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QUESTION

\begin{description}

\item[(a)]
Obtain the two complex roots ($z_1$ and $z_2$) of the quadratic
equation

$$(1+j)z^2-2jz-(1-j)=0$$

and calculate

(i)\ \ $z_1^2+z_2^2$,\hspace{1.5cm}(ii)\ \
$\ds\frac{1}{z_1}+\frac{1}{z_2}$.

\item[(b)]
Find the solution of the differential equation

$$x\frac{dx}{dt}=(1+x^2)t^2e^{3t},$$

which satisfies $x=0$ when $t=0$.

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(a)]
$(1+j)z^2-2jz-(1-j)=0$
\begin{eqnarray*}
z&=&\frac{-(-2j)\pm\sqrt{(-2j)^2-4(1+j)(-1+j)}}{2(1+j)}\\
&=&\frac{2j\pm\sqrt{-4-4(-1+j-j+j^2)}}{2(1+j)}\\
&=&\frac{2j\pm\sqrt{4}}{2(1+j)}\\&=&\frac{2j\pm2}{2(1+j)}\\&=&\frac{\pm1+j}{1+j}
\end{eqnarray*}

With the positive sign, $\ds z_1=\frac{1+j}{1+j}=1$\\ With the
negative sign, $\ds
z_2=\left(\frac{-1+j}{1+j}\right)\left(\frac{1-j}{1-j}\right)=\frac{-1+j+j-j^2}{1-j^2}=\frac{2j}{1+1}=j$

\begin{description}

\item[(i)]
$z_1^2+z_2^2 =1^2+j^2=1-1=0$

\item[(ii)]
$\ds\frac{1}{z_1}+\frac{1}{z_2}=\frac{1}{1}+\frac{1}{j}=1+\frac{1(-j)}{j(-j)}=1-\frac{j}{1}=1-j$

\end{description}

\item[(b)]
$\ds x\frac{dx}{dt}=\left(1+x^2\right)t^2e^{3t};\ \ x(0)=0\\
\int\frac{x}{1+x^2}\,dx=\int t^2e^{3t}\,dt.$

Therefore
\begin{eqnarray*}
\frac{1}{2}\ln(1+x^2) &=&
t^2\frac{e^{3t}}{3}-\int\frac{e^{3t}}{3}(2t)\,dt\\
&=&\frac{1}{3}t^2e^{3t}-\frac{2}{3}\left\{t\frac{e^{3t}}{3}-\int\frac{e^{3t}}{3}.1\,dt\right\}\\
&=&\frac{1}{3}t^2e^{3t}-\frac{2}{9}te^{3t}+\frac{2}{9}\left(\frac{e^{3t}}{3}\right)+c
\end{eqnarray*}

$\ds x(0)=0\Rightarrow\frac{1}{2}\ln1=0-0+\frac{2}{27}(1)+c,\
c=-\frac{2}{27}$\\ Therefore
$$\ln(1+x^2)=\frac{2}{3}t^2e^{3t}-\frac{4}{9}te^{3t}+\frac{4}{27}e^{3t}-\frac{4}{27}$$
or
$$x^2=\exp{\left(\frac{2}{3}t^2e^{3t}-\frac{4}{9}te^{3t}+\frac{4}{27}e^{3t}-\frac{4}{27}\right)}-1$$

(Take the square root for $x$.)

\end{description}




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