<s>[INST] Solve the following homogeneous differential equation Initial value problem: $\displaystyle \frac{dy}{dx}= \frac{y}{x}+2 \frac{x^2}{y^2}$, $y(1)=1$.  Can anyone help
 [/INST] Let, $y=ux$
$\frac{dy}{dx}=u+x\frac{du}{dx}=u+\frac{2}{u^2}$
$x\frac{du}{dx}=\frac{2}{u^2}$
$\frac{dx}{x}=\frac{u^2}{2}du$
$\ln x+\ln c=\frac{u^3}{2*3}$
$6(\ln  cx)=\frac{y^3}{x^3}$
$6{x^3}(\ln cx)=y^3$
$y=(6{x^3}(\ln cx))^{1/3}$
$y=(6{x^3}(\ln cx))^{1/3}$ -- (|)
$y(1)=(6(\ln c))^{1/3}=1$
$\ln c=\frac{1}{6}$ -- (||)
From, (|) & (||) we have,
$y=(6{x^3}(\ln c+\ln x))^{1/3}$
$y=(x^3(1+6\ln x))^{1/3}$
$y^3=x^3(1+6\ln x)$
$y^3=x^3(1+1*(3-1)*3\ln x)$
</s>