Solve the differential equation yy'-x=0

Solve the differential equation $yy'-x=0$.

Explanation of Method

The equation is already separable in $y\,dy$ and $x\,dx$; integrate both sides directly.

Step 1: Separate variables

\[yy'=x\quad\Longrightarrow\quad y\,dy=x\,dx.\]

Step 2: Integrate

\[\int y\,dy=\int x\,dx \quad\Longrightarrow\quad \frac{y^2}{2}=\frac{x^2}{2}+C \quad\Longrightarrow\quad y^2-x^2=c.\]

Step 3: Domain remarks

If $c\ge0$, $y=\pm\sqrt{x^2+c}$ is defined for all $x$; if $c<0$, the domain is $|x|\ge\sqrt{-c}$. The family $y^2-x^2=c$ consists of hyperbolas ($c\neq0$) or the lines $y=\pm x$ ($c=0$), and is the family of orthogonal trajectories to $xy=k$.

Result

Final answer:

\[\boxed{\,y^2-x^2=c,\qquad c\in\mathbb{R}.}\]