Problem 116
Find the general integral of the total differential equation…
Find the general integral of the total differential equation $(x^3+3xy^2)dx+(y^3+3x^2y)dy=0$.
Explanation of Method
Check exactness with $M=x^3+3xy^2$, $N=y^3+3x^2y$, and integrate.
Step 1: Check exactness
$M_y=6xy=N_x$, so the equation is exact.
Step 2: Find the potential function
$F=\int(x^3+3xy^2)dx=\dfrac{x^4}{4}+\dfrac{3x^2y^2}{2}+g(y)$. Matching $F_y=3x^2y+g'(y)=y^3+3x^2y$ gives $g'(y)=y^3$, so $g(y)=\dfrac{y^4}{4}$.
Result
Final answer:
\[\boxed{\,x^4+6x^2y^2+y^4=C,\qquad C\in\mathbb{R}}\]