Problem 118
Find the general integral of the total differential equation…
Find the general integral of the total differential equation $(2x^3-xy^2)dx+(2y^3-x^2y)dy=0$.
Explanation of Method
Check exactness with $M=2x^3-xy^2$, $N=2y^3-x^2y$, and integrate.
Step 1: Check exactness
$M_y=-2xy=N_x$, so the equation is exact.
Step 2: Find the potential function
$F=\int(2x^3-xy^2)dx=\dfrac{x^4}{2}-\dfrac{x^2y^2}{2}+g(y)$. Matching $F_y=-x^2y+g'(y)=2y^3-x^2y$ gives $g'(y)=2y^3$, so $g(y)=\dfrac{y^4}{2}$.
Result
Final answer:
\[\boxed{\,x^4-x^2y^2+y^4=C,\qquad C\in\mathbb{R}}\]