Solve the differential equation 2xy(y′)² + 2(x² + y²)y′ + 2xy = 0, first solving for y′ and then integrating each resulting first-order equation.

Step-by-step solution

1. 1

   Recognise the equation as algebraic in $y'$, not differential yet. Divide through by $2$ and write $p=y'$:

   $xy\,p^{2}+(x^{2}+y^{2})\,p+xy=0.$

   Nothing here involves a derivative of $p$, so this is an ordinary quadratic equation whose "coefficients" happen to depend on $x$ and $y$. Solving it turns one messy equation into two clean separable ones.

2. 2

   Compute the discriminant and notice it is a perfect square.

   $(x^{2}+y^{2})^{2}-4(xy)(xy)=x^{4}+2x^{2}y^{2}+y^{4}-4x^{2}y^{2}=(x^{2}-y^{2})^{2}.$

   This is why the problem is solvable in closed form: the square root comes out rational, $\sqrt{(x^{2}-y^{2})^{2}}=|x^{2}-y^{2}|$, and either sign is absorbed by the $\pm$ in the quadratic formula.

3. 3

   Read off the two branches.

   $p=\frac{-(x^{2}+y^{2})\pm(x^{2}-y^{2})}{2xy}.$

   The $+$ branch gives $p=\dfrac{-2y^{2}}{2xy}=-\dfrac{y}{x}$; the $-$ branch gives $p=\dfrac{-2x^{2}}{2xy}=-\dfrac{x}{y}$. Each is separable.

4. 4

   Integrate the first branch $y'=-y/x$. Separating,

   $\frac{dy}{y}=-\frac{dx}{x}\;\Longrightarrow\;\ln|y|=-\ln|x|+c\;\Longrightarrow\;\ln|xy|=c,$

   so $xy=C_{1}$: a family of rectangular hyperbolas.

5. 5

   Integrate the second branch $y'=-x/y$. Separating,

   $y\,dy=-x\,dx\;\Longrightarrow\;\frac{y^{2}}{2}=-\frac{x^{2}}{2}+c\;\Longrightarrow\;x^{2}+y^{2}=C_{2}:$

   circles centred at the origin.

6. 6

   Check both families in the original equation. For $xy=C_1$, $y'=-y/x$; substituting, $xy\frac{y^{2}}{x^{2}}-(x^{2}+y^{2})\frac{y}{x}+xy=\frac{y^{3}}{x}-xy-\frac{y^{3}}{x}+xy=0$. For $x^{2}+y^{2}=C_2$, $y'=-x/y$; substituting, $xy\frac{x^{2}}{y^{2}}-(x^{2}+y^{2})\frac{x}{y}+xy=\frac{x^{3}}{y}-\frac{x^{3}}{y}-xy+xy=0$. Both vanish identically. Geometrically the two families are not orthogonal: the slopes $-y/x$ and $-x/y$ multiply to $+1$, not $-1$. They are the two roots of the same quadratic, and they coincide exactly where $|x|=|y|$ — there the discriminant vanishes, the two branches merge, and the hyperbola is tangent to the circle. (The orthogonal trajectories of $xy=C$ are the hyperbolas $x^2-y^2=C$.)