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\centerline{{\sc Higher order linear ODEs}}

\b

Much of what we did for second-order ODEs carries through with higher-order ODEs.  In particular, there	 theorems on existence and uniqueness, there are notions of fundamental sets of solutions, and so on.

\b

An n$^{th}$-order linear ODE has the form $$P_0(x)\dn{y}{x}{n} +  P_1(x)\dn{y}{x}{n-1} \cdots P_{n-1}(x)\d{y}{x} + P_n(x)y = Q(x).$$	If $P_0(x) \ne 0$ for all $x$ in some interval $I$, then we can divide by $P_n(x)$ for $x \in I$ to get $$ \dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x).$$

\b

\textbf{Theorem.}  Suppose each of the functions $p_1, \cdots , p_{n}, q$ is continuous on an open interval $I$.	 Then the ODE $ \dn{y}{x}{n-1}+	 p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x)$ has a solution on $I$.	 Furthermore, if $x_0 \in I$, the initial value problem $ \dn{y}{x}{n-1}+	 p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x)$, $y(x_0) = y_0$, $y'(x_0) = y_0', \cdots$, $y^{(n-1)}(x_0) = y^{(n-1)}_0$ has a unique solution on the interval $I$.

\b

\textbf{Example} The initial value problem $\dn{y}{x}{3} - (\frac{1}{x - 2})\dd{y}{x} + ( \frac{1}{x})\d{y}{x} + x^2 y = \tan x$, $y(1) = 3$, $y'(1) = 5$, $y''(1) = -7$ has a unique solution on the interval $(0, \frac{\pi}{2}	)$.	 We get the interval $(0,\frac{\pi}{2})$ from the fact that that is the largest interval containing $x = 1$ on which each of the functions $\frac{1}{x - 2}$, $\frac{1}{x} $, and $\tan x$ is defined and continuous.

\b

A set $\{f_1, f_2, \cdots , f_n \}$ of functions defined on an interval $I$ is \textit{linearly independent} on $I$ if $c_1f_1(x) + \cdots + c_nf_n(x) = 0$ for all $x \in I$ implies $c_1 = \cdots = c_n = 0$. Here $c_0, \cdots, c_n$ are constants.	Otherwise, the set is \textit{linearly dependent}.

\b

We sometimes say $f_1, f_2, \cdots , f_n$ are linearly independent if the set $\{f_1, f_2, \cdots , f_n \}$ is linearly independent.	Similar language is used for linear dependence.

\b

\textbf{Example.}  Show that the functions $f_1(x) = x$, $f_2(x) = x^2$, $f_3(x) = x^3$ are linearly independent on $(-\infty,\infty)$.

\b

Suppose $c_1x + c_2x^2 + c_3x^3 = 0$ for all $x$. Differentiating, we get $c_1 + 2c_2x + 3c_3x^2 = 0$.	 Differentiating again, we get $2c_2 + 6c_3 x = 0$.	 Therefore, we get a system of three equations:

%\b

\begin{eqnarray*} c_1x + c_2x^2 + c_3x^3 & = & 0.\\
                  c_1 + 2c_2x + 3c_3x^2	 & = & 0. \\
                        2c_2  + 6c_3x	 & = & 0. \\
\end{eqnarray*}


These equations hold for all $x$.  If we let $x = 0$, we get from the last two equations that $c_1 = 0$ and $c_2 = 0$.	 Then the first equation becomes $c_1x = 0$.  Letting $x = 1$	in any of the equations then gives that $c_3 = 0$.

There are other ways of doing this problem.	 For example, we could let $x = 1$, then $x = 2$, then $x = 3$ in the original equation to get the system

\begin{eqnarray*} c_1 + c_2 + c_3 & = & 0.\\
                  2c_1 + 4c_2 + 8c_3  & = & 0. \\
                  3c_1 + 9c_2  + 27c_3x	   & = & 0. \\
\end{eqnarray*}

The only solution of this system is $c_1 = c_2 = c_3$.

\b

\textbf{Definition.} Given functions $f_1, \cdots , f_n$, the \textit{Wronskian} $W(f_1, \cdots , f_n)$ is defined by $$W(f_1, \cdots ,f_n)(x) = \left| \begin{array}{ccc} f_1(x) & \cdots & f_n(x) \\
                                     f_1'(x) & \cdots & f_n'(x) \\
                                     \vdots	 & \vdots & \vdots \\
                                     f_1^{(n-1)} & \cdots & f_n^{(n-1)} \\
                  \end{array} \right|.$$ Here we assume that all derivatives exist.


\b


\textbf{Example.} $W(x,x^2,x^3) = \left| \begin{array}{ccc} x & x^2 & x^3 \\
                                     1 & 2x & 3x^2 \\
                                     0 & 2 & 6x \\
                  \end{array} \right| = 2x^3.$


\b

\textbf{Theorem.}  If $f_1, \cdots , f_n$ are functions	 defined on an open interval $I$ and if $W(f_1, \cdots , f_n)(x_0) \ne 0$ for at least one $x_0 \in I$, then $f_1, \cdots, f_n$ are linearly independent on $I$.

\b

Note that it is possible that the Wronskian $W(f_1, \cdots , f_n)(x)$ is $0$ for some values of $x$ and non-zero for other values of $x$.  If there is just one where value where the Wronskian is non-zero, the functions are linearly independent.

\b

\textbf{Theorem.} Suppose that $p_1, \cdots , p_n$ are continuous functions defined on an interval $I$. If $y_1, \cdots, y_n$ are linearly independent solutions of the homogeneous linear ODE $ \dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = 0$ on $I$, then every solution of the ODE has the form $y = C_1 y_1 + \cdots + C_n y_n$.	

\b

If If $y_1, \cdots, y_n$ are linearly independent solutions of the homogeneous linear ODE $ \dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = 0$ on $I$, then we say that the set $\{y_1, \cdots , y_n\}$ is a \textit{fundamental set of solutions} of the ODE, and we say that $y = C_1 y_1 + \cdots + C_n y_n$ is the \textit{general solution} of the ODE.

\b

\textbf{Corollary.} Suppose that $p_1, \cdots , p_n$ are continuous functions defined on an interval $I$. If $y_1, \cdots, y_n$ are solutions of the ODE $ \dn{y}{x}{n-1}+	 p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = 0$ such that $W(y_1, \cdots , y_n)(x_0) \ne 0$ for some $x_0 \in I$, then $\{y_1, \cdots , y_n\}$ is a fundamental set of solutions, that is,	 every solution of the ODE has the form $y = C_1 y_1 + \cdots + C_n y_n$.

\b

\textbf{Example.}  Each of the functions $y_1(x) = 1$, $y_2(x) = e^x$, $y_3(x) = e^{-x}$ is a solution of the ODE $\dn{y}{x}{3} - \d{y}{x} = 0$.  $$W(y_1,y_2,y_3)(x) = \left| \begin{array}{ccc} 1 & e^x & e^{-x} \\
                                     0 & e^x & -e^{-x} \\
                                     0 & e^x & e^{-x} \\
                  \end{array} \right|  = 2 \ne 0,$$
so every solution of ODE has the form $y = C_1 + C_2e^x + C_3e^{-x}$.

\b

There is also a version of Abel's Theorem that holds for higher-order ODEs.

\b

\textbf{Theorem.} (Abel) Suppose that $p_1, \cdots , p_n$ are continuous functions defined on an interval $I$, and suppose that $y_1, \cdots, y_n$ are solutions of the ODE $ \dn{y}{x}{n-1}+	p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = 0$.	Then $W(y_1, \cdots , y_n)(x) = Ce^{-\int p_1(x) dx}$, where $C$ is a constant that depends on $y_1, \cdots , y_n$, but not on $x$.	 In particular, if $W(y_1, \cdots ,y_n)(x_0) = 0$ for some $x \in I$, then $W(y_1, \cdots ,y_n)(x_0) = 0$ for all $x \in I$, and	if $W(y_1, \cdots ,y_n)(x_0) \ne 0$ for some $x \in I$, then $W(y_1, \cdots ,y_n)(x_0) \ne 0$ for all $x \in I$.

\b

We can also deal with the non-homogeneous case.	 Again, the situation is analogous to the case of second-order linear ODEs.

\b

\textbf{Theorem.}  Suppose that $\{ y_1, \cdots , y_n \}$ is a fundamental set of solutions of the homogeneous ODE $\dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = 0$ and $y_{par}$ is a solution of the non-homogeneous ODE $\dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x)$.  Then every solution of the non-homogeneous ODE $\dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x)$ has the form $$y = C_1y_1 + \cdots + C_ny_n + y_{par}.$$

\b

The expression	$y = C_1y_1 + \cdots + C_ny_n + y_{par}$ is called the \textit{general solution} of the ODE $\dn{y}{x}{n-1}+  p_1(x)\dn{y}{x}{n-1} \cdots p_{n-1}(x)\d{y}{x} + p_n(x)y = q(x)$.

\b

\{ Page 224 \hspace{1in} 1-4, 7-10, 11-14, 26, 27 \\

\pagebreak

\centerline{{\sc Higher order homogeneous linear ODEs with constant coefficients}}

\b

The solution of higher order homogeneous linear ODEs with constant coefficients is analogous to what we did for second order homogeneous linear ODEs with constant coefficients.	Given the ODE
$\dn{y}{x}{n-1}+  a_1(x)\dn{y}{x}{n-1} \cdots a_{n-1}(x)\d{y}{x} + a_n(x)y = 0$, where each $a_i$ is a constant, the \textit{characteristic equation} is $$r^n + a_1r^{n-1} + \cdots + a_{n-1}r + a_n = 0,$$ and the \textit{characteristic polynomial} is the polynomial $$p(r) =
r^n + a_1r^{n-1} + \cdots + a_{n-1}r + a_n.$$

\b

\textbf{Example.} The characteristic polynomial of the ODE $\dn{y}{x}{4} - 7\dd{y}{x} + 3\d{y}{x} + 5y = 0$ is $r^4 -7r^2 + 3r + 5$.

\b


A polynomial $p(r)$ with real coefficients factors into terms of the form $(r - b)^k$, where $b$ is a real number, and terms of the form $[(r - (\alpha + i\beta))(r - (\alpha - i\beta)]^k$, where $\alpha$ and $\beta$ are real numbers.  If $p(r)$ is the characteristic polynomial of the ODE $\dn{y}{x}{n-1}+  a_1(x)\dn{y}{x}{n-1} \cdots a_{n-1}(x)\d{y}{x} + a_n(x)y = 0$, then the general solution will be a sum of expressions of the following types:

\b

For the term $(r - b)^k$, the general solution will include an expression of the form $q(x)e^{bx}$, where $q(x)$ is an arbitrary polynomial of degree $k-1$.

\b

For the term $[(r - (\alpha + i\beta))(r + (\alpha + i\beta)]^k$, the general solution will include an expression of the form $q_1(x)e^{\alpha x}\cos (\beta x) + q_2(x)e^{\alpha x}\sin (\beta x)$, where $q_1$ and $q_2$ are arbitrary polynomials of degree $k-1$.

\b


If all roots are distinct, we can replace the polynomials by constants.

\b

\textbf{Examples.}

\begin{enumerate}

\item Solve $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = 0$.

\b

The characteristic polynomial is $r^3 - 5r^2 -r + 5 = (r - 1)(r + 1)(r - 5)$.  Here there are three distinct real roots, so the general solution is $$y = C_1e^x + C_2e^{-x} + C_3e^{5x}.$$



\b

\item Solve	 $\dn{x}{t}{4} + 8\dd{x}{t} + 16x = 0$.


\b

The characteristic polynomial  is $$p(r) = r^4 + 8r^2 + 16 = (r^2 + 4)^2 = [(r - 2i)(r+2i)]^2 = (x -2i)^2(x+2i)^2.$$  The roots are $\pm2i$ and each has multiplicity $2$.  Therefore, the general solution is $y = (A_1t + B_1) \cos (2t) + (A_2t + B_2) \sin (2t).$  Note that there are $4$ constants here.



\item Solve $\dn{y}{x}{3} - 7\dd{y}{x}+19\d{y}{x}-13y = 0$.

\b

The characteristic polynomial is $p(r) = r^3 - 7r^2 + 19r - 13$ which factors as $(r-1)(r^2 - 6r + 13)$.	The roots of $r^2 - 6r + 13$ are $r = 3 \pm 2i$ so the roots of $p(r)$, are $1$ and $3 \pm 2i$.	 Therefore, the general solution is $$y = C_1e^x + C_2e^{3x}\cos (2x) + C_3 e^{3x} \sin (2x).$$


\item Solve the initial value problem $\dn{y}{x}{3} - 2 \dd{y}{x} + \d{y}{x} - 2y = 0$, $y(0) = 4$, $y'(0) = 3$, $y''(0) =6$.

\b

The characteristic polynomial is $p(r) = r^3 - 2r^2 + r - 2 = (r - 2)(r^2 + 1)$.  The roots of this polynomial are $r = 2$ and $r = \pm i$.	 Therefore, the general solution is $$y = C_1e^{2x} + C_2 \cos x + C_3 \sin x.$$  It follows that $$y' = 2C_1e^{2x} - C_2 \sin x + C_3 \cos x,$$ and $$y'' = 4C_1e^2x - C_2 \cos x - C_3 \sin x.$$  From the initial conditions we get

\begin{eqnarray*}
C_1 + C_2 & = & 4\\
2C_1 + C_3 & = & 3 \\
4C_1 - C_2 & = & 6 \\
\end{eqnarray*}

Adding the first and third equations and solving gives	$C_1 = 2$.	Then from the first equation we get $C_2 = 2$ and from the second equation we get $C_3 = -1$.	 This gives the solution $$y = 2 e^{2x} + 2 \cos x - \sin x.$$





\item Solve the ODE $\dn{x}{t}{4}-5\dn{x}{t}{3}+9\dd{x}{t} -7\d{y}{x} + 2 = 0$.

\b

The characteristic polynomial is $p(r) = r^4 - 5r^3 +9r^2 -7r + 2 = (r-1)^3(r-2)$.	The roots are $1$ and $2$.	Since the factor $(r-1)$ is raised to the third power, we will multiply the corresponding expression in the solution by a polynomial of degree $2$ (which is one less than the exponent $3$).	Therefore, we get the general solution $$y = (C_1 + C_2x + C_3x^2)e^x + C_4 e^{2x}.$$	Notice that there are $4$ constants in the solution.

In general, the general solution of an $n^{th}$-order linear  ODE should have $n$ arbitrary constants.


\end{enumerate}



\b

\{ Page 231 \hspace{1in} 11-22, 37 \\

\pagebreak

\centerline{{\sc Non-homogeneous higher order homogeneous linear ODEs}}

\b

Here we will work with an ODE of the form $$\dn{y}{x}{n} + a_1\dn{y}{x}{n-1} + \cdots + a_n \d{y}{x} = g(x),$$ where each	$a_i$ is a constant.	 We know how to solve $\dn{y}{x}{n} + a_1\dn{y}{x}{n-1} + \cdots + a_n \d{y}{x} = 0$, and we know that to solve the non-homogeneous equation we add the solution of the homogeneous equation to any solution of the non-homogeneous equation. The method of undetermined coefficients can sometimes be used to find a solution to the non-homogeneous equation.  The principles are the same as for the second-order ODEs, but the algebra might be harder.

\b

\textbf{Examples.}

\begin{enumerate}

\item Solve $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = x$.

\b

We saw in an earlier example that the general solution of the ODE $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = 0$ is $y = C_1e^x + C_2e^{-x} + C_3e^{5x}.$  To find a particular solution of the non-homogeneous ODE, we look for a solution of the form $y = Ax + B.$  Then $y' = A$ and $y'' = y^{(3)} = 0$, so substituting in the ODE gives $-A + 5(Ax + B) = x$, or $5Ax + (5B-A) = x$.	 Comparing coefficients gives $5A = 1$ and $5B - A = 0$.	 Therefore, $A = \frac{1}{5}$ and $B = \frac{1}{25}$.  This gives the particular solution $y = (\frac{1}{5})x + \frac{1}{25}$, so the general solution is $$y = C_1e^x + C_2e^{-x} + C_3e^{5x} + (\frac{1}{5})x + \frac{1}{25}.$$

\item Solve $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = e^x$.


\b

Again the general solution of the homogeneous ODE $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = 0$ is $y = C_1e^x + C_2e^{-x} + C_3e^{5x}.$ To find a solution of the non-homogeneous ODE, we would look for a particular solution of the form $y = Ae^x$, but we know that won't work, so we look for a solution of the form $y = Axe^x$.	 Then $y' = Ae^x + Axe^x$, $y'' = 2Ae^x + Axe^x$, and $y^{(3)} = 3Ae^x + Axe^x$.	Substituting into the non-homogeneous equation gives $$3Ae^x + Axe^x - 5(2Ae^x + Axe^x) - (Ae^x + Axe^x) + 5Axe^x = e^x.$$  This simplifies to $-8Ae^x = e^x$, so $A = - \frac{1}{8}$.  Therefore, we get the particular solution $y = - (\frac{1}{8})xe^x$, so the general solution is $$y = C_1e^x + C_2e^{-x} + C_3e^{5x} - (\frac{1}{8})xe^x.$$


\item Solve the initial value problem $\dn{y}{x}{3} - 5\dd{y}{x} - \d{y}{x} + 5y = e^x$, $y(0) = 1$, $y'(0) =  \frac{7}{8} $, $y''(0) = -\frac{1}{4}  $.

\b

From the last problem, we know that the general solution of the ODE is	$$y = C_1e^x + C_2e^{-x} + C_3e^{5x} - (\frac{1}{8})xe^x.$$	 Differentiating gives $$y' = C_1e^x -C_2e^{-x} + 5C_3e^{5x} - (\frac{1}{8})xe^x - (\frac{1}{8})e^x,$$ and differentiating again gives
$$y'' = C_1e^x +C_2e^{-x} + 25C_3e^{5x} - (\frac{1}{8})xe^x - (\frac{1}{4})e^x.$$	 From the initial conditions, we get the system
\begin{eqnarray*}
C_1 + C_2 + C_3 & = & 1\\
C_1 - C_2 + 5C_3 & = & 1 \\
C_1 + C_2 + 25C_3 & = 0 \\
\end{eqnarray*}


Solving this system gives $C_1 = \frac{9}{8}$, $C_2 = - \frac{1}{12}$, $C_3 = - \frac{1}{24}$.	Therefore, the solution is
 $$y = (\frac{9}{8})e^x -(\frac{1}{12})e^{-x} -(\frac{1}{24})e^{5x} - (\frac{1}{8})xe^x.$$



\item Solve the ODE $\dn{x}{t}{4} + 8\dd{x}{t} + 16x =128 e^{2t}$.

\b

By an earlier example, the general solution of the homogeneous ODE $\dn{x}{t}{4} + 8\dd{x}{t} + 16x = 0$ is  $y = (A_1t + B_1) \cos (2t) + (A_2t + B_2) \sin (2t).$  We look for a particular solution of the non-homogeneous ODE of the form $x = Ae^{2t}$.	Then $x'(t) = 2Ae^{2t}$, $x''(t) = 4Ae^{2t}$, $x^{(3)}(t) = 8Ae^{2t}$, and $x^{(4)}(t) = 16Ae^{2t}$.	 Substituting into the non-homogeneous ODE gives $16Ae^{2t} + 8(4Ae^{2t}) + 16Ae^{2t} = 128 e^{2t}$.  Solving for $A$ gives $A = 2$, so a particular solution is $x = 2e^{2t}$.	Therefore, the general solution of the non-homogeneous ODE is $$y = (A_1t + B_1) \cos (2t) + (A_2t + B_2) \sin (2t) + 2e^{2t}.$$

\b





\end{enumerate}

\b

\{ Page 237 \hspace{1in} 1-6, 13-16 \\

\pagebreak

\centerline{{\sc The Laplace Transform}}

\b

A real-valued function $f$ defined on an interval $I = [a,b]$  is \textit{piecewise continuous} on $I$ if $f$ is continuous except for finitely many points of $I$, and for each point of discontinuity, the left and right limits of $f$ exist.  (Only the right limit need exist at $a$,	and only the left limit need exist at $b$.)	$f$ is piecewise continuous on an interval $[a, \infty)$ if it is piecewise continuous on each interval $[a,b]$ where $b > a$. A similar definition works for intervals of the form $(-\infty,b]$.

\b

Note that any continuous function is piecewise continuous on any interval on which it is defined.

\b

\textbf{Examples.}

\begin{enumerate}

\item The function $f(x) = \cases{ x^2 & If $x \leq 0$, \cr 1 - x^2 & If $x > 0$,}$ is piecewise continuous on $[-3, 3]$ (or on any interval).

\item The function $f(x) = \cases{ \frac{\sin x}{x} & If $x \ne 0$, \cr 0 & If $x = 0$, }$ is piecewise continuous on any interval.

\item  The function $f(x) = \cases{ \sin (\frac{1}{x}) & If $x \ne 0$, \cr 0 & If $x = 0$, }$ is not piecewise continuous on any interval containing $0$.	 (However, it is continuous on any interval not containing $0$.)

\item The function $f(x) = \cases{0 & If $2n \leq x < 2n + 1$ for some integer $n$, \cr 1 & If $2n-1 \leq x < 2n$ for some integer $n$,}$ is piecewise continuous on $(-\infty, \infty)$.

\end{enumerate}


\b

\textbf{Recall:} Suppose that $f$ is a real-valued function defined on an interval of the form $[a,\infty)$.	Then we say that the \textit{improper integral $\int_a^\infty f(x) dx$ converges} if $\lim_{t \to \infty}\int_a^t f(x) dx$ exists, and if $\lim_{t \to \infty}\int_a^t f(x) dx = L$, we write $\int_a^\infty f(x) dx = L$.  If the improper integral does not converge, it diverges.

\b

\textbf{Examples:}

\begin{enumerate}

\item $\int_0^\infty \frac{1}{1 + x^2} dx = \lim_{t \to \infty}\int_0^t \frac{1}{1 + x^2} dx = \lim_{t \to \infty}(\arctan t - \arctan 0) = \frac{\pi}{2}$.

\item $\int_1^\infty \frac{1}{\sqrt{t} } dt =  \lim_{u \to \infty}\int_1^u \frac{1}{\sqrt{t} } dt = \lim_{u \to \infty}(\frac{2}{3})u^{\frac{3}{2}} - 1 = \infty$.	Therefore, the improper integral diverges.

\end{enumerate}

\b

\textbf{Theorem.} (Comparison Theorem) Suppose that $f$ is piecewise continuous on an interval $[a, \infty)$.

\begin{enumerate}

\item If $g$ is a function such that for some $M \geq a$, $|f(x)| \leq g(x)$ whenever $x \geq M$, and if $\int_a^{\infty}g(x)dx$ converges, then $\int_a^{\infty}f(x)dx$ converges.

\item If $g$ is a function such that for some $M \geq a$, $f(x) \geq g(x) \geq 0$ whenever $x \geq M$, and if $\int_a^{\infty}g(x)dx$ diverges, then $\int_a^{\infty}f(x)dx$ diverges.


\end{enumerate}



\b

\textbf{Definition.} Suppose the function $f(t)$ is defined for $t > 0$.  Then the \textit{Laplace transform} $\L(f(t))$ is defined by $\L(f(t))(s) = \int_0^\infty e^{-st}f(t) dt$, for all $s$ such that the improper integral converges.

\b

Note that the Laplace transform is a function of $s$.  We will see later how to use the Laplace transform to trade an initial value problem for a problem of solving a regular algebraic equation of the sort we did in high school.	

\b


The Laplace transform of a function does not always exist.	However, in certain cases it does.

\b

\textbf{Definition.} A piecewise continuous function $f(t)$ defined on an interval $[a,\infty)$ has \textit{exponential order as $t \to \infty$} if there are positive constants $\lambda$, $K$, and $M$ such that $f(t) < Ke^{\lambda t}$ for all $t > M$.  (Roughly, this says that eventually, $f$ is smaller than some exponential function.)

\b

\textbf{Theorem.}  Suppose $f(t)$ is a piecewise continuous function which has exponential order as $t \to \infty$.  Then $\L(f(t))(s)$ is defined for $s > \lambda$, where $\lambda$ is as in the definition.

\b

\textbf{Examples.}


\begin{enumerate}

\item $\L(t)(s) = \int_0^\infty e^{-st}t dt$.  Using integration by parts, we get $\int e^{-st}tdt = (\frac{-t}{s})e^{-st} - \frac{e^{-st}}{s^2}$, so $\L(t)(s) = \lim_{u \to \infty}[(\frac{-u}{s})e^{-su} - \frac{e^{-su}}{s^2}] - [(\frac{0}{s})e^{0} - \frac{e^{0}}{s^2}] = \frac{1}{s^2}$ for $s > 0$.

\item $\L(e^t)(s) = \int_0^\infty e^{-st}e^t dt = \int_0^\infty e^{(1-s)t} dt = \frac{1}{s - 1}	$ if $s > 1$.

\item Let $f(t) = \cases{1 & If $1 \leq t \leq 4$. \cr 0 & Otherwise.}$.  Find $\L(f(t))$.

    \b

By definition, $\L(f(t))(s) = \int_0^\infty e^{-st}f(t) dt$.  Since $f(t) = 0$ if $t$ is not in the interval $[1,4]$, and $f(t) = 1$ if $1 \leq t \leq 4$, so	 $\L(f(t))(s) = \int_1^4e^{-st} dt = \frac{e^{-st}}{-s}|_{t=1}^4 = \frac{ e^{-s} - e^{-4s}}{s}$.

\end{enumerate}

\b

It is worth noting in the last example, if we had defined the function $f$ to be, for example, $f(t) = \cases{1 & If $1 < t < 4$, \cr 0 & Otherwise,}$ we would have got the same Laplace transform.	In fact, changing finitely many values of a function does not change the Laplace transform, so many functions can have the same Laplace transform.

\b

The next theorem says that the Laplace transform is linear.	 It follows from the linearity of the integral.

\b

\textbf{Theorem.}  If $f_1$ and $f_2$ are functions having Laplace transforms, and $C_1$ and $C_2$ are constants, then $\L(C_1f_1 + C_2f_2) = C_1\L(f_1) + C_2\L(f_2)$.  This equation holds for all $s$ where both Laplace transforms are defined.

\b

\textbf{Example.} From the linearity of the Laplace transform and our earlier examples, we get $\L(5t + 7e^t)(s) = 5 \L(t) + 7 \L(e^t) = \frac{5}{s^2} + \frac{7}{s-1}.$

\b

\{ Page 311 \hspace{1in} 1-3, 5-8, 15, 16 \\

\pagebreak

\centerline{{\sc Using the Laplace Transform for Initial Value Problems}}

\b

The value of the Laplace Transform is that it makes an initial value problem into a problem of solving a simpler equation.  To see how this works, we need a basic fact about the Laplace transform.

\b

Suppose that $f$ is a function with derivative $f'$.  If we use integration by parts, with $u = e^{-st}$ and $dv = f'(t)dt$, we get
\[ \int f'(t)e^{-st} dt = f(t)e^{-st} + s\int f(t)e^{-st} dt.\]

Now suppose that $f$ also suppose that $\lim_{t \to \infty}f(t)e^{-st} = 0$.  Then we get
\[ \int_0^\infty f'(t)e^{-st} dt = -f(0) + s\int_0^\infty f(t)e^{-st} dt,\] that is, \[ \L(f'(t))(s) = s \L(f(t))(s) - f(0).\]

Now we can replace $f'$ with $f''$ to get
\[\L(f''(t))(s) = s \L(f'(t))(s) - f'(0) = s( s \L(f(t))(s) - f(0)) - f'(0) = s^2 \L(f(t))(s) - sf(0) - f'(0).\]

We can continue in this way, to get the following.

\b

\textbf{Theorem.} Suppose that $f$ is a function having an $n^{th}$ derivative such that $f^{(n)}$ is piecewise continuous, and suppose that each derivative is of exponential order.  Then \[ \L (f^{(n)}(t))(s) = s^n\L(f(t))(s) -s^{n-1}f(0) - s^{n-2}f'(0) - sf^{(n-2)}(0) - f^{(n-1)}(0).\]


\b


This theorem can be used to find Laplace transforms of particular functions.

\b

\textbf{Example.}  Find $\L(\sin t)$. 

\b

Let $f(t) = \sin t$. Then $f'(t) = \cos t$ and $f''(t) = -\sin t$.   By the theorem, $\L(f''(t))(s) = s^2\L(f(t))(s) - sf(0) - f'(0)$.  Therefore, $\L(-\sin t)(s) = s^2 \L(\sin t)(s) - s\cdot 0 - 1$, or $1 = (1 + s^2)\L(\sin t)(s)$.  Therefore, $\L(\sin t)(s) = \frac{1}{1 + s^2}$.





\b

The \textit{inverse Laplace transform} of  a function $F(s)$ is the differentiable function $\L^{-1}(F(s))(t)$ whose Laplace transform is $F(s)$.

\b

The inverse Laplace transform is linear: $\L^{-1}(AF(s) + BG(s))(t) = A\L^{-1}(F(s))(t) + B\L^{-1}(G(s))(t)$. The table on page 317 will help find inverse Laplace transforms.

\b

\textbf{Example.}  Find $\L^{-1}(\frac{3}{s-2} + \frac{2}{s^2+9}).$

\b

$\L^{-1}(\frac{3}{s-2} + \frac{2}{s^2+9})(t) = 3\L^{-1}(\frac{1}{s-2})(t) + (\frac{2}{3})\frac{3}{s^2+9}) = 3e^{2t} + (\frac{2}{3})\sin (3t).$

\b

\textbf{Examples.}


\begin{enumerate}

\item Solve the initial value problem $\dn{x}{t}{2} - 4\d{x}{t} + 4x = 0$, $x(0) = 1$, $x'(0) = 2$.

\b

We take Laplace transforms of both sides of the ODE and use the linearity of the Laplace transform to get \[ \L(x''(t))(s) - 4 \L(x'(t))(s) + 4 \L(x(t)(s) = \L(0).\]  Now using the theorem we get
\[(s^2\L(x(t))(s) - sx(0) - x'(0)) - 4(s\L(x(t))(s)- x(0)) + 4\L(x(t))(s) = 0.\]  This can be written as \[(s^2 - 4s + 4)\L(x(t))(s) -sx(0) - x'(0) + 4x(0) = 0.\]  From the initial conditions, this becomes \[ (s^2 - 4s + 4)\L(x(t))(s) - s - 2 + 4 = 0,\] or \[\L(x(t))(s) = \frac{s - 2}{s^2 - 4s + 4} = \frac{1}{s-2}.\]  Now we need to find a function whose Laplace transform is $\frac{1}{s-2}$.  There are tables that give this information. In particular, there is such a table on page 317 of the text.  According to formula 2 of that table, $x(t) = e^{2t}$ is such a function.  It is differentiable, and it is easy to check that it satisfies both differential equation and the initial conditions.





\item Solve the initial value problem $\dd{x}{t} - 3\d{x}{t} + 2x = \cos t$, $x(0) = 1$, $x'(0) = 0$.

\b

Taking Laplace transforms of both sides gives \[ \L(x'')(s) - 3 \L(x')(s) + 2 \L(x)(s) = \L(\cos t)(s).\]  Then from the theorem we get
\[(s^2\L(x)(s) - sx(0) - x'(0)) - 3(s\L(x)(s)- x(0)) + 2\L(x)(s) = \L(\cos t)(s),\] which simplifies to \[(s^2 - 3s + 2)\L(x)(s) - (sx(0)+x'(0)
-3x(0)) = \L(\cos t)(s).\] From the initial conditions, we get \[ (s^2 - 3s + 2)\L(x)(s) - (s - 3) = \L(\cos t)(s).\]	 From the table on page 317 of the text, we get $\L (\cos t)(s) = \frac{s}{s^2 +1}$, so \[ (s^2 - 3s + 2)\L(x)(s) - (s - 3) = \frac{s}{s^2 +1}.\] Solving gives \[\L(x)(s) = \frac{s - 3}{s^2-3s+2} + \frac{s}{(s^2 +1)(s^2-3s+2)} = \frac{s^3-3s^2+2s-3}{(s^2+1)(s-1)(s-2)} .\]

Now convert the expression on the right to partial fractions.  (We can do this with MAPLE by using the commands:

\begin{verbatim}f:=s->(s^3-3*s^2+2*s-3)/((s^2+1)*(s-1)*(s-2));
convert(f(s),parfrac);\end{verbatim}

We get $\L(x)(s) = \frac{s-3}{10(s^2+1)} - \frac{3}{2(s-1)} - \frac{3}{5(s-2)}$ which we write as
\[ \L(x)(s) = (\frac{1}{10})\frac{s}{s^2 + 1} - (\frac{3}{10})\frac{1}{s^2 + 1} + (\frac{3}{2})\frac{1}{s-1} - (\frac{3}{5})\frac{1}{s-2}.\]	Next we use the table on page 317 again to see which functions have the Laplace transforms on the right to get
\[x = (\frac{1}{10})\cos t - (\frac{3}{10})\sin t + (\frac{3}{2})e^t - (\frac{3}{5})e^{2t}.\]




\end{enumerate}


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