\documentclass[a4paper,12pt]{article}

\begin{document}

\parindent=0pt

QUESTION

Show that

$$\left(\frac{y^2+1}{2y}\right)^2=\frac{y^2}{4}+\frac{1}{2}+\frac{1}{4y^2}.$$

A particle is projected along the positive $y$- axis so that,
initially, at time $t=0$, its position is $y=1$ and its velocity
is $\frac{dy}{dt}=1$. The equation of motion of the particle is
described by

$$\frac{d^2y}{dt^2}=\frac{y}{4}-\frac{1}{4y^3}.$$

Find the following:

\begin{description}

\item[(i)]
an expression for the velocity as a function of position;

\item[(ii)]
an expression for the position as a function of time;

\item[(iii)]
an expression for the velocity as a function of time.

\end{description}




ANSWER


$$\left(\frac{y^2+1}{2y}\right)^2=\frac{y^4+2y^2+1}{4y^2}=\frac{y^2}{4}+\frac{1}{2}+\frac{1}{4y^2}$$

\begin{description}

\item[(i)]
Let $v=\frac{dy}{dt}$. Then $\frac{d^2y}{dt^2}=\frac{dv}{dt}$ and
using the chain rule:
$\frac{dv}{dt}=\frac{dv}{dy}\frac{dy}{dt}=\frac{dv}{dy}v$. Hence
the equation of motion becomes:
$\frac{dv}{dy}v=\frac{y}{4}-\frac{1}{4y^3}$ which is a first order
separable differential equation. Integrating gives:

\begin{eqnarray*}
\int v\,dv&=&\int\frac{y}{4}-\frac{1}{4y^3}\,dy+C\\
\frac{1}{2}v^2&=&\frac{y^2}{8}+\frac{1}{8y^2}+C
\end{eqnarray*}

Using initial condition $y=1$ and $v=1$ when %t=0$ gives

$$\frac{1}{2}=\frac{1}{8}+\frac{1}{8}+C\Rightarrow C=\frac{1}{4}$$

So $\frac{1}{2}v^2=\frac{y^2}{8}+\frac{1}{8y^2}+\frac{1}{4}$ or
$v^2=\frac{y^2}{4}+\frac{1}{4y^2}+\frac{1}{2}$

By first part of question : $v^2=\left(\frac{y^2+1}{2y}\right)^2$
and so velocity as a function of position:
$\left(\frac{dy}{dt}=\right)v=\frac{y^2+1}{2y}$ (choose positive
square root to agree with initial condition $v=1$ when $y=1$)

\item[(ii)]
Since $v=\frac{dy}{dt}$ we have $\frac{dy}{dt}=\frac{y^2+1}{2y}$
which is first order separable. Integrating gives

$$\int\frac{2y}{y^2+1}\,dy=\int\,dt+k$$

Either substitute $u=y^2+1$ or note that
$\int\frac{f'(y)}{f(y)}\,dy=\ln|f(y)|$ gives

$$\ln|y^2+1|=t+K$$

Using initial condition $y=1$ when $t=0$ gives

$$\ln2=0+k\Rightarrow k=\ln 2$$

so $\ln(y^2+1)=t+\ln2$

Taking exponentials: $y^2+1=e^{t+\ln2}=e^te^{\ln2}=2e^t$

So position as a function of time: $y=\sqrt{2e^t-1}$

(Take positive square root to agree with initial condition $y=1$
when $t=0$.)

\item[(iii)]
Since $\frac{dy}{dt}=\frac{y^2+1}{2y}$, velocity as a function of
time:

$$\frac{dy}{dt}=\frac{(2e^t-1)+1}{2(2e^t-1)^\frac{1}{2}}=e^t(2e^t-1)^{-\frac{1}{2}}$$

\end{description}





\end{document}