt{1+x^2}}. \]

Multiplying through by $\mu$ collapses the equation to an exact derivative, $\frac{d}{dx}\!\left[\frac{y\,y'}{x+\sqrt{1+x^2}}\right] = 0$.

so the bracket is a constant. Writing that constant as $C_1/2$ (a convenient relabeling for the next step), $y\,y' = \frac{C_1}{2}\Bigl(x+\sqrt{1+x^2}\Bigr)$.

Step 3: Integrating once more

Since $y\,y'=\tfrac12(y^2)'$, this reads $(y^2)' = C_1\bigl(x+\sqrt{1+x^2}\bigr)$. Using

\[ \int\sqrt{1+x^2}\,dx = \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\!\bigl(x+\sqrt{1+x^2}\bigr) + \mathrm{const}, \]

integration in $x$ gives

\[ y^2 = C_1\left[\frac{x^2}{2} + \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\!\bigl(x+\sqrt{1+x^2}\bigr)\right] + C_2. \tag{A} \]

Step 4: 2° — The substitution $y'=y\,z(x)$

With $y'=yz$, differentiating gives $y''=y'z+yz'=yz^2+yz'$. Substituting both into $(*)$ and dividing throughout by $y^2$ ($y\not\equiv0$): $2z^2 + z' - \frac{z}{\sqrt{1+x^2}} = 0$.

a Bernoulli equation in $z$, of degree $n=2$.

Step 5: Bernoulli reduction: $w=1/z$

Dividing by $z^2$ and setting $w=1/z$ (so $w'=-z'/z^2$) turns the equation into the linear one

\[ w' + \frac{w}{\sqrt{1+x^2}} = 2. \]

Its integrating factor is $\exp\!\left(\int dx/\sqrt{1+x^2}\right)=x+\sqrt{1+x^2}$, so

\[ \frac{d}{dx}\Bigl[w\bigl(x+\sqrt{1+x^2}\bigr)\Bigr] = 2\bigl(x+\sqrt{1+x^2}\bigr). \]

Integrating the right side (same antiderivative as in Step 3) gives

\[ w\bigl(x+\sqrt{1+x^2}\bigr) = x^2 + x\sqrt{1+x^2} + \ln\!\bigl(x+\sqrt{1+x^2}\bigr) + \widetilde C. \]

Step 6: Recovering $y$ from $w$

Since $y'=yz=y/w$, we have $dy/y = dx/w$. Introduce

\[ \Phi(x) := \frac{x^2}{2} + \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\!\bigl(x+\sqrt{1+x^2}\bigr), \]

which is exactly the bracket in $(A)$; differentiating term by term confirms $\Phi'(x)=x+\sqrt{1+x^2}$, and the boxed denominator above is precisely $2\Phi(x)+\widetilde C$. Hence

\[ \frac{dy}{y} = \frac{\Phi'(x)\,dx}{2\Phi(x)+\widetilde C} = \frac12\,\frac{d\bigl(2\Phi+\widetilde C\bigr)}{2\Phi+\widetilde C}. \]

Integrating, $\ln|y| = \tfrac12\ln|2\Phi+\widetilde C| + \mathrm{const}$, i.e. $y^2 = A\bigl(2\Phi(x)+\widetilde C\bigr)$ for constants $A,\widetilde C$. Renaming $2A\to C_1$, $A\widetilde C\to C_2$, this is

\[ y^2 = C_1\left[\frac{x^2}{2} + \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\!\bigl(x+\sqrt{1+x^2}\bigr)\right] + C_2, \]

identical to $(A)$.

Step 7: Verification

Both routes produce the same one-parameter-squared family; write it as $y^2=C_1\Phi(x)+C_2$, so $2y\,y'=C_1\Phi'(x)=C_1(x+\sqrt{1+x^2})$, i.e. $y\,y'=\tfrac{C_1}{2}(x+\sqrt{1+x^2})$, exactly the relation from Step 2. Since $\Phi'(x)=x+\sqrt{1+x^2}$, differentiating gives $\Phi''(x)=1+x/\sqrt{1+x^2}=(x+\sqrt{1+x^2})/\sqrt{1+x^2}=\Phi'(x)/\sqrt{1+x^2}$. Differentiating $y\,y'=\tfrac{C_1}{2}\Phi'(x)$ once more, using $y\,y''+(y')^2=(y\,y')'$,

\[ y\,y''+(y')^2 = \frac{C_1}{2}\Phi''(x) = \frac{C_1}{2}\cdot\frac{\Phi'(x)}{\sqrt{1+x^2}} = \frac{y\,y'}{\sqrt{1+x^2}}, \]

which is exactly $(*)$. The integrating-factor method and the $y'=yz$ substitution therefore agree exactly.

\[y^2 = C_1\left[\frac{x^2}{2} + \frac{x}{2}\sqrt{1+x^2} + \frac12\ln\!\bigl(x+\sqrt{1+x^2}\bigr)\right] + C_2\]