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\begin{center}
\textbf{Ordinary Differential Equations}

\textit{\textbf{Classification}}
\end{center}

\textbf{Question}

Show that $y=\cos x$ and $y=\sin x$ are solutions of $y''+y=0$.

Which of the following are solutions? Justify your answer.

\begin{description}
\item{(a)}
$\sin x -\cos x$ 

\item{(b)}
$\sin(x+3)$

\item{(c)}
$\sin 2x$ 
\end{description}

\textbf{Answer}

\begin{eqnarray*}
\textrm{If  } y & = & \cos x\\
\Rightarrow y'' + y & = & -\cos x + \cos x = 0\\
\textrm{If  } y & = & \sin x\\
\Rightarrow y''+ y & = & -\sin x + \sin x = 0
\end{eqnarray*}
So $y=\cos x$ and $y=\sin x$ are both solutions.

As the DE is linear and homogeneous, any function of the form
$$y = A \cos x + B \sin x $$
is also a solution.

\begin{description}
\item{(a)}
$\sin x -\cos x$ fits with $A=-1$, $B=1$ and so is a solution.

\item{(b)}
$\sin(x+3) = \sin 3 \cos x + \cos 3 \sin x$ fits with $A= \sin 3$, $B=
\cos 3$ and so is also a solution.

\item{(c)}
$\sin 2x$ cannot be represented in the form $A\cos x + B\sin x$ and
therefore is not a solution.
\end{description}
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