Differential equation $2x^4yy'+y^4 = 4x^6$

Question

I have differential equation

$2x^4yy'+y^4 = 4x^6$

How to find real parameter $m$ for which, when we introduce substitution $y=z^m$, given equation becomes first order homogeneous differential equation?

Answer 1

$$ 2x^4y y'+y^4=4x^6\tag 1 $$ We write $$ x^4(y^2)'+(y^2)^2=4x^6 $$ Set $w=y^2$. Hence $$ x^4w'+w^2=4x^6\tag 2 $$ This equation has partial solution $w=-4x^3$. So we set $w=-4x^3z$ and (2) becomes $$ x(1+3z-4z^2+xz')=0 $$ Hence $$ z'=(4z^2-3z-1)/x $$ or $$ \frac{dz}{4z^2-3z-1}=\frac{dx}{x} $$ or integrating $$ \frac{1}{5}\log(1-z)-\frac{1}{5}\log(1+4z)=c+\log x $$ or $$ z=\frac{1-Ax^5}{1+4Ax^5}\textrm{, }A=const $$ Hence $$ w=-4x^3\frac{1-Ax^5}{1+4Ax^5} $$ and finaly $$ y=y(x)=2\sqrt{x^3\frac{Ax^5-1}{4x^5+1}}\textrm{, }A=const $$

Answer 2

Ans: $\frac{3}{2}$

we have, $2x^4yy'+y^4 = 4x^6$

rearranging, $y'=\frac{4x^6-y^4}{2x^4y}$

now substituting $$y=z^m$$ $$dy=mz^{m-1}dz$$ $$z'=\frac{4x^6-z^{4m}}{mz^{2m-1}x^4}$$

Now for $m=\frac32$ equation is homogeneous.