Question 1188306
<pre>
{{{3x^2y + (y^5 - x^3)*"y'" = 0}}}

{{{3x^2y + (y^5 - x^3)*expr(dy/dx) = 0}}}

{{{3x^2y + y^5*expr(dy/dx) - x^3*expr(dy/dx) = 0}}}

{{{3x^2y*dx + y^5*dy - x^3*dy = 0}}}

The trick here is to get the left side into the expression of either the product expression of 
the differential of uv, which is:

{{{u*dv + v*du}}}

or the quotient expression for the differential of {{{u/v}}}, which is:

{{{(v*du-u*dv)/v^2}}}

We notice by inspection that the third term has x<sup>3</sup> and the first term has 3x<sup>2</sup>dx, which is the 
differential of x<sup>3</sup>.  So we will get those two terms together and the other term
on the other side:

{{{3x^2y*dx - x^3*dy = y^5*dy}}}

The product rule isn't going to work because there's a MINUS sign between the terms, not a 
PLUS sign.  So, we think of trying to make the left side into the form 

{{{(v*du-u*dv)/v^2}}}, the differential form for {{{u/v}}}

The du can be the 3x<sup>2</sup>dx (with u as x<sup>3</sup>), and the v can be the y. So we'll write the y first 
in the first term:  

{{{y*3x^2*dx - x^3*dy = y^5*dy}}}

Now on the left we have the numerator of the differential form for {{{x^3/y}}}.

Now all we need to make the left side into the quotient differential form of {{{x^3/y^""}}} is to 
divide both sides of the equation through by y<sup>2</sup>:

{{{(y*3x^2*dx - x^3*dy)/y^2 = expr(y^5/y^2)*dy}}}

So the integrating factor used here is {{{1/y^2}}}.

{{{(y*3x^2*dx - x^3*dy)/y^2}}}{{{""=""}}}{{{y^3*dy}}}

Now we can integrate both sides of the equation:

{{{int((y*3x^2*dx - x^3*dy)/y^2)}}}{{{""=""}}}{{{int(y^3*dy)}}}

The whole left side integrates all together as the quotient {{{x^3/y}}}, and, the right side 
integrates as {{{y^4/4}}} and we must add an arbitrary constant:

{{{x^3/y}}}{{{""=""}}}{{{y^4/4}}}{{{""+""}}}{{{c[1]}}}

Multiply through by 4y

{{{4x^3}}}{{{""=""}}}{{{y^5}}}{{{""+""}}}{{{Cy}}}

[We just write C because 4 times an arbitrary constant is just another arbitrary
constant].

Edwin</pre>