Homogeneous equation I am trying to solve the following homogeneous equation:
Thanks for your tips 
$xy^3y′=2(y^4+x^4)$

I think this isHomogeneous of order4

=> $xy^3dy/dx=2(y^4+x^4/1)$
=> $xy^3dy=2(x^4+y^4)dx$
=> $xy^3dy-2(x^4+y^4)dx=0$
I do not know how to continued

Make the substitution $v=y^4$. Then by the chain rule we have $v'=4y^3 y'$. Now your DE turns into:
$$x \frac{v'}{4}=2(v+x^4)$$
Then can be simplified to:
$$v'-8\frac{v}{x}=8x^3$$
We first solve the homogeneous part:
$$v_h' -8\frac{v_h}{x}=0$$
This leads to $v_h=c\cdot x^8$ so that $v_h'=c\cdot 8x^7$. This is our homogeneous solution. Now to find a particular solution we guess it will look something like $$v_p=ax^4+bx^3+dx^2+3x+f$$
Filling this into the DE we get:
$$(4ax^3 +3bx^2+2dx+3)-\frac{8}{x}(ax^4+bx^3+dx^2+3x+f)=8x^3$$
the only terms with $x^3$ in it are the $4ax^3$ and the $-8ax^3$. Conclusion: $-4a=8$ so $a=-2$. All the other terms are zero. We now have our particular solution $v_p=-2x^4$. The general solution is the sum of the particular and homogeneous solution so
$$v_g=c\cdot x^8 -2x^4$$
Now backsubstitute $v=y^4$ or $y=v^{\frac{1}{4}}$ to obtain the final answer.
