Find the general integral of x^2(y+x)y'' - (xy'-y)^2 = 0

Find the general integral of $x^2(y+x)y'' - (xy'-y)^2 = 0$.

Step 1: The homogeneous substitution $y=vx$

Both $(y+x)$ and $(xy'-y)$ scale like $x$ under $y\to vx$, suggesting the standard substitution for homogeneous-type equations. With $y=vx$, $y'=v+xv'$, $y''=2v'+xv''$: \[xy'-y = x(v+xv')-vx = x^2v', \qquad y+x = x(v+1).\]

Substituting into the equation: $x^2\cdot x(v+1)\cdot(2v'+xv'') - x^4v'^2 = 0$.

Dividing through by $x^3$: $x(v+1)v'' + 2(v+1)v' - xv'^2 = 0$.

Step 2: Simplifying with $w=v+1$

Since $w'=v'$ and $w''=v''$, this is just

\[ xww'' + 2ww' - xw'^2 = 0. \]

Dividing by $w^2$ (excluding the degenerate case $w=0$, treated separately below): \[x\frac{w''}{w} + 2\frac{w'}{w} - x\left(\frac{w'}{w}\right)^2 = 0.\]

Step 3: Linearizing with $u=\ln w$

Since $u'=w'/w$ and $u''=w''/w-(w'/w)^2$, i.e. $w''/w = u''+u'^2$, the equation becomes

\[ x(u''+u'^2) + 2u' - xu'^2 = 0 \;\Longrightarrow\; xu'' + 2u' = 0. \]

The nonlinear $u'^2$ terms cancel completely, leaving a linear equation for $u(x)$ — the chain of substitutions $y\to v\to w\to u$ has fully linearized the problem.

Step 4: Solving the linear equation $xu''+2u'=0$

This is again an exact derivative, exactly as in Problem 181: $\dfrac{d}{dx}\bigl(x^2u'\bigr) = x^2u''+2xu' = x(xu''+2u')=0$. So

\[ x^2u' = C_1 \;\Longrightarrow\; u' = \frac{C_1}{x^2}. \]

Integrating once more: $u = -\frac{C_1}{x} + C_2$.

Step 5: Unwinding the substitution chain

Back-substitute $w=e^u$: \[w = e^{-C_1/x+C_2} = Ae^{-C_1/x}, \qquad A=e^{C_2}.\]

Then $v+1=w$ gives $\dfrac{y}{x}+1 = Ae^{-C_1/x}$, i.e.

\[ y+x = Axe^{-C_1/x}. \]

Renaming the constants ($C_1\to-C_1$, $A\to C_2$) to present them in the most natural sign:

Step 6: The singular solution $y=-x$

Step 7: Verification of the general family

With $y+x=C_2xe^{C_1/x}$, differentiate: $y' = C_2e^{C_1/x} + C_2x e^{C_1/x}\!\left(-\dfrac{C_1}{x^2}\right) = C_2e^{C_1/x}\left(1-\dfrac{C_1}{x}\right)$. Then

\[ xy'-y = xy' - (C_2xe^{C_1/x}-x) = x\left[C_2e^{C_1/x}\left(1-\frac{C_1}{x}\right) - C_2e^{C_1/x} + 1\right] - x \]

Simplifying the bracket, $C_2e^{C_1/x}\bigl(1-\tfrac{C_1}{x}-1\bigr) = -\dfrac{C_1C_2}{x}e^{C_1/x}$, so

\[ xy'-y = x\left(-\frac{C_1C_2}{x}e^{C_1/x}+1\right) - x = -C_1C_2\,e^{C_1/x}. \]

Differentiating $y'$ once more (product/chain rule) gives $y'' = C_2e^{C_1/x}\cdot\dfrac{C_1^2}{x^3}$. Substituting both results into the original equation, using $y+x=C_2xe^{C_1/x}$:

\[ x^2(y+x)y'' - (xy'-y)^2 = x^2\cdot C_2xe^{C_1/x}\cdot\frac{C_1^2C_2e^{C_1/x}}{x^3} - \bigl(-C_1C_2e^{C_1/x}\bigr)^2 = C_1^2C_2^2e^{2C_1/x} - C_1^2C_2^2e^{2C_1/x} = 0. \]

The two terms match exactly and cancel, confirming the boxed family satisfies the original nonlinear equation identically for all $C_1,C_2$.\ $\checkmark$