# 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 $$F(x,y) = c$$

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

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 :

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

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

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.

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