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QUESTION

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\item[(a)]
Express $-16$ in exponential form and hence find all the complex
values of $(-16)^\frac{1}{4}$, writing your answers in the form
$x+jy$.

Display these values on an Argand diagram.

\item[(b)]
Using Laplace transforms find, with the aid of tables, the
solution of the ordinary differential equation

$$\frac{dx}{dt}+2x=1$$

which satisfies the condition $x=2$ when $t=0$.

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ANSWER

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\item[(a)]

${}$

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$\theta=\pi$ therefore $-16=16e^{j\pi}=16e^{j(\pi+2k\pi)}\
k=0,\pm1,\pm2,\ldots$\\ so
$(-16)^\frac{1}{4}=(16)^\frac{1}{4}e^{j\frac{(\pi+2k\pi)}{4}}$

$k=0,\
2e^{j\frac{\pi}{4}}=2\left(\cos\frac{\pi}{4}+j\sin\frac{\pi}{4}\right)=
2\left(\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}}\right)=\sqrt{2}(1+j)\\
k=1,\ 2e^{j\frac{3\pi}{4}}
=2\left(\cos\frac{3\pi}{4}+j\sin\frac{3\pi}{4}\right)
=2\left(-\frac{1}{\sqrt{2}}+j\frac{1}{\sqrt{2}}\right)
=\sqrt{2}(-1+j)\\ k=2,\ 2e^{j\frac{5\pi}{4}}
=2\left(\cos\frac{5\pi}{4}+j\sin\frac{5\pi}{4}\right)
=2\left(-\frac{1}{\sqrt{2}}-j\frac{1}{\sqrt{2}}\right)
=\sqrt{2}(-1-j)\\ k=3,\ 2e^{j\frac{7\pi}{4}}
=2\left(\cos\frac{7\pi}{4}+j\sin\frac{7\pi}{4}\right)
=2\left(\frac{1}{\sqrt{2}}-j\frac{1}{\sqrt{2}}\right)
=\sqrt{2}(1-j)$

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\item[(b)]
$\ds\frac{dx}{dt}+2x=1,\ x=2$ when $t=0$

Taking Laplace transforms
$\ds{\cal{L}}\left\{\frac{dx}{dt}+2x\right\}
={\cal{L}}\left\{\frac{dx}{dt}\right\}+2{\cal{L}}\{x\}={\cal{L}}\{1\}$

$\ds\{sX-x(0)\}+2X=\frac{1}{s}$

$\ds(s+2)X=\frac{1}{s}+2,\ X=\frac{1}{(s+2)s}+\frac{2}{s+2}\\
\frac{1}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}=\frac{A(s+2)+Bs}{s(s+2)}\\
1=A(s+2)+Bs\\ s=0,\ 1=2A,\ A=\frac{1}{2}\\ s=-2,\ 1=-2B,\
B=-\frac{1}{2}$

Therefore $\ds
X=\frac{\frac{1}{2}}{s}-\frac{\frac{1}{2}}{s+2}+\frac{2}{s+2}=
\frac{\frac{1}{2}}{s}+\frac{\frac{3}{2}}{s+2}$\\ Taking the
inverse Laplace transform gives $\ds
x(t)=\frac{1}{2}+\frac{3}{2}e^{-2t}$.

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