Differentiate $(x^3 + xy^2 + a^2y) dx + (y^3 + yx^2 – a^2x) dy =0$

Question

Differentiate $$(x^3 + xy^2 + a^2y) dx + (y^3 + yx^2 – a^2x) dy =0$$

Is the above equation an exact differential equation? because it doesn't follow the necessary condition of exact differential equation

But if we divide the equation by $(x^2+y^2)$ it follows the necessary condition. Can you please explain the reason behind this?

Answer 1

$$(x^3 + xy^2 + a^2y) dx + (y^3 + yx^2 – a^2x) dy =0$$ Divide by $x^2+y^2$: $$(x + a^2\dfrac y{x^2+y^2}) dx + (y – a^2\dfrac x{x^2+y^2}) dy =0$$ $$xdx+ydy -a^2(\dfrac {xdy-ydx}{x^2+y^2}) =0$$ It's now exact: $$\frac 12 d(x^2+y^2) -a^2d(\arctan (y/x))=0$$ Integrate: $$x^2+y^2 -2a^2\arctan (y/x)=C$$

Answer 2

It is not an exact d.e. By the definition, the differential equation $$I(x,y) \mathrm{d}x + J(x,y) \mathrm{d}y = 0$$ is called exact, if there exists a continuously differentiable function $F$, called the potential function, so that $$\frac{\partial F}{\partial x} = I \quad \text{and} \quad \frac{\partial F}{\partial y} = J.$$ You can double check by integration that you do not get such an $F$.

Moreover, dividing by $(x^2 + y^2)$ also does not make the equation exact, as:

$$\int \frac{x^3 + xy^2 + a^2y}{x^2 + y^2}\mathrm{d}x = \frac{x^2}{2} - a^2\arctan\left(\frac{y}{x}\right) + c_1(y)$$ $$\int \frac{x^3 + xy^2 - a^2x}{x^2 + y^2}\mathrm{d}x = \frac{y^2}{2} + a^2\arctan\left(\frac{x}{y}\right) + c_2(x)$$ You could say that $c_1(y) = \dfrac{y^2}{2}$ and $c_2(x) = \dfrac{x^2}{2}$ but the $\arctan$ terms mess things up.