Find the general integral of (x^2+y^2)y'' - yy'^2 + xy'^3 +…

Find the general integral of $(x^2+y^2)y'' - yy'^2 + xy'^3 + xy' - y = 0$.

Step 1: Regroup the lower-order terms

Expand the product $(y'^2+1)(xy'-y)$: $(y'^2+1)(xy'-y) = xy'^3 - yy'^2 + xy' - y$.

which matches the last four terms of the equation exactly. So the equation is equivalent to

\[ (x^2+y^2)y'' + (y'^2+1)(xy'-y) = 0. \]

Step 2: Divide to isolate two recognizable pieces

Dividing through by $(x^2+y^2)(y'^2+1)$ separates the equation into a term built purely from $y'$ and a term built purely from $x,y$: $\frac{y''}{1+y'^2} + \frac{xy'-y}{x^2+y^2} = 0.$

Step 3: The key insight: both terms are derivatives of an angle

The whole point of the regrouping in Step 1 is to expose this: each fraction is exactly the derivative of an arctangent. Recall

\[ \frac{d}{dx}\arctan y' = \frac{y''}{1+y'^2}, \qquad \frac{d}{dx}\arctan\!\frac{y}{x} = \frac{1}{1+(y/x)^2}\cdot\frac{y'x-y}{x^2} = \frac{xy'-y}{x^2+y^2}. \] So with $\varphi = \arctan y'$ (the angle $y'$ makes as a slope) and $\theta=\arctan(y/x)$ (the polar angle of the point $(x,y)$), the equation collapses to

\[ \frac{d\varphi}{dx} + \frac{d\theta}{dx} = 0. \]

Step 4: Integrate directly

Since the sum of the two derivatives vanishes identically, $\varphi+\theta$ is constant: $\arctan y' + \arctan\frac{y}{x} = C_1.$

Step 5: Take the tangent of both sides

Using $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ with $A=\arctan y'$, $B=\arctan(y/x)$: $\frac{y' + y/x}{1 - y'\,y/x} = \tan C_1 =: A$.

a new arbitrary constant. Multiplying numerator and denominator by $x$: $\frac{xy'+y}{x-yy'} = A.$

Step 6: Solve for $y'$: a first-order homogeneous equation

Cross-multiplying, $xy'+y = A(x-yy') = Ax - Ayy'$, so $y'(x+Ay) = Ax-y$, giving

\[ y' = \frac{Ax-y}{x+Ay}, \]

which is homogeneous of degree zero in $x,y$ — the angle-sum trick has traded the original nonlinear second-order equation for a first-order equation solvable by the standard $y=vx$ substitution.

Step 7: Homogeneous substitution $y=vx$

With $y=vx$, $y'=v+xv'$. Dividing numerator and denominator of $\dfrac{Ax-vx}{x+Avx}$ by $x$: \[v + xv' = \frac{A-v}{1+Av} \implies xv' = \frac{A-v}{1+Av} - v = \frac{A-2v-Av^2}{1+Av}.\]

Step 8: Separate variables and integrate

Separating, $\frac{(1+Av)\,dv}{A-2v-Av^2} = \frac{dx}{x}.$

Since $\dfrac{d}{dv}(A-2v-Av^2) = -2-2Av = -2(1+Av)$, the left side is $-\tfrac12$ times the logarithmic derivative of its own denominator: \[-\frac12\ln|A-2v-Av^2| = \ln|x| + \ln|C_2| \implies A-2v-Av^2 = \frac{B}{x^2},\] where $B=1/C_2^2$ is a new arbitrary constant.

Step 9: Return to $x,y$

Substituting back $v=y/x$: \[A - \frac{2y}{x} - \frac{Ay^2}{x^2} = \frac{B}{x^2}.\]

Multiplying through by $x^2$: $Ax^2 - 2xy - Ay^2 = B.$ $A(x^2-y^2) - 2xy = B$.

where $A$ and $B$ are arbitrary constants — a two-parameter family of conics.

Step 10: Verification

Differentiating $A(x^2-y^2)-2xy=B$ implicitly with respect to $x$: \[A(2x-2yy') - 2y - 2xy' = 0 \implies Ax - Ayy' - y - xy' = 0 \implies y'(x+Ay) = Ax-y,\]

which reproduces $y'=\dfrac{Ax-y}{x+Ay}$ exactly — Step 6's relation. Differentiating this once more (implicitly, treating $y'$ as a function of $x$) and substituting back into $(x^2+y^2)y''-yy'^2+xy'^3+xy'-y$ reproduces the original equation identically, confirming the boxed relation is the general integral. $\checkmark$