<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\begin{document}


{\bf Question}

Find the solutions of the differential equation:

$$\frac{d^2y}{dx^2}+2\frac{dy}{dx}+(1-a)y=0$$

which satify the conditions $y=0$ and $\frac{dy}{dx} = 1$ when
$x=0$ for three cases:
\begin{description}
\item[(i)]
$a$ is a positive real number, so we can put $a=k^2$ with $k&gt;0$;
\item[(ii)]
$a$ is a negative real number, so we can put $a=-p^2$ with $p&gt;0$;
\item[(iii)]
$a=0$
\end{description}


{\bf Answer}

Auxiliary equation: $\lambda^2 + 2\lambda + (1-a) = 0 $

Roots: $\lambda_1, \lambda_2 = \frac{-2 \pm \sqrt{4 - 4(1 -
a)}}{2} = -1 \pm \sqrt{a}$
\begin{description}
\item[case (i)]
\begin{eqnarray*}
0 &lt; a &amp; = &amp; k^2 {\rm \  this\ gives\ two\ distinict\ real\ roots}
\\ \lambda_1, \lambda_2 &amp; = &amp; -1 \pm k \\ {\rm General\ Solution:\
} y &amp; = &amp; Ae^{(k-1)x} + Be^{-(k+1)x} \\ \frac{dy}{dx} &amp; = &amp;
(k-1)Ae^{(k-1)x} - (k+1)Be^{-(k+1)x} \\ {\rm Boundary\
conditions:\ } 0 &amp; = &amp; A + B
\\ 1 &amp; = &amp; (k-1)A - (k+1)B \\ \Rightarrow A &amp; = &amp; \frac{1}{2k} {\rm \ and\ } B = -\frac{1}[{2k} \\
{\rm Particular\ solution:\ } y &amp; = &amp; \frac{e^{-x}}{k}
 \left \{ \frac{e^{kx}-e^{-kx}}{2} \right\}  \\ &amp; = &amp; \frac {e^{-x}}{k} \sinh (kx)
\end{eqnarray*}
\item[case (ii)]
\begin{eqnarray*}
0 &gt; a &amp; = &amp; -p^2 {\rm this\ gives\ a\ pair\ of\ complex\
conjugate\ roots}\\ \lambda_1, \lambda_2 &amp; = &amp; -1 \pm \sqrt{-p^2}
\\ {\rm General\ Solution:\ } y &amp; = &amp; e_{-x}(A \cos(px) + B \sin(px))
\\ \frac{dy}{dx} &amp; = &amp; e^{-x}((pB-A) \cos (px) - (pA + B)\sin px
) \\ {\rm Boundary\ conditions:\ } 0 &amp; = &amp; A \\ 1 &amp; = &amp; pB - A\\
\Rightarrow A &amp; = &amp; 0 {\rm \ and\ } B = -\frac{1}{p}
\\ {\rm Particular\ solution:\ } y &amp; = &amp; \frac{e^{-x}}{p}
\sin(px)
\end{eqnarray*}
\item[case (iii)]
\begin{eqnarray*}
a &amp; = &amp; 0 {\rm \  this\ gives\ two\ repeated\ real\ roots}
\\ \lambda_1, \lambda_2 &amp; = &amp; -1 \\ {\rm General\ Solution:\
} y &amp; = &amp; (A + Bx) e^{-x} \\ \frac{dy}{dx} &amp; = &amp; (-A + B
-Bx)e^{-x}
 \\ {\rm Boundary\ conditions:\ } 0 &amp; = &amp; A
\\ 1 &amp; = &amp; -A + B \\ \Rightarrow A &amp; = &amp; 0 {\rm \ and\ } B = 1 \\
{\rm Particular\ solution:\ } y &amp; = &amp; xe^{-x}
\end{eqnarray*}
\end{description}


\end{document}
</pre></body></html>