\docname{sess4.5}
\doctitle{Superposition Principle}

\mylhead
\myrhead

\title{\doctitle}
\maketitle

\section{Superposition Principle for Inputs}

One of the most important properties of first-order linear equations
is
the {\bf superposition principle}. In fact, we will see that superposition
is
the defining characteristic of linear equations or any order.

The superposition principle allows us to break up a problem into
simpler problems and then at the end assemble the answer from its
simpler pieces.  For the ODE $y'+p(t)y=q(t)$, let $q_1$ and $q_2$ be
inputs and $c_1$ and $c_2$ be constants. Then:
\[q_1\rightarrow y_1, q_2\rightarrow y_2 \Rightarrow c_1q_1+c_2q_2\rightarrow c_1y_1+c_2y_2.\]
This is true because the ODE is linear. The proof takes two lines:
\begin{align*}
(c_1y_1+c_2y_2)'+p(c_1y_1+c_2y_2)&=(c_1y'_1+pc_1y_1)+(c_2y'_2+pc_2y_2)\\
&=c_1q_1+c_2q_2.
\end{align*}

We present some easy examples below.

\section{Examples}

We will learn later how to find the following solutions. For this
example you can check the following solutions by substitution:
\begin{enumerate}[i.]
\item $\overdot x+2x=1$ has a solution $x(t)=\frac 12$
\item $\overdot x+2x=e^{-2t}$ has a solution $x(t)=te^{-2t}$
\item $\overdot x+2x=0$ has a solution $x(t)=e^{-2t}$.
\end{enumerate}

\bigskip 

Using the solutions above as a basis, we can solve more complicated equations.

\examp{1}
Use superposition to find a solution to $\overdot x+2x=1+e^{-2t}$

\ans The input is a superposition of the inputs from (i) and (ii). Therefore a solution is $x(t)=\frac 12+te^{-2t}$.

\examp{2}
Find a solution to $\overdot x+2x=2+3e^{-2t}$.

\ans The input is $2\cdot(1)+3\cdot(e^{-2t})$; it is a superposition (with coefficients) of the inputs from (i) and (ii). Therefore, $x(t)=2\cdot\frac 12+3(te^{-2t})=1+3te^{-2t}$ is a solution.

\examp{3} 
Find \emph{lots} of solutions to $\overdot x+2x=1$

\ans We can write the input as: \[1=1+c\cdot 0.\]
That is, as a superposition of the input from (i) and the homogeneous equation (iii). Therefore $x(t)=\frac 12+ce^{-2t}$ is a solution for any value of the parameter $c$.

\examp{4} Find \emph{lots} of solutions to $\overdot x+2x=1+e^{-2t}$.

\ans Use superposition: $x(t)=\frac 12+te^{-2t}+ce^{-2t}$.