How to reduce diffrential equation below using simple diffrentiation skills??

Question

How to recognise a suitable value for $\alpha$ to reduce the diffrential equation into this form. I found derivative of y with respect to w to express derivative of w with respect to x using chain rule.

Answer 1

One way to proceed would be to substitute $y = x^\alpha w$ in the first equation and then compare it with the target equation to try and deduce a value for $\alpha$. We start by calculating the first and second derivatives of y

$$ y = x^\alpha w \Rightarrow y' = \alpha x^{\alpha-1} w + x^\alpha w' $$ and $$y'' = \alpha(\alpha-1)x^{\alpha-2}w + 2ax^{\alpha -1} w' + x^\alpha w'' $$ Now we substitute this values for $y, y'$ and $y''$ in the first equation. This gives us

\begin{align*} f(x) &= 2 x^2 \Big( \alpha(\alpha-1)x^{\alpha-2}w + 2ax^{\alpha -1} w' + x^\alpha w'' \Big) \\ &+ (3x^2 + 8x) \Big( \alpha x^{\alpha-1} w + x^\alpha w' \Big) \\ &+(x^2 + 6x +4)x^\alpha w \end{align*}

Notice how this only gives us one term in $w''$ this term is $2x^2 x^\alpha w''$ This means we must have $$ 2x^2 x^\alpha w'' = 2w''$$ This gives us the value $\alpha = -2$. We develop are expression further and then set $\alpha = -2$

\begin{align*} f(x) &= 2 x^2 \Big( \alpha(\alpha-1)x^{\alpha-2}w + 2ax^{\alpha -1} w' + x^\alpha w'' \Big) \\ &+ (3x^2 + 8x) \Big( \alpha x^{\alpha-1} w + x^\alpha w' \Big) \\ &+(x^2 + 6x +4)x^\alpha w \\ &= 2x^2 x^\alpha w'' \\ &+ \Big(4x^2\alpha x^{\alpha-1} + (3x^2 + 8x)x^\alpha \Big)w'\\ &+ \Big(2\alpha (\alpha-1) x^2 x^{\alpha - 2} +\alpha(3x^2+8x)x^{\alpha-1} +(x^2 + 6x +4)x^\alpha \Big) w \\ &= 2 w'' + 3 w' + w \end{align*}