A difficult differential equation $ y(2x^4+y)\frac{dy}{dx} = (1-4xy^2)x^2$ How to solve the following differential equation?
$$ y(2x^4+y)\dfrac{dy}{dx} = (1-4xy^2)x^2$$
No clue as to how to even begin. Hints?
 A: Hint: it's an exact differential equation.
A: $$y(2x^4+y)\frac{dy}{dx}=(1-4xy^2)x^2$$
$$2x^4y\frac{dy}{dx}+y^2\frac{dy}{dx}=x^2-4x^3 y^2$$
$$2x^4y\frac{dy}{dx}+4x^3y^2+y^2\frac{dy}{dx}=x^2$$
$$\frac{d}{dx}(x^4 y^2)+y^2\frac{dy}{dx}=x^2$$
$$x^4y^2+\frac{y^3}{3}=\frac{x^3}{3}+C$$
A: Hint:
Apply $M(x,y)=\frac{du}{dx}$   and    $ N(x,y)=\frac{du}{dy}$
A: Since this is an exact differential equation, it is written in the form $N(x, y)+y'M(x, y)=0$ Rewriting the equation, we get$$(4x^3y^2-1)+\frac{dy}{dx}(2x^4y+y^2)=0$$From here, we can see clearly that $N(x, y)=4x^3y^2-1$ and $M(x, y)=2x^4y+y^2$. Now let $\Psi_x=N$ and $\Psi_y=M$. Therefore, $$\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\cdot\frac{dy}{dx}=0$$Recall that this is similar to the multivariable chain rule: $$\frac{d\Psi(x, y(x))}{dx}=\frac{\partial\Psi}{\partial x}\cdot\frac{dx}{dx}+\frac{\partial\Psi}{\partial y}\cdot\frac{dy}{dx}=\frac{\partial\Psi}{\partial x}+\frac{\partial\Psi}{\partial y}\cdot\frac{dy}{dx}=0$$This means that $\Psi$ is the solution to the differential equation. $$\Psi=\int\Psi_x\;dx=\int N\;dx\;\;OR\;\;\int\Psi_y\;dy=\int M\;dy$$
$$\int N\;dx=\int4x^3y^2-1\;dx=x^4y^2-x+h(y)+C_1$$$$\int M\;dx=\int2x^4y+y^2\;dy=x^4y^2+\frac{1}{3}y^3+g(x)+C_2$$
Combining, we get$$\Psi=x^4y^2-x+\frac{1}{3}y^3+C$$ Because the derivative of $\Psi$ is $0$, $\Psi$ must evaluate to a constant $$\boxed{x^4y^2-x+\frac{1}{3}y^3=C}$$
