Solve $y’ = \frac{2-xy^3}{3x^2y^2}$ using integration factor

Question

Show that the equation:

$$ y’ = \frac{2-xy^3}{3x^2y^2} $$

Has an integration factor that depends on $x$ And solve it that way.

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Already we got to:

$$ y’ + \frac{xy^3}{3x^2y^2} = \frac{2}{3x^2y^2} $$

Therefore:

$$ y’ + \frac{1}{3x}y = \frac{2}{3x^2}y^{-2} $$

But, in order to get an integration factor, shouldn’t we have a linear equation? Of the form:

$$ y‘ + p(x)y = g(x) $$

That way getting the integration factor:

$$ \mu = ke^{\int p(x)}, k \in R $$

But what we have is a non-linear equation, so how could an integration factor exists?

Thanks.

Answer 1

Make life easier letting $$y=\frac z {\sqrt[3]x}\implies 3 x z'(x)=\frac{2}{z(x)^2}$$ which is simpler

Answer 2

$$y’ = \frac{2-xy^3}{3x^2y^2}$$ Is linear if you substitute $w=y^3$ and $w'=3y^2y'$ $$3y^2y’ = \frac{2-xy^3}{x^2}$$ $$w' = \frac{2-xw}{x^2}$$ $$xw'+w = \dfrac 2x$$ $$(wx)'=\dfrac 2x$$ Integrate.

Answer 3

You can manipulate the expression and obtain $(xy^3)' = 2/x$

Answer 4

$$ y’=\frac{2-xy^3}{3x^2y^2}\tag1 $$ Multiply $(1)$ by $3y^2g$ and shuffle some terms: $$ \left(y^3\right)'g+y^3\frac{g}x=\frac{2g}{x^2}\tag2 $$ We want the integrating factor to satisfy $$ g'=\frac{g}x\tag3 $$ So we can use $g(x)=x$. Then $(2)$ becomes $$ \left(xy^3\right)'=\frac2x\tag4 $$ which is much easier to solve.