Find solution of differential equation
$$ 4x^2y''+8xy'+y=2xy^3, \quad y(1)=1, \ y'(1)=0 $$
I see that $ 4x^2y''+8xy' = (4x^2y')' $, but don't know, if it is useful or not
If we divide by $ y^3 $ ($y \equiv 0$ is a solution) we get
$$ 4x^2 \frac{y''}{y^3}+8x \frac{y'}{y^3}+\frac{1}{y^2}=2x $$
I've tried such substitutions:
- $ y=z^{-1/2}: \quad 3x^2 \frac{z'^2}{z}-2x^2z''-4xz'+z = 2x $
- $ y=e^{-z/2}: \quad (-2x^2z''+x^2z'^2-4xz'+1)e^z=2x $
- $ y=tx^{-1/2}: \quad xt''+t'=2x^{-1}t^3 $
For the last one I found answer in parametrical form:
$$ x=C_2 \exp(\int(C_1 \pm \tau^4)^{-1/2} d\tau) $$
$$ y=\frac{\pm \tau}{\sqrt{x}} $$
I understand that these substitutions aren't fully correct and want to find solution in ordinary form
We try the substitution $y(x)=\frac{f\! \left(x\right)}{\sqrt{x}}$:
$$4 \left(\frac{d^{2}}{d x^{2}}f\! \left(x\right)\right) x^{2}-2 f\! \left(x\right)^{3}+4 \left(\frac{d}{d x}f\! \left(x\right)\right) x=0$$
Next we substitute $f(x)=g\! \left(\ln\! \left(x\right)\right)$:
$$4 \mathrm{D}^{\left(2\right)}\! \left(g\right)\! \left(\ln\! \left(x\right)\right)-2 g\! \left(\ln\! \left(x\right)\right)^{3}=0$$
Variable change $x=\exp\left(z\right)$ leads to:
$$4 \frac{d^{2}}{d z^{2}}g\! \left(z\right)-2 g\! \left(z\right)^{3}=0$$
The initial conditions $[y(1)=1, y'(1)=0]$ transform to $[g(0)=1, g'(0)=\frac{1}{2}]$
Hence we get the solution $$g\! \left(z\right)=\frac{2}{2-z}$$
$$g(\ln(x))=f(x)=\frac{2}{2-\ln\! \left(x\right)}$$
and finally
$$y(x)=\frac{f(x)}{\sqrt{x}}=\frac{2}{\sqrt{x}\, \left(2-\ln\! \left(x\right)\right)}$$