Given 2x^4yy'+y^4=4x^6 , use the substitution y=z^m to…

Given $2x^4yy'+y^4=4x^6$, use the substitution $y=z^m$ to determine $m$ so the equation becomes homogeneous, then find its general integral.

Step 1: Determine $m$

With $y=z^m$, $y'=mz^{m-1}z'$, the equation becomes $2mx^4z^{2m-1}z'+z^{4m}=4x^6$. Matching total degrees ($3+2m$ for the first term, $4m$ for the second, $6$ for the third) requires $4m=6$, so $m=\dfrac32$, and indeed $3+2(\tfrac32)=6$ too.

Step 2: Rewrite and substitute $z=vx$

With $m=\tfrac32$: $3x^4z^2z'+z^6=4x^6$, i.e. $z'=\dfrac{4x^6-z^6}{3x^4z^2}$. With $z=vx$,

\[x\frac{dv}{dx}=\frac{4-v^6}{3v^2}-v=\frac{4-v^6-3v^3}{3v^2}.\]

Step 3: Separate and integrate

With $u=v^3$ ($du=3v^2dv$), $\dfrac{3v^2\,dv}{4-3v^3-v^6}=\dfrac{du}{-(u^2+3u-4)}=\dfrac{du}{-(u+4)(u-1)}$. Partial fractions give $\dfrac15\ln\left|\dfrac{u+4}{u-1}\right|=\ln|x|+C$, so

\[\frac{v^3+4}{v^3-1}=Kx^5.\]

Step 4: Back-substitute

Since $v=z/x$ and $z^3=y^2$ (from $y=z^{3/2}$), $v^3=y^2/x^3$, so $\dfrac{v^3+4}{v^3-1}=\dfrac{y^2+4x^3}{y^2-x^3}$:

\[\boxed{\,\frac{y^2+4x^3}{y^2-x^3}=Cx^5.}\]