Find the general integral of 2x(1+√(x^2-y))dx-√(x^2-y) dy=0

Find the general integral of $2x\left(1+\sqrt{x^2-y}\right)dx-\sqrt{x^2-y}\,dy=0$.

Explanation of Method

The radical suggests the substitution $u=x^2-y$, which eliminates $x$ and $y$ separately in favor of $u$ and $x$, reducing the equation to a separable one.

Step 1: Substitute $u=x^2-y$

Then $dy=2x\,dx-du$. Substituting: $2x(1+\sqrt{u})\,dx-\sqrt{u}(2x\,dx-du)=0$, and the $2x\sqrt{u}\,dx$ terms cancel, leaving

\[2x\,dx+\sqrt{u}\,du=0.\]

Step 2: Integrate

$x^2+\dfrac23u^{3/2}=C_1$. Multiplying by $\tfrac32$ and relabeling the constant:

Step 3: Back-substitute

With $u=x^2-y$:

Result

Final answer:

\[\boxed{\,(x^2-y)^{3/2}+\frac32x^2=C,\qquad C\in\mathbb{R}}\]