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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 $F(x,y) = c$

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  • $\begingroup$ Try integrating factor in the form $x^a y^b$ and find $a,b$ that make the equation exact, if they exist $\endgroup$ – Yuriy S Nov 11 '18 at 10:07
  • $\begingroup$ @YuriyS I get this thing: $ \frac{\mu'}{\mu(F(x,y))} = - \frac{2}{yx} $ $\endgroup$ – Software_t Nov 11 '18 at 10:09
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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 :

$$\boxed{\frac{3}{y^4(x) + c_1y(x)} = x}$$

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    $\begingroup$ You could have just used the original equation to write down $dx/dy$ in the first place. $\endgroup$ – J.G. Nov 11 '18 at 11:37
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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.

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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 $$ y^4-Cy=\frac3x. $$

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