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