How can I solve $ y dx + (x+x^2y^4)dy = 0$? How can I find a general solution to the following equation:
$$ y dx + (x+x^2y^4)dy = 0$$ 
I thought to solve it by integration factor $\mu = \mu(F(x,y))$ while $∇F= \space(y,x)$ and $F(x,y) = c$
 A: It is :
$$y \mathrm{d}x + (x+x^2y^4)\mathrm{d}y = 0 \Leftrightarrow y + (x+x^2y^4)\frac{\mathrm{d}y}{\mathrm{d}x} = 0 $$
$$\Leftrightarrow$$
$$\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{y(x)}{x(xy^4 + 1)}$$
Now, you can use a trick here to re-write the differential equation in terms of $x(y)$. Note that :
$$\frac{\mathrm{d}y}{\mathrm{d}x} \cdot \frac{\mathrm{d}x}{\mathrm{d}y} = 1 \quad \text{and} \quad \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$$
The equation then becomes :
$$\frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} = - \frac{y}{(y^4x(y) + 1)x(y)} \Leftrightarrow \frac{\mathrm{d}x(y)}{\mathrm{d}y} = -y^3x^2(y) - \frac{x(y)}{y}$$
$$\Leftrightarrow$$
$$-\frac{\frac{\mathrm{d}x(y)}{\mathrm{d}y}}{x^2(y)}-\frac{1}{yx(y)} = y^3$$
Now, let the following be :
$$v(y) = \frac{1}{x(y)} \implies \frac{\mathrm{d}v(y)}{\mathrm{d}y}=-\frac{\frac{\mathrm{d}x(y)}{\mathrm{d}y}}{x^2(y)}$$
Thus the equation becomes :
$$\frac{\mathrm{d}v(y)}{\mathrm{d}y} - \frac{v(y)}{y} = y^3$$
Now, let an integrating factor be :
$$\mu(y) = e^{\int -1/y \mathrm{d}y} = \frac{1}{y}$$
Multiplying both sides with $\mu(y)$, we yield :
$$\frac{\frac{\mathrm{d}v(y)}{\mathrm{d}y}}{y} - \frac{v(y)}{y^2} = y^2 \Leftrightarrow 
 \frac{\frac{\mathrm{d}v(y)}{\mathrm{d}y}}{y} + \frac{\mathrm{d}}{\mathrm{d}y}\bigg(\frac{1}{y}\bigg)v(y) = y^2 $$
$$\Leftrightarrow$$
$$\frac{\mathrm{d}}{\mathrm{d}y}\bigg(\frac{v(y)}{y}\bigg) = y^2 \implies \frac{v(y)}{y} = y^3 + c_1 $$
Now dividing by $\mu(y)$ and substituting back for $v(y)$ while rewritting for $y(x)$, one yields the final result :
$$\boxed{\frac{3}{y^4(x) + c_1y(x)} = x}$$
A: Try to see this equation as
$$
(xdy+ydx)+(xy)^2y^2dy
$$
to identify a useful change in variables resp. the separation into exact differentials.
$$
\frac{d(xy)}{(xy)^2}+y^2dy=0\implies -\frac3{xy}+y^3=C,
$$
which can be solved directly for $x$,
$$
x=\frac{3}{y^4-Cy},
$$
or while the solution for $y$ can only be given implicitly
$$
y^4-Cy=\frac3x.
$$
A: Hint:
Consider that for
$$M=y~~~,~~~N=x+x^2y^4$$
we have
$$\dfrac{M_y-N_x}{Ny-Mx}=\dfrac{-2}{xy}=\dfrac{-2}{z}$$
then $\mu=e^{\int p(z)dz}=\dfrac{1}{x^2y^2}$ is integrating factor.
