# The Laplace Transform: Differential Equations with Variable Coefficients

In this note we study how to solve differential equations with variable coefficients using the Laplace transform. For this we need Derivatives of Transforms. Differentiating
$$f(s)=\int_0^\infty e^{-st}F(t)dt$$
with respect to $s$ we obtain
\begin{align*}
f'(s)&=\int_0^\infty e^{-st}(-tF(t))dt\\
&=\mathcal{L}\{-tF(t)\}
\end{align*}
Continue differentiating to find

\label{eq:laplace14}
f^{(n)}(s)=\mathcal{L}\{(-t)^nF(t)\}

for $n=1,2,\cdots$.

Example. Given that $\mathcal{L}\{\sin kt\}=\frac{k}{s^2+k^2}$, find $\mathcal{L}\{t\sin kt\}$.

Solution. Using \eqref{eq:laplace14}
\begin{align*}
\mathcal{L}\{t\sin kt\}&=-\frac{d}{ds}\frac{k}{s^2+k^2}\\
&=\frac{2ks}{(s^2+k^2)^2}.
\end{align*}

The equation \eqref{eq:laplace14} together with transform of derivative formula allows us to transform differential equations with variable coefficients. For example,
\begin{align*}
\mathcal{L}\{t^nX(t)\}&=(-1)^nx^{(n)}(s)\\
\mathcal{L}\{t^2\dot{X}(t)\}&=\frac{d^2}{ds^2}[sx(s)-X(0)]\\
&=\frac{d}{ds}[x(s)+sx'(s)]\\
&=sx^{\prime\prime}(s)+2x'(s)\\
\mathcal{L}\{t\ddot{X}(t)\}&=-\frac{d}{ds}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=-s^2x'(s)-2sx(s)+X(0)
\end{align*}

We are now ready to solve differential equations with variable coefficients.

Example. Find the solution of the problem
$$\ddot{X}(t)+t\dot{X}(t)-X(t)=0,\ X(0)=0,\ \dot{X}(0)=1$$

Solution. The transformed equation is
$$s^2x(s)-1-\frac{d}{ds}[sx(s)]-x(s)=0$$
which can be written as the first-order linear differential equation
$$\frac{d}{ds}x(s)+\left(\frac{2}{s}-s\right)x(s)=-\frac{1}{s}$$
The integrating factor is
$$\mu(s)=e^{\int\left(\frac{2}{s}-s\right)ds}=s^2e^{-\frac{1}{2}s^2}$$
and hence the solution $x(s)$ is
\begin{align*}
x(s)&=\frac{\int\mu(s)\left(-\frac{1}{2}\right)ds}{\mu(s)}\\
&=\frac{-\int se^{-\frac{1}{2}s^2}ds}{s^2e^{-\frac{1}{2}s^2}}\\
&=\frac{1}{s^2}+\frac{C}{s^2}e^{\frac{1}{2}s^2}
\end{align*}
where $C$ is a constant. Since $x(s)\to 0$ as $s\to\infty$, $C$ must be $0$. Therefore $x(s)=\frac{1}{s^2}$ and consequently $X(t)=t$.

Example. Solve Bessel’s equation with index zero
$$t\ddot{X}(t)+\dot{X}(t)+tX(t)=0$$
with the initial condition $X(0)=1$.

Solution. The transformed equation is
$$-\frac{d}{ds}[s^2x(s)-s-\dot{X}(0)]+sx(s)-1-\frac{d}{ds}x(s)=0$$
which simplifies to the first-order separable differential equation
$$(s^2+1)x'(s)+sx(s)=0$$
Performing the integrals
$$\int\frac{dx}{x}=-\int\frac{sds}{s^2+1}$$
we find
\begin{align*}
x(s)&=\frac{C}{\sqrt{s^2+1}}\\
&=\frac{C}{s}\left(1+\frac{1}{s^2}\right)^{-\frac{1}{2}}\\
&=\frac{C}{2}\sum_{n=0}^\infty\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}\left(\frac{1}{s^2}\right)^n,
\end{align*}
where $C$ is a constant and $s>1$, by the Binomial Theorem.
\begin{align*}
\begin{pmatrix}
-\frac{1}{2}\\
n
\end{pmatrix}&=\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)\cdots\left(-\frac{1}{2}-n+1\right)}{n!}\\
&=\frac{(-1)^n1\cdot 2\cdot 3\cdot 5\cdots(2n-1)}{2^nn!}\\
&=\frac{(-1)^n(2n)!}{(2^nn!)^2}
\end{align*}
So we have
$$x(s)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}$$
Since $x(s)$ is an infinite sum we cannot directly use the linearity of $\mathcal{L}^{-1}$ to obtain $X(t)$. Nonetheless we can show that
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
\begin{align*}
\mathcal{L}\left\{C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}&=C\mathcal{L}\left\{\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\right\}\\
&=C\int_0^\infty e^{-st}\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\int_0^\infty e^{-st}\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\int_0^\infty e^{-st}t^{2n}dt\\
&=C\lim_{k\to\infty}\sum_{n=0}^k\frac{(-1)^n}{(2^nn!)^2}\mathcal{L}\{t^{2n}\}\\
&=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}\frac{(2n)!}{s^{2n+1}}\\
&=x(s)
\end{align*}
By the uniqueness of $\mathcal{L}^{-1}$, we have
$$X(t)=C\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
Since $X(0)=1$, we obtain $C=1$ and hence
$$X(t)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}$$
This series is denoted by $J_0(t)$ i.e.

\begin{aligned}
J_0(t&)=\sum_{n=0}^\infty\frac{(-1)^n}{(2^nn!)^2}t^{2n}\\
&=1-\frac{t^2}{2^2}+\frac{t^4}{2^2\times 4^2}-\frac{t^6}{2^2\times 4^2\times 6^2}+\cdots
\end{aligned}\label{eq:bessel0}

One can easily show using, for example, the ratio test that the series in \eqref{eq:bessel0} converges for all $t$. We now have the Laplace transform

\mathcal{L}\{J_0(t)\}=\frac{1}{\sqrt{s^2+1}}\ (s>1)

The differential equation

\label{eq:besseleqn}
t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-n^2)X(t)=0

is called the Bessel’s equation of index $n$. The solution $X(t)$ is
$$X(t)=CJ_n(t),\ n=0,1,2,\cdots$$
where

\label{eq:besseln}
J_n(t)=\sum_{k=0}^\infty\frac{(-1)^k}{k!(n+k)!}\left(\frac{t}{2}\right)^{n+2k}

$J_n(t)$, $n=0,1,2,\cdots$ is called the Bessel function of the first kind. There is another solution of the Bessel’s equation in \eqref{eq:besseleqn} which is linearly independent from $J_n(t)$. It is in a pretty horrible form

\begin{aligned}
N_n(x)=&\frac{2}{\pi}\left[\ln\left(\frac{x}{2}\right)+\gamma-\frac{1}{2}\sum_{p=1}^n\frac{1}{p}\right]J_n(x)\\
&-\frac{1}{\pi}\sum_{r=0}^\infty\frac{(-1)^r}{r!(n+r)!}\left(\frac{x}{2}\right)^{n+2r}\sum_{p=1}^r\left[\frac{1}{p}+\frac{1}{p+n}\right]\\
&-\frac{1}{\pi}\sum_{r=0}^{n-1}\frac{(n-r-1)!}{r!}\left(\frac{x}{2}\right)^{-n+2r}
\end{aligned}\label{eq:neumann}

where $\gamma$ is the Euler-Mascheroni constant defined by
\begin{align*}
\gamma&=\lim_{n\to\infty}\left(\sum_{m=1}^n\frac{1}{m}-\ln n\right)\\
&\approx 0.57721566\cdots
\end{align*}
$N_n(x)$, $n=0,1,2,\cdots$ is called the Bessel function of the second kind or the Neumann function. Hence the general solution of the Bessel’s equation is given by
$$X(t)=AJ_n(x)+BN_n(x)$$
Those who wish to know more about Bessel functions and Neumann functions may refer to the reference [1] below.

Let us now consider $n=1$ case.
$$t^2\ddot{X}(t)+t\dot{X}(t)+(t^2-1)X(t)=0$$
Using \eqref{eq:laplace14} we obtain
\begin{align*}
\mathcal{L}\{t^2\ddot{X}(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{\ddot{X}(t)\}\\
&=\frac{d^2}{ds^2}[s^2x(s)-sX(0)-\dot{X}(0)]\\
&=s^2x^{\prime\prime}(s)+4sx'(s)+2x(s)\\
\mathcal{L}\{t\dot{X}(t)\}&=-\frac{d}{ds}\mathcal{L}\{\dot{X}(t)\}\\
&=-\frac{d}{ds}[sx(s)-X(0)]\\
&=-x(s)-sx'(s)\\
\mathcal{L}\{t^2X(t)\}&=\frac{d^2}{ds^2}\mathcal{L}\{X(t)\}\\
&=x^{\prime\prime}(s)
\end{align*}
Hence the transformed equation is

\label{eq:laplace15}
(s^2+1)x^{\prime\prime}(s)+3sx'(s)=0

Let $y(s)=x'(s)$. Then \eqref{eq:laplace15} becomes the separable first-order differential equation
$$(s^2+1)y'(s)+3sy(s)=0$$
which can then be written as
$$\frac{dy}{y}=-\frac{3s}{s^2+1}ds$$
Integrating this we find
$$y(s)=\frac{dx}{ds}=\frac{C_1}{(s^2+1)^{\frac{3}{2}}}$$
Using the trigonometric substitution $s=\tan\theta$, we find
$$x(s)=\int\frac{C_1}{(s^2+1)^{\frac{3}{2}}}ds=\frac{C_1s}{\sqrt{s^2+1}}+C_2$$
Since $\lim_{s\to\infty}x(s)=0$, $C_2=-C_1$. Setting $C_1=C$, we have

\begin{aligned}
x(s)&=C\left[\frac{s}{\sqrt{s^2+1}}-1\right]\\
&=C[s\mathcal{L}\{J_0(t)\}-J_0(0)]\\
&=C\mathcal{L}\{J_0′(t)\}
\end{aligned}\label{eq:laplace16}

Hence,
$$X(t)=CJ_0′(t)$$
From \eqref{eq:besseln} we find $J_1(t)=-J_0′(t)$ so

\label{eq:laplace17}
X(t)=-CJ_1(t)

We can obtain the Laplace transform of $J_1(t)$ using \eqref{eq:laplace16} and \eqref{eq:laplace17}.
\begin{align*}
\mathcal{L}\{J_1(t)\}&=1-\frac{s}{\sqrt{s^2+1}}\\
&=\frac{1}{\sqrt{s^2+1}(\sqrt{s^2+1}+s)}
\end{align*}

References:

[1] Mathematical Methods for Physicists, George Arfken, Third Edition, Academic Press, 1985

# The Laplace Transform: Convolution

Definition. The convolution $F\ast G$ of $F(t)$ and $G(t)$ is defined by
$$F(t)\ast G(t)=\int_0^t F(\tau)G(t-\tau)d\tau$$

We introduce Convolution Theorem without a proof. For those who are interested a proof can be found in [1] of the References below.

Theorem. Let $\mathcal{L}\{F(t)\}=f(s)$ and $\mathcal{L}\{G(t)\}=g(s)$. Suppose that $F(t)$ and $G(t)$ are piecewise continuous and are of order $e^{\alpha t}$ as $t\to\infty$. Then $\mathcal{L}\{F(t)\ast G(t)\}$ exists when $s>\alpha$ and it is $f(s)g(s)$. Equivalently,
$$\mathcal{L}^{-1}\{f(s)g(s)\}=F(t)\ast G(t)$$

Example. Let $F(t)=t$ and $G(t)=e^{at}$. Then $\mathcal{L}\{F(t)\}=\frac{1}{s^2}$ and $\mathcal{L}\{G(t)\}=\frac{1}{s-a}$. By Convolution Theorem
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{s^2}\frac{1}{s-a}\right\}&=t\ast e^{at}\\
&=\int_0^t\tau e^{a(t-\tau)}d\tau\\
&=\frac{1}{a^2}(e^{at}-at-1)
\end{align*}
Note that partial fractions can be also used to obtain the result.

When $G(t)=F(t)$, we have the formula
$$[f(s)]^2=\mathcal{L}\{F(t)\ast F(t)\}$$

Example.
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{1}{(s^2+k^2)^2}\right\}&=\frac{1}{k^2}\mathcal{L}^{-1}\left\{\frac{k^2}{(s^2+k^2)^2}\right\}\\
&=\frac{1}{k^2}\int_0^t\sin kt\ast \sin kt\\
&=\frac{1}{k^2}\int_0^t\sin k\tau\sin k(t-\tau)d\tau\\
&=\frac{1}{k^3}\{\sin kt-kt\cos kt\}
\end{align*}

Theorem (Properties of Convolution).

1. $F(t)\ast G(t)=G(t)\ast F(t)$
2. $F(t)\ast[G(t)+H(t)]=F(t)\ast G(t)+F(t)\ast H(t)$
3. $F(t)\ast [kG(t)]=k[F(t)\ast G(t)]$ where $k$ is a constant.
4. $F(t)\ast[G(t)\ast H(t)]=[F(t)\ast G(t)]\ast H(t)$

Proof. We prove only the part 1.
\begin{align*}
F(t)\ast G(t)&=\int_0^t F(\tau)G(t-\tau)d\tau\\
&=\int_0^tF(t-\lambda)G(\lambda)d\lambda\ (\lambda=t-\tau)\\
&=G(t)\ast F(t)
\end{align*}

\begin{align*}
\frac{1}{s}f(s)&=\mathcal{L}\{F(t)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t F(\tau)d\tau\right\}\\
\frac{1}{s^2}f(s)&=\mathcal{L}\{F(t)\ast 1\ast 1\}\\
&=\mathcal{L}\{(F(t)\ast 1)\ast 1\}\\
&=\mathcal{L}\left\{\int_0^t(F(\tau)\ast 1)d\tau\right\}\\
&=\mathcal{L}\left\{\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau\right\}
\end{align*}
Therefore we have the following theorem.

Theorem.
\begin{align}\label{eq:conv}\mathcal{L}^{-1}\left\{\frac{1}{s}f(s)\right\}&=\int_0^tF(\tau)d\tau\\
\label{eq:conv2}\mathcal{L}^{-1}\left\{\frac{1}{s^2}f(s)\right\}&=\int_0^t\int_0^{\tau}F(\lambda)d\lambda d\tau
\end{align}

Convolution Theorem can be used for solving non-homogeneous linear differential equations.

Example. Find the general solution of the differential equations
$$\ddot{X}(t)+k^2X(t)=F(t)$$

Solution. Let $\mathcal{L}\{X(t)\}=x(s)$. Then the transformed equation is
$$s^2x(s)-sX(0)-\dot{X}(0)+k^2x(s)=f(s)$$
So we have
$$x(s)=\frac{1}{k}\frac{k}{s^2+k^2}f(s)+\frac{s}{s^2+k^2}X(0)+\frac{1}{k}\frac{k}{s^2+k^2}\dot{X}(0)$$
Therefore by Convolution Theorem the general solution is given by

\begin{aligned}
X(t)&=\frac{1}{k}(\sin kt)\ast F(t)+X(0)\cos kt+\frac{\dot{X}(0)}{k}\sin kt\\
&=\frac{1}{k}\int_0^t\sin k(t-\tau)F(\tau)d\tau+C_1\cos kt+C_2\sin kt
\end{aligned}\label{eq:conv3}

Let us redo the problem in the last example in here using \eqref{eq:conv3}.

Example. Find the general solution of the differential equation
$$\ddot{x}+4x=8\tan t,\ -\frac{\pi}{2}<x<\frac{\pi}{2}$$

Solution. Using the formula \eqref{eq:conv3} with $k=2$
\begin{align*}
x(t)&=\frac{1}{2}\int_0^t\sin 2(t-\tau)(8\tan\tau)d\tau+C_1\cos 2t+C_2\sin 2t\\
&=4\int_0^t\sin 2(t-\tau)\tan\tau d\tau+C_1\cos 2t+C_2\sin 2t\\
&=C_1\cos 2t+C_2\sin 2t-4t\cos 2t+4\sin 2t\ln(\cos t)
\end{align*}

Convolution Theorem can be also used for solving integral equations as shown in the following example.

Example. Solve the integral equation
$$X(t)=at+\int_0^tX(\tau)\sin(t-\tau)d\tau$$

Solution. $X(t)=at+X(t)\ast \sin t$ so its transformed equation is
$$x(s)=\frac{a}{s^2}+x(s)\frac{1}{s^2+1}$$
Hence we have
\begin{align*}
x(s)&=\frac{a}{s^2}\frac{s^2+1}{s^2}\\
&=a\left(\frac{1}{s^2}+\frac{1}{s^4}\right)
\end{align*}
Therefore the solution is given by
$$X(t)=a\left(t+\frac{1}{6}t^3\right)$$

References:

[1] Ruel V. Churchill, Operational Mathematics, McGraw-Hill, 1958

# The Laplace Transform: Further Properties

Let $\mathcal{L}\{F(t)\}=f(s)$ i.e. $f(s)=\int_0^\infty e^{-st}F(t)dt$. Then

\label{eq:laplace11}
e^{-bs}f(s)=\int_0^\infty e^{-s(t+b)}F(t)dt

where $b$ is a positive constant. Using the substitution $\tau=t+b$ \eqref{eq:laplace11} can be written as
$$e^{-bs}f(s)=\int_b^\infty e^{-s\tau}F(\tau-b)d\tau$$
Define a function $F_b(\tau)$ by

F_b(\tau)=\left\{\begin{aligned}
0,\ &0<\tau<b\\
F(\tau-b),\ &\tau>b
\end{aligned}\right.\label{eq:laplace12}

Then
$$e^{-bs}f(s)=\int_0^\infty e^{-s\tau}F_b(tau)d\tau$$

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ then for any positive constant $b$,
$$e^{-bs}f(s)=\mathcal{L}\{F_b(t)\}$$
where $F_b(t)$ is defined by \eqref{eq:laplace12}.

Let
S_b(t)=\left\{\begin{aligned} 0,\ &0<t<b\\ 1,\ &t>b \end{aligned}\right.
for any positive constant $b$. Define $S_0(t)=1$ for $t>0$. Then $S_b(t)$ is called a \emph{unit step function}. Using this unit step function, $F_b(t)$ can be written as
$$F_b(t)=S_b(t)F(t-b)$$

Example.
\begin{align*}
F_b(t)&=S_b(t)\sin k(t-b)\\
&=\mathcal{L}^{-1}\left\{\frac{ke^{-bs}}{s^2+k^2}\right\}
\end{align*}

\begin{align*}
\mathcal{L}\{F(at)\}&=\int_0^\infty e^{-st}F(at)dt\ (a>0)\\
&=\frac{1}{a}\int_0^\infty e^{-\frac{s}{a}\tau}F(\tau)d\tau\ (\tau=at)\\
&=\frac{1}{a}f\left(\frac{s}{a}\right)
\end{align*}

Theorem. If $\mathcal{L}\{F(t)\}=f(s)$ when $s>\alpha$, then
$$\mathcal{L}\{F(at)\}=\frac{1}{a}f\left(\frac{s}{a}\right)\ (s>a\alpha,\ a>0)$$

Example. Given that
$$\frac{s}{s^2+1}=\mathcal{L}\{\cos t\}$$
\begin{align*}
\frac{s}{s^2+k^2}&=\frac{1}{k^2}\frac{s}{\left(\frac{s}{k}\right)^2+1}\\
&=\frac{1}{k}\frac{\frac{s}{k}}{\left(\frac{s}{k}\right)^2+1}\\
&=\mathcal{L}\{\cos kt\}
\end{align*}

With the translation of transform (equation (1) in here) and the above theorem the effect of a general linear substitution for $s$ can be shown in the following formula.
\begin{align*}
f(as-b)&=f\left[a\left(s-\frac{b}{a}\right)\right]\\
&=\mathcal{L}\left\{\frac{1}{a}e^{\frac{b}{a}t}F\left(\frac{t}{a}\right)\right\}\ (a>0)
\end{align*}

# The Laplace Transform: Solving Differential Equations

In this note we study how the Laplace transform can be used to solve differential equations. For now we only consider linear differential equations with constant coefficients. But later on we will see that the Laplace transform is also a powerful tool to solve linear differential equations with variable coefficients.

Example. Find the general solution of the differential equation
$$\ddot{X}(t)+k^2X(t)=0$$

Solution. Using the linearity transform the differential equation.

\label{eq:laplace10}
\mathcal{L}\{\ddot{X}(t)\}+k^2\mathcal{L}\{X(t)\}=0

By the transform of derivative formula
$$\mathcal{L}\{\ddot{X}(t)\}=s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)$$
Let $X(0)=A$ and $\dot{X}(0)=B$. Then \eqref{eq:laplace10} becomes an algebraic equation of $\mathcal{L}\{X(t)\}$
$$(s^2+k^2)\mathcal{L}\{X(t)\}-As-B=0$$
Thus
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{As+B}{s^2+k^2}\\
&=A\frac{s}{s^2+k^2}+\frac{B}{k}\frac{k}{s^2+k^2}
\end{align*}
Hence,
$$X(t)=A\cos kt+B’\sin kt$$
where $B’=\frac{B}{k}$.

Example. Find the solution of the initial value problem
$$\ddot{X}(t)-\dot{X}(t)-6X(t)=2$$
with $X(0)=1$ and $\dot{X}(0)=0$.

Solution. The transform of the differential equation, with the aid of the transform of derivative formula, is
$$[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]-[s\mathcal{L}\{X(t)\}-X(0)]-6\mathcal{L}\{X(t)\}=\frac{2}{s}$$
This simplifies to
\begin{align*}
(s^2-s-6)\mathcal{L}\{X(t)\}&=\frac{2}{s}+s-1\\
&=\frac{s^2-s+2}{s}
\end{align*}
So
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{s^2-s+2}{s(s^2-s-6)}\\
&=\frac{s^2-s+2}{s(s-3)(s+2)}\\
&=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s+2}
\end{align*}
This results in the equation
$$A(s-3)(s+2)+Bs(s+2)+Cs(s-3)=s^2-s+2$$
For $s=0$, we obtain $-6A=2$ i.e $A=-\frac{1}{3}$. For $s=3$, we obtain $15B=8$ i.e. $B=\frac{15}{8}$. For $s=-2$, we obtain $10C=8$ i.e $C=\frac{4}{5}$.
Hence,
$$\mathcal{L}\{X(t)\}=-\frac{1}{3}\frac{1}{s}+\frac{15}{8}\frac{1}{s-3}+\frac{4}{5}\frac{1}{s+2}$$
and therefore
$$X(t)=-\frac{1}{3}+\frac{15}{8}e^{3t}+\frac{4}{5}e^{-2t}$$

Example. Solve the problem
$$\dddot{X}(t)-2\ddot{X}(t)+5\dot{X}(t)=0$$
with $X(0)=0$, $\dot{X}(0)=1$, and $X\left(\frac{\pi}{8}\right)=1$.

Solution. The transformed equation, with the aid ofthe transform of derivative formula, is
\begin{align*}
[s^3\mathcal{L}\{X(t)\}-s^2X(0)-s\dot{X}(0)-\ddot{X}(0)]&-2[s^2\mathcal{L}\{X(t)\}-sX(0)-\dot{X}(0)]\\&
+5[s\mathcal{L}\{X(t)\}-X(0)]=0
\end{align*}
This simplifies, with the initial conditions $X(0)=0$ and $\dot{X}(0)=1$, to the equation
$$(s^3-2s^2+5s)\mathcal{L}\{X(t)\}-s+2-\ddot{X}(0)=0$$
Let $\ddot{X}(0)=C$. Then
\begin{align*}
\mathcal{L}\{X(t)\}&=\frac{C-2+s}{s^3-2s^2+5s}\\
&=\frac{C-2+s}{s(s^2-2s^+5)}\\
&=\frac{C-2+s}{s[(s-1)^2+4]}\\
&=\frac{C-2}{5}\left[\frac{1}{s}-\frac{s-1}{(s-1)^2+4}\right]+\frac{C+3}{10}\frac{2}{(s-1)^2+4}
\end{align*}
Hence
$$X(t)=\frac{C-2}{5}+e^t\left(\frac{C+3}{10}\sin 2t-\frac{C-2}{5}\cos 2t\right)$$
The condition $X\left(\frac{\pi}{8}\right)=1$ results in the equation
$$1=\frac{C-2}{5}+\frac{e^{\frac{\pi}{8}}}{10\sqrt{2}}(-C+7)$$ whose solution is $C=7$. Therefore,
$$X(t)=1+e^t(\sin 2t-\cos 2t)$$

# The Laplace Transform: The Inverse Transforms

Let $\mathcal{L}^{-1}\{f(s)\}$ denote a function whose Laplace transform is $f(s)$ i.e. if $\mathcal{L}\{F(t)\}=f(s)$ then $F(t)=\mathcal{L}^{-1}\{f(s)\}$. For example,
$$\mathcal{L}^{-1}\left\{\frac{1}{s-k}\right\}=e^{kt},\ \mathcal{L}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin kt$$
$F(t)$ is called the inverse transform of $f(s)$. There is a question that must be addressed. Is the inverse transform of $f(s)$ unique? The answer is not really. For example, we know that $F_1(t)=e^{kt}$ is an inverse transform of $\frac{1}{s-k}$. Define
F_2(t)=\left\{\begin{aligned} e^{kt},\ &0<t<2,\ t>2\\ 1,\ &t=2 \end{aligned}\right.
The transform of $F_2(t)$ is
\begin{align*}
\mathcal{L}\{F_2(t)\}&=\int_0^\infty e^{-st}F_2(t)dt\\
&=\int_0^2 e^{-st}e^{kt}dt+\int_2^\infty e^{-st}e^{kt}dt\\
&=\frac{1}{s-k}
\end{align*}
So are we in trouble then? Not really. We introduce the following theorem without proof.

Theorem. If two functions $F_1(t)$ and $F_2(t)$ have the same Laplace transform, then $F_2(t)=F_1(t)+N(t)$ where $N(t)$ satisfies
$$\int_0^T N(t)dt=0$$
for every positive $T$.

Such a function $N(t)$ is called a null function. In the above example,
N(t)=\left\{\begin{aligned} 1-e^{2k},\ &=t=2\\ 0,\ &t\ne 2 \end{aligned}\right.
In view of this theorem the inverse transform is essentially unique because a null function is not important in the applications. For this reason this theorem is called the uniqueness of the inverse transform.

Due to the linearity of $\mathcal{L}$, we have
\begin{align*}
\mathcal{L}\{AF(t)+BG(t)\}&=A\mathcal{L}\{F(t)\}+B\mathcal{L}\{G(t)\}\\
&=Af(s)+Bg(s)
\end{align*}
Now
\begin{align*}
\mathcal{L}^{-1}\{Af(s)+Bg(s)\}&=AF(t)+BG(t)\\
&=A\mathcal{L}^{-1}\{f(s)\}+B\mathcal{L}^{-1}\{g(s)\}
\end{align*}
so $\mathcal{L}^{-1}$ is also linear.

Suppose that the Laplace transform of $F(t)$ converges when $s>k$.
$$f(s)=\int_0^\infty e^{-st}F(t)dt\ (s>k)$$
Substitute $s-a$ for $s$. Then
\begin{align*}
f(s-a)&=\int_0^\infty e^{-(s-a)t}F(t)dt\\
&=\int_0^\infty e^{-st}e^{at}F(t)dt\\
&=\mathcal{L}\{e^{at}F(t)\}
\end{align*}
Therefore

\mathcal{L}\{e^{at}F(t)\}=f(s-a),\ s>a+k

Example.
$$\mathcal{L}\{e^{at}t^n\}=\frac{n!}{(s-a)^{n+1}},\ s>a$$

Example.
$$\mathcal{L}\{e^{-at}\cos kt\}=\frac{s+a}{(s+a)^2+k^2},\ s>-a$$

Partial fractions often play a useful role in finding $\mathcal{L}^{-1}$ as seen in the following examples.

Example. Find $\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}$

Solution. $s^2+2s=s(s+1)$ so let
$$\frac{s+1}{s^2+2s}=\frac{A}{s}+\frac{B}{s+2}$$
Then we have
\begin{align*}
s+1&=A(s+2)+Bs\\
&=(A+B)s+2A
\end{align*}
This means $A+B=1$ and $2A=1$ i.e. $A=B=\frac{1}{2}$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{s+1}{s^2+2s}\right\}&=\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}+\frac{1}{2}\mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\}\\
&=\frac{1}{2}+\frac{1}{2}e^{-2t}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}$ ($a\ne 0$).

Solution. Let
$$\frac{a^2}{s(s+a)^2}=\frac{A}{s}+\frac{B}{s+a}+\frac{C}{(s+a)^2}$$
Then we have
$$a^2=A(s+a)^2+Bs(s+a)+Cs$$
For $s=0$, $a^2=Aa^2$ i.e. $A=1$. For $s=-a$, $a^2=-Ca$ i.e. $C=-a$. For $s=a$, $a^2=3a^2+2a^2B$ which is reduced to $-2a^2=2a^2B$ i.e. $B=-1$. Hence
\begin{align*}
\mathcal{L}^{-1}\left\{\frac{a^2}{s(s+a)^2}\right\}&=\mathcal{L}^{-1}\left\{\frac{1}{2}\right\}-\mathcal{L}^{-1}\left\{\frac{1}{s+a}\right\}-a\mathcal{L}^{-1}\left\{\frac{1}{(s+a)^2}\right\}\\
&=1-e^{-at}-ate^{-at}
\end{align*}

Example. Find $\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2}\right\}$ ($a^2\ne b^2$).

$$\frac{s}{(s^2+a^2)(s^2+b^2)}=\frac{As+B}{s^2+a^2}+\frac{Cs+D}{s^2+b^2}$$
However one can write
\begin{align*}
\frac{s}{(s^2+a^2)(s^2+b^2)}&=\frac{s}{a^2-b^2}\left[\frac{s^2+a^2-(s^2+b^2)}{(s^2+a^2)(s^2+b^2)}\right]\\
&=\frac{1}{b^2-a^2}\left[\frac{s}{s^2+a^2}-\frac{s}{s^2+b^2}\right]
\end{align*}
Hence
$$\mathcal{L}^{-1}\left\{\frac{s}{(s^2+a^2)(s^2+b^2)}\right\}=\frac{1}{b^2-a^2}(\cos at-\cos bt)$$

Example. Find $F(t)$ is $f(s)=\frac{5s+3}{(s-1)(s^2+2s+5)}$.

Solution. Let
$$\frac{5s+3}{(s-1)(s^2+2s+5)}=\frac{A}{s-1}+\frac{Bs+C}{s^2+2s+5}$$
Then we obtain $A=1$, $B=-1$ and $C=2$.
\begin{align*}
f(s)&=\frac{1}{s-1}-\frac{s-2}{s^2+2s+5}\\
&=\frac{1}{s-1}-\frac{s-2}{(s+1)^2+4}\\
&=\frac{1}{s-1}-\frac{s+1-3}{(s+1)^2+2^2}\\
&=\frac{1}{s-1}-\frac{s+1}{(s+1)^2+2^2}+\frac{3}{2}\frac{2}{(s+1)^2+2^2}
\end{align*}
Hence
$$F(t)=e^t-e^{-t}\cos 2t+\frac{3}{2}e^{-t}\sin 2t$$