Problem source

By means of the substitution $x^{\alpha},$ where $\alpha$ is a suitably chosen constant, find the general solution for $x>0$ of the differential equation 
	\[
	x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y=0,
	\]
	where $b$ is a constant and $b>-1$. 
	Show that, if $b>0$, there exist solutions which satisfy $y\rightarrow1$ and $\mathrm{d}y/\mathrm{d}x\rightarrow0$ as $x\rightarrow0$, but that these conditions do not determine a unique solution. For what values of $b$ do these conditions determine a unique solution?

Solution source

Let $z = x^\alpha, \frac{\d z}{\d x}=\alpha x^{\alpha-1} $, then 
\begin{align*}
\frac{\d y}{\d x} &= \frac{\d y}{\d z} \frac{\d z}{\d x}  \\
&= \alpha x^{\alpha-1}\frac{\d y}{\d z} \\
\\
\frac{\d^2 y}{\d x^2} &= \frac{\d }{\d x} \left  ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) \\
&= \alpha (\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha x^{\alpha-1} \frac{\d ^2 y}{\d z^2} \frac{\d z}{\d x} \\
&= \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2}
\end{align*}
\begin{align*}
&& 0 &=x\frac{\mathrm{d}^{2}y}{\mathrm{d}x^{2}}-b\frac{\mathrm{d}y}{\mathrm{d}x}+x^{2b+1}y \\
&&&= x \left ( \alpha(\alpha-1)x^{\alpha-2} \frac{\d y}{\d z} + \alpha^2 x^{2\alpha-2} \frac{\d ^2y}{\d z^2}\right) - b \left ( \alpha x^{\alpha-1}\frac{\d y}{\d z} \right) + x^{2b+1}y \\
&&&= \alpha^2 x^{2\alpha-1} \frac{\d^2 y}{\d z^2} +\left (\alpha(\alpha-1)x^{\alpha-1}-b\alpha  x^{\alpha-1} \right) \frac{\d y}{\d z} + x^{2b+1} y \\
\end{align*}

If we set $\alpha = b +1$ the middle term disappears, so we get

\begin{align*}
&& 0 &= (b+1)^2 x^{2b+1} \frac{\d^2 y}{\d z^2} + x^{2b+1} y  \\
\Rightarrow && 0 &= (b+1)^2 \frac{\d^2 y}{\d z^2} + y \\
\Rightarrow && y &= A \sin \left (\frac{z}{b+1} \right) + B \cos \left (\frac{z}{b+1} \right) \\
&&&= \boxed{A \sin \left (\frac{x^{b+1}}{b+1} \right) + B \cos \left (\frac{x^{b+1}}{b+1} \right)}
\\
\\
\lim_{x \to 0}: && y &\to B \\
&& \frac{\d y}{\d x} &= A x^b \cos\left (\frac{x^{b+1}}{b+1} \right) - B x^b \sin\left (\frac{x^{b+1}}{b+1} \right) \\
b>0: && \frac{\d y}{\d x} &\to 0 \\
\end{align*}

So there are infinitely many different solutions with $B = 1$ and $A$ is anything it wants to be.

If $b = 0$ $y' \to A$ so $A  =0 $ and unique.

If $b < 0$ $x^b \to \infty$ so we need $A = 0$, unique. However, we also need $y' \to 0$, so we need to check $y' = -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right) \to 0$,

\begin{align*}
y' &=  -x^b \sin \left ( \frac{x^{b+1}}{b+1}\right)  \\
&\approx  -x^b \left ( \frac{x^{b+1}}{b+1}\right)  \\
&= - \frac{x^{2b+1}}{b+1}
\end{align*}

so we need $2b+1>0 \Rightarrow b > -\frac12$.

Therefore the solution is unique on $(-\frac12,0]$