Hello,

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**posix_memalign** = 2tv'' - v' = 0, *(32)*

Note that the coefficient of v is zero, as it should be, this provides a useful check on our algebra.

Separating the variables in Eq. (32) and solving for v'(t), we find that

v'(t) = ct^(1/2), *(33)*

...

I have difficulty following the step from **32** to **33**.

Could someone perform the steps in between 32 and 33 in detail for me so I can see what is actually taking place there?

Let $\displaystyle u=v'$. (32) can be written $\displaystyle 2t\cdot \frac{\mathrm{d}u}{\mathrm{d}t}=u(t)$.

Separating the variables : $\displaystyle \frac{\mathrm{d}u}{u}=\frac12\cdot \frac{\mathrm{d}t}{t}$

Integrating both sides :

$\displaystyle \ln |u(t)|=\frac{1}{2}\ln t + K=\ln \sqrt{t}+K \implies |u(t)|=\underbrace{\exp K}_{\text{constant}} \sqrt{t}\implies u(t)=v'(t)=C\sqrt{t}$

The last implication comes from that for $\displaystyle t>0$, $\displaystyle \sqrt{t}>0$ so $\displaystyle u(t)$ never equals 0 hence $\displaystyle u$ (which is continuous) does not change sign on $\displaystyle (0,\infty)$. Is it clearer ?