Solve the equation:
$y(x^3 – y) dx – x(x^3 + y) dy = 0$
Solution:
Regroup the terms:
$x^3(y dx – x dy) = y(y dx + x dy)$
Then, substitute $d(xy)$ to $(y dx + x dy)$ since it is the exact differential form of $(y dx + x dy)$:
$x^3(y dx – x dy) = yd(xy)$
However, according to the book (Rainville, E. D., Bedient, P. E. (1989). Elementary Differential Equation. 7th Edition. New York: MacMillan Publishing Company), there are two exact differential form on the equation. The terms of the equation must be divided by $y^2$ to have an exact differential form, $d(\frac{x}{y})$, for $(y dx – x dy)$.
I don't know how $y^2$ obtained from the equation, $y(x^3 – y) dx – x(x^3 + y) dy = 0$, which can be used to have an exact differential equation for $(y dx – x dy)$. Any help is highly appreciated. Thank You.
Best Answer
$$x^3(y dx - x dy) = yd(xy)$$ Divide by $y^2$ both sides. You have the right to do so. Multiply or divide both sides of an equation by the same quantity. The equality still holds. $$x^3d\frac x y = \frac {d(xy)}{y}$$ Divide by $x^2y$ : $$ \frac x y d\frac x y = \frac {d(xy)}{(xy)^2}$$ Integrate. $$ \frac 12 \frac {x^2 }{y^2} =- \frac {1}{xy}+C$$ $$ {x^3 } +{2y}+C_1{xy^2}=0$$
Edit1
$$x^3(y dx - x dy) = yd(xy)$$ You can also choose to divide by $x^2$ $$-x^3d\frac y x = \frac {yd(xy)}{x^2}$$ Divide by $y^3$ : $$-\frac {x^3}{y^3}d\frac y x = \frac {d(xy)}{x^2y^2}$$ Now you can integrate.