With that solution the equation you ask for has to be :
y' + (4/x) y = x³ y²
This is a Bernoulli equation.
Substitute u = y⁻¹ ==> y = u⁻¹ ==> dy/dx = dy/du du/dx = - u⁻² du/dx .
Then the equation becomes
- u⁻² du/dx + (4/x) u⁻¹ = x³ u⁻²
Multiplying the equation by - u² yields
u' - (4/x) u = - x³
Integrating factor :
μ = e^∫(-4/x) dx = e^(-4ln x) = x⁻⁴
Multiplying the equation by μ:
u' x⁻⁴ - 4x⁻⁵ u = - x⁻¹
Notice the LHS is the differential d(u x⁻⁴)/dx
d(u x⁻⁴)/dx = - x⁻¹
d(u x⁻⁴) = - x⁻¹ dx
Integrating :
u x⁻⁴ = - ln x + C
u = x⁴ ( C - ln x )
y = 1 / ( x⁴ ( C - ln x ) )
http://www.wolframalpha.com/input/?i=y'+++(4/x)+y+=+x³+y² Edit.
@Rapidfire:I am totally overwhelmed by your prowess and perception ! Wow !
On the other hand, if you were able to read a little further, you could see the answer he was given using an accurate program - which is the result of plugging in the First Order Equation.
Plugging into Mathematica the Second Order Equation would have not yielded that explicit form he has received as result.
In conclusion, there is a typo : the equation should be of First Order and the answer is correct ;
If you need more explanation, ask and maybe I'll even draw for you.
And now for something completely different : keep up the good job and try to think harder !
Please, keep your thumbs to yourself, you deserve them more than others do.