Solve the differential equation xy(y')^2+(x^2+y^2)y'+xy=0
Solve the differential equation $xy(y')^2+(x^2+y^2)y'+xy=0$.
Explanation of Method
Treat the equation as quadratic in $y'$ and apply the quadratic formula; the discriminant simplifies to a perfect square, splitting the problem into two separable first-order equations.
Step 1: Solve the quadratic in $y'$
Dividing by $xy$ gives $(y')^2+\dfrac{x^2+y^2}{xy}y'+1=0$. The discriminant simplifies:
\[\Delta=\left(\frac{x^2+y^2}{xy}\right)^2-4=\frac{(x^2-y^2)^2}{x^2y^2},\]
so the quadratic formula yields the two cases $y'=-\dfrac{y}{x}$ and $y'=-\dfrac{x}{y}$.
Step 2: Solve each case
Case 1: $\dfrac{dy}{y}=-\dfrac{dx}{x}$ integrates to $\ln|xy|=$ const, i.e. $xy=C_1$.
Case 2: $y\,dy=-x\,dx$ integrates to $x^2+y^2=C_2$.
Step 3: Verify
Substituting $y'=-y/x$ and $y'=-x/y$ back into the original equation confirms both families satisfy it identically, so the general solution consists of two families of curves — rectangular hyperbolas and circles — that are orthogonal trajectories of each other.
Result
Final answer:
\[\boxed{\,xy=C_1\quad\text{(hyperbolas)}\qquad\text{or}\qquad x^2+y^2=C_2\quad\text{(circles)}.}\]