Find the general integral of the total differential equation…

Find the general integral of the total differential equation $\dfrac{2x}{y^3}dx+\dfrac{y^2-3x^2}{y^4}dy=0$.

Explanation of Method

Check exactness with $M=2x/y^3$, $N=(y^2-3x^2)/y^4$, and integrate.

Step 1: Check exactness

$M_y=-\dfrac{6x}{y^4}=N_x$, so the equation is exact.

Step 2: Find the potential function

$F=\int\dfrac{2x}{y^3}dx=\dfrac{x^2}{y^3}+g(y)$. Matching $F_y=-\dfrac{3x^2}{y^4}+g'(y)=\dfrac{y^2-3x^2}{y^4}$ gives $g'(y)=\dfrac{1}{y^2}$, so $g(y)=-\dfrac1y$.

Result

Final answer:

\[\boxed{\,x^2-y^2=Cy^3,\qquad C\in\mathbb{R}}\]