Math Help - DE Tutorial - Part IV: Laplace Transforms

1. DE Tutorial - Part IV: Laplace Transforms

The DE Tutorial is currently being split up into different threads to make editing these posts easier.

Laplace Transforms (Part I - Introduction, IVPs and Partial Fraction Techniques)

There are many types of transformations out there. For example, differentiation and integration are types of linear transformations. However, there is one particular transform that we would like to analyze. This transform is of the form:

$\int_0^{\infty}K\!\left(s,t\right)f\!\left(t\right )\,dt$

where $K\!\left(s,t\right)$is called the kernel of the transformation.

In this case, we are interested in the transform with a kernel of $K\!\left(s,t\right)=e^{-st}$. With this kernel, we take $f\!\left(t\right)$ and transform it into another function $F\!\left(s\right)$. This transformation described by $\int_0^{\infty}e^{-st}f\!\left(t\right)\,dt=F\!\left(s\right)$ is called the Laplace Transform. It is denoted by $\mathcal{L}\left\{f\!\left(t\right)\right\}$.

Before we go and derive all the common Laplace Transforms (we will derive many more as we get futher into later posts), let us take a look at a familar function to some of us (this may also be totally knew to some of you out there).

Given $x\in\mathbb{R}$, where $x>0$, we define the Gamma Function $\Gamma\!\left(x\right)=\int_0^{\infty}e^{-t}t^{x-1}\,dt$. It has the property $\Gamma\!\left(1\right)=1$ and $\Gamma\!\left(x+1\right)=x\Gamma\!\left(x\right)$.

Now, if $n\in\mathbb{N}$, then it follows by a similar idea that $\Gamma\left(n+1\right)=n\Gamma\!\left(n\right)$. If we continue simplifying, we have

\begin{aligned}n\Gamma\!\left(n\right) & = n\left(n-1\right)\Gamma\!\left(n-1\right)\\ &= n\left(n-1\right)\left(n-2\right)\Gamma\!\left(n-2\right)\\ &\vdots\\ &=n\left(n-1\right)\left(n-2\right)\dots2\cdot\Gamma\!\left(2\right)\\ &= n\left(n-1\right)\left(n-2\right)\dots2\cdot1\cdot\Gamma\!\left(1\right)\en d{aligned}

This implies that when $n\in\mathbb{N}$, $\Gamma\!\left(n+1\right)=n!$.

(Thus it is interesting to point out that since $1\in\mathbb{N},\,\Gamma\!\left(1\right)=\color{red }\boxed{1=0!}$, an identity for factorials.)

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Common Laplace Transforms

In this part, I will list the common Laplace Transforms, and leave the derivation of each in a spoiler for you to look at if you decide too.

$\mathcal{L}\left\{1\right\}=\frac{1}{s},\,s>0$

Spoiler:

$\mathcal{L}\left\{1\right\}=\int_0^{\infty}e^{-st}\,dt=\lim_{b\to\infty}\left.\left[-\frac{e^{-st}}{s}\right]\right|_0^{b}$ $=-\frac{1}{s}\left[\lim_{b\to\infty}e^{-bs}-e^{0}\right]=\frac{1}{s}$ where $s>0$

$\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a},\,s>a$ (This will pop up again, when we talk about translation theorems)

Spoiler:

$\mathcal{L}\left\{e^{at}\right\}=\int_0^{\infty}e^ {-st}e^{at}\,dt=\int_0^{\infty}e^{-\left(s-a\right)t}\,dt=\lim_{b\to\infty}\left.\left[-\frac{e^{-\left(s-a\right)t}}{s-a}\right]\right|_0^{b}$ $=-\frac{1}{s-a}\left[\lim_{b\to\infty}e^{-b\left(s-a\right)}-e^{0}\right]=\frac{1}{s-a}$ where $s>a$

$\mathcal{L}\left\{t^a\right\}=\frac{\Gamma\left(a+ 1\right)}{s^{a+1}},\,s>0$; If $n\in\mathbb{N},\,\mathcal{L}\left\{t^n\right\}=\fr ac{\Gamma\!\left(n+1\right)}{s^{n+1}}=\frac{n!}{s^ {n+1}}$

Spoiler:

$\mathcal{L}\left\{t^a\right\}=\int_0^{\infty}e^{-st}t^a\,dt$

Let $u=st\implies t=\frac{u}{s}\implies \,dt=\frac{\,du}{s}$

Thus, $\int_0^{\infty}e^{-st}t^a\,dt\xrightarrow{u=st}{}\int_0^{\infty}e^{-u}\left(\frac{u}{s}\right)^a\left(\frac{\,du}{s}\r ight)=\frac{1}{s^{a+1}}\int_0^{\infty}e^{-u}u^a\,du=\frac{\Gamma\!\left(a+1\right)}{s^{a+1}}$ where $s>0$

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\cosh\!\left(kt \right)\right\}=\frac{s}{s^2-k^2},\,s>k>0$

Spoiler:

Recall that $\cosh\!\left(kt\right)=\frac{e^{kt}+e^{-kt}}{2}$. Therefore,

$\mathcal{L}\left\{\cosh\!\left(kt\right)\right\}=\ mathcal{L}\left\{\frac{e^{kt}+e^{-kt}}{2}\right\}$

Since the Laplace Transform is linear (i.e. $\mathcal{L}\left\{af\!\left(t\right)+bg\!\left(t\r ight)\right\}=a\mathcal{L}\left\{f\!\left(t\right) \right\}+b\mathcal{L}\left\{g\!\left(t\right)\righ t\}$), we have

$\mathcal{L}\left\{\frac{e^{kt}+e^{-kt}}{2}\right\}=\frac{1}{2}\left(\mathcal{L}\left\ {e^{kt}\right\}+\mathcal{L}\left\{e^{-kt}\right\}\right)$ $=\frac{1}{2}\left(\frac{1}{s-k}+\frac{1}{s+k}\right)=\frac{1}{2}\left(\frac{2s} {s^2-k^2}\right)=\frac{s}{s^2-k^2}$ where $s>k>0$

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\sinh\!\left(kt \right)\right\}=\frac{k}{s^2-k^2},\,s>k>0$

Spoiler:

Recall that $\sinh\!\left(kt\right)=\frac{e^{kt}-e^{-kt}}{2}$. Therefore,

$\mathcal{L}\left\{\sinh\!\left(kt\right)\right\}=\ mathcal{L}\left\{\frac{e^{kt}-e^{-kt}}{2}\right\}$

Since the Laplace Transform is linear, we have

$\mathcal{L}\left\{\frac{e^{kt}-e^{-kt}}{2}\right\}=\frac{1}{2}\left(\mathcal{L}\left\ {e^{kt}\right\}-\mathcal{L}\left\{e^{-kt}\right\}\right)$ $=\frac{1}{2}\left(\frac{1}{s-k}-\frac{1}{s+k}\right)=\frac{1}{2}\left(\frac{2k}{s^ 2-k^2}\right)=\frac{k}{s^2-k^2}$ where $s>k>0$

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\cos\!\left(kt\ right)\right\}=\frac{s}{s^2+k^2},\,s>0$

Spoiler:

In exponential form, $\cos\!\left(kt\right)=\frac{e^{ikt}+e^{-ikt}}{2}$. Therefore,

$\mathcal{L}\left\{\cos\!\left(kt\right)\right\}=\m athcal{L}\left\{\frac{e^{ikt}+e^{-ikt}}{2}\right\}$

Since the Laplace Transform is linear, we have

$\mathcal{L}\left\{\frac{e^{ikt}+e^{-ikt}}{2}\right\}=\frac{1}{2}\left(\mathcal{L}\left \{e^{ikt}\right\}+\mathcal{L}\left\{e^{-ikt}\right\}\right)$ $=\frac{1}{2}\left(\frac{1}{s-ik}+\frac{1}{s+ik}\right)=\frac{1}{2}\left(\frac{2 s}{s^2-(ik)^2}\right)=\frac{s}{s^2+k^2}$ where $s>0$

Given $k\in\mathbb{R},\,\mathcal{L}\left\{\sin\!\left(kt\ right)\right\}=\frac{k}{s^2+k^2},\,s>0$

Spoiler:

In exponential form, $\sin\!\left(kt\right)=\frac{e^{ikt}-e^{-ikt}}{2i}$. Therefore,

$\mathcal{L}\left\{\sin\!\left(kt\right)\right\}=\m athcal{L}\left\{\frac{e^{ikt}-e^{-ikt}}{2i}\right\}$

Since the Laplace Transform is linear, we have

$\mathcal{L}\left\{\frac{e^{ikt}-e^{-ikt}}{2i}\right\}=\frac{1}{2i}\left(\mathcal{L}\le ft\{e^{ikt}\right\}-\mathcal{L}\left\{e^{-ikt}\right\}\right)$ $=\frac{1}{2i}\left(\frac{1}{s-ik}-\frac{1}{s+ik}\right)=\frac{1}{2i}\left(\frac{2ik} {s^2-(ik)^2}\right)=\frac{k}{s^2+k^2}$ where $s>0$

Given $a\in\mathbb{R},\,\mathcal{L}\left\{u\!\left(t-a\right)\right\}=\frac{e^{-as}}{s},\,s>0,\,a>0$

Spoiler:

Since $u\!\left(t-a\right)=\left\{\begin{matrix}0 & \text{for }t, it follows that

$\mathcal{L}\left\{u\!\left(t-a\right)\right\}=\int_0^{\infty}e^{-st}u\!\left(t-a\right)\,dt=\int_a^{\infty}e^{-st}\,dt=\lim_{b\to\infty}\left.\left[-\frac{e^{-st}}{s}\right]\right|_a^b$ $=-\frac{1}{s}\lim_{b\to\infty}e^{-st}+\frac{e^{-as}}{s}=\frac{e^{-as}}{s}$ where $s>0,\, a>0$

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Let us go through some examples on how to apply linearity and some of these formulas.

Example 31

Find the Laplace Transform of $f\!\left(t\right)=3t^{5/2}-4t^3$

By linearity, we have $\mathcal{L}\!\left\{3t^{5/2}-4t^3\right\}=3\mathcal{L}\left\{t^{5/2}\right\}-4\mathcal{L}\left\{t^3\right\}=3\frac{\Gamma\!\lef t(\tfrac{7}{2}\right)}{s^{7/2}}-4\frac{3!}{s^4}$

Taking into consideration Gamma function properties, we have $\Gamma\!\left(\tfrac{7}{2}\right)=\tfrac{5}{2}\Gam ma\!\left(\tfrac{5}{2}\right)=\tfrac{15}{4}\Gamma\ !\left(\tfrac{3}{2}\right)=\tfrac{15}{8}\Gamma\!\l eft(\tfrac{1}{2}\right)$.

Its not hard to show that $\Gamma\!\left(\tfrac{1}{2}\right)=\sqrt{\pi}$. Therefore, $\Gamma\!\left(\tfrac{7}{2}\right)=\frac{15\sqrt{\p i}}{8}$.

Thus, $\color{red}\boxed{\mathcal{L}\!\left\{3t^{5/2}-4t^3\right\}=\frac{15\sqrt{\pi}}{8s^{7/2}}-\frac{24}{s^4}=\frac{15\sqrt{\pi s}-192}{8s^4}}$

Example 32

Find the Laplace Transform of $f\!\left(t\right)=\sin\!\left(3t\right)\cos\!\left (3t\right)$

Note that $\sin\!\left(3t\right)\cos\!\left(3t\right)=\tfrac{ 1}{2}\sin\!\left(6t\right)$

Therefore, $\mathcal{L}\left\{\sin\!\left(3t\right)\cos\!\left (3t\right)\right\}=\mathcal{L}\left\{\tfrac{1}{2}\ sin\!\left(6t\right)\right\}=\tfrac{1}{2}\mathcal{ L}\left\{\sin\!\left(6t\right)\right\}=\color{red} \boxed{\frac{3}{s^2+36}}$

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Inverse Laplace Transforms

As the name suggests, the Inverse Laplace Transform applied to a function $F(s)$ will give you the original $f\!\left(t\right)$:

$\mathcal{L}\left\{f\!\left(t\right)\right\}=F\!\le ft(s\right)\implies \mathcal{L}^{-1}\left\{\mathcal{L}\left\{f\!\left(t\right)\right \}\right\}=\mathcal{L}^{-1}\left\{F\!\left(s\right)\right\}\implies \mathcal{L}^{-1}\left\{F\!\left(s\right)\right\}=f\!\left(t\righ t)$

We now list the common inverse Laplace Transforms:

$\mathcal{L}^{-1}\left\{\frac{1}{s}\right\}=1$

$\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{at}$

$\mathcal{L}^{-1}\left\{\frac{\Gamma\!\left(a+1\right)}{s^{a+1}}\ right\}=t^a$

$\mathcal{L}^{-1}\left\{\frac{n!}{s^{n+1}}\right\}=t^n$

$\mathcal{L}^{-1}\left\{\frac{s}{s^2-k^2}\right\}=\cosh\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{k}{s^2-k^2}\right\}=\sinh\!\left(kt\right)$

$\mathcal{L}^{-1}\left\{\frac{s}{s^2+k^2}\right\}=\cos\!\left(kt\ right)$

$\mathcal{L}^{-1}\left\{\frac{k}{s^2+k^2}\right\}=\sin\!\left(kt\ right)$

$\mathcal{L}^{-1}\left\{\frac{e^{-as}}{s}\right\}=u\!\left(t-a\right)$

It is also worth mentioning that the Inverse Laplace Transform is linear.

Let us now go through a couple examples.

Example 33

Find the Inverse Laplace Transform of $F\!\left(s\right)=2s^{-1}e^{-3s}$

$\mathcal{L}^{-1}\left\{\frac{2e^{-3s}}{s}\right\}=2\mathcal{L}^{-1}\left\{\frac{e^{-3s}}{s}\right\}=\color{red}\boxed{2u\!\left(t-3\right)}$

Example 34

Find the Inverse Laplace Transform of $F\!\left(s\right)=s^{-3/2}$

$\mathcal{L}^{-1}\left\{\frac{1}{s^{3/2}}\right\}=\frac{1}{\Gamma\!\left(\tfrac{3}{2}\ri ght)}\mathcal{L}^{-1}\left\{\frac{\Gamma\!\left(\tfrac{3}{2}\right)}{ s^{3/2}}\right\}=\color{red}\boxed{2\sqrt{\frac{t}{\pi} }}$

Example 35

Find the Inverse Laplace Transform of $F\!\left(s\right)=\frac{10s-3}{25-s^2}$

$\mathcal{L}^{-1}\left\{\frac{10s-3}{25-s^2}\right\}=-10\mathcal{L}^{-1}\left\{\frac{s}{s^2-25}\right\}+3\mathcal{L}\left\{\frac{1}{s^2-25}\right\}$ $=-10\mathcal{L}^{-1}\left\{\frac{s}{s^2-25}\right\}+\tfrac{3}{5}\mathcal{L}\left\{\frac{5} {s^2-25}\right\}=\color{red}\boxed{-10\cosh\!\left(5t\right)+\tfrac{3}{5}\sinh\!\left( 5t\right)}$

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Laplace Transforms and IVPs (involving Partial Fraction Techniques)

We now introduce a method of solving initial value problems with Laplace Transforms. Before we go through this method, we first need to find the Laplace Transforms for $f^{\prime}\!\left(t\right)$, $f^{\prime\prime}\!\left(t\right)$, and in general $f^{\left(n\right)}\!\left(t\right)$

I will leave the derivation of each in a spoiler.

$\mathcal{L}\left\{f^{\prime}\!\left(t\right)\right \}=sF\!\left(s\right)-f\!\left(0\right)$

Spoiler:

$\mathcal{L}\left\{f^{\prime}\!\left(t\right)\right \}=\int_0^{\infty}e^{-st}f^{\prime}\!\left(t\right)\,dt$

Let $u=e^{-st}$ and $\,dv=f^{\prime}\!\left(t\right)\,dt$. Thus, $\,du=-se^{-st}\,dt$ and $v=f\!\left(t\right)$.

Therefore, $\int_0^{\infty}e^{-st}f^{\prime}\!\left(t\right)\,dt=\lim_{b\to\infty }\left.\left[e^{-st}f\!\left(t\right)\right]\right|_0^{\infty}+s\int_0^{\infty}e^{-st}f\!\left(t\right)\,dt$ $=-f\!\left(0\right)+s\mathcal{L}\left\{f\!\left(t\ri ght)\right\}=sF\!\left(s\right)-f\!\left(0\right)$

$\mathcal{L}\left\{f^{\prime\prime}\!\left(t\right) \right\}=s^2F\!\left(s\right)-sf\!\left(0\right)-f^{\prime}\!\left(0\right)$

Spoiler:

$\mathcal{L}\left\{f^{\prime\prime}\!\left(t\right) \right\}=\int_0^{\infty}e^{-st}f^{\prime\prime}\!\left(t\right)\,dt$

Let $u=e^{-st}$ and $\,dv=f^{\prime\prime}\!\left(t\right)\,dt$. Thus, $\,du=-se^{-st}\,dt$ and $v=f^{\prime}\!\left(t\right)$.

Therefore, $\int_0^{\infty}e^{-st}f^{\prime}\!\left(t\right)\,dt=\lim_{b\to\infty }\left.\left[e^{-st}f^{\prime}\!\left(t\right)\right]\right|_0^{\infty}+s\int_0^{\infty}e^{-st}f^{\prime}\!\left(t\right)\,dt$ $=-f^{\prime}\!\left(0\right)+s\mathcal{L}\left\{f^{\ prime}\!\left(t\right)\right\}=s\left(sF\!\left(s\ right)-f\!\left(0\right)\right)-f^{\prime}\!\left(0\right)=s^2F\!\left(s\right)-sf\!\left(0\right)-f^{\prime}\!\left(0\right)$

$\mathcal{L}\left\{f^{\left(n\right)}\right\}=s^nF\ !\left(s\right)-s^{n-1}f\!\left(0\right)-s^{n-2}f^{\prime}\!\left(0\right)-\dots-sf^{\left(n-2\right)}\!\left(0\right)-f^{\left(n-1\right)}\!\left(0\right)$

Spoiler:

$\mathcal{L}\left\{f^{\left(n\right)}\!\left(t\righ t)\right\}=\int_0^{\infty}e^{-st}f^{\left(n\right)}\!\left(t\right)\,dt$

Let $u=e^{-st}$ and $\,dv=f^{\left(n\right)}\!\left(t\right)$. Thus, $\,du=-se^{-st}\,dt$ and $v=f^{\left(n-1\right)}\!\left(t\right)$

Therefore, $\int_0^{\infty}e^{-st}f^{\left(n\right)}\!\left(t\right)\,dt=\lim_{b\ to\infty}\left.\left[e^{-st}f^{\left(n-1\right)}\!\left(t\right)\right]\right|_0^b+s\int_0^{\infty}e^{-st}f^{\left(n-1\right)}\!\left(t\right)\,dt$ $=s\mathcal{L}\left\{f^{\left(n-1\right)}\!\left(t\right)\right\}-f^{\left(n-1\right)}\!\left(0\right)$.

Similarly, we see that $\mathcal{L}\left\{f^{\left(n-1\right)}\!\left(t\right)\right\}=s\mathcal{L}\lef t\{f^{\left(n-2\right)}\!\left(t\right)\right\}-f^{\left(n-2\right)}\!\left(0\right)$

So, $\mathcal{L}\left\{f^{\left(n\right)}\!\left(t\righ t)\right\}=s\left(s\mathcal{L}\left\{f^{\left(n-2\right)}\!\left(t\right)\right\}-f^{\left(n-2\right)}\!\left(0\right)\right)-f^{\left(n-1\right)}\!\left(0\right)$ $=s^2\mathcal{L}\left\{f^{\left(n-2\right)}\!\left(t\right)\right\}-sf^{\left(n-2\right)}\!\left(0\right)-f^{\left(n-1\right)}\!\left(0\right)$

After you do this $n$ times, we have

$\mathcal{L}\left\{f^{\left(n\right)}\!\left(t\righ t)\right\}=s^nF\!\left(s\right)-s^{n-1}f\!\left(0\right)-s^{n-2}f^{\prime}\!\left(0\right)-\dots-sf^{\left(n-2\right)}\!\left(0\right)-f^{\left(n-1\right)}\!\left(0\right)$

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Translation Theorem

We will discuss an important translation theorem:

Theorem: If $F\!\left(s\right)=\mathcal{L}\left\{f\!\left(t\rig ht)\right\}$ exists for $s>c$, then $\mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}$ exists for $s>a+c$ and $\mathcal{L}\left\{e^{at}f\!\left(t\right)\right\}= F\!\left(s-a\right)\implies \mathcal{L}^{-1}\left\{F\!\left(s-a\right)\right\}=e^{at}f\!\left(t\right)$.

Pf: Its obvious that $F\!\left(s-a\right)=\int_0^{\infty}e^{-\left(s-a\right)t}f\!\left(t\right)\,dt=\int_0^{\infty}e^{-st}\left[e^{at}f\!\left(t\right)\right]\,dt=\mathcal{L}\left\{e^{at}f\!\left(t\right)\rig ht\}.\quad\square$

As a result of this translation theorem, we have six more Laplace Transforms to add to the list (I leave it for you to verify them):

$\mathcal{L}\left\{e^{at}t^{k}\right\}=\frac{\Gamma \!\left(k+1\right)}{\left(s-a\right)^{k+1}}$

$\mathcal{L}\left\{e^{at}t^{n}\right\}=\frac{n!}{\l eft(s-a\right)^{n+1}}$

$\mathcal{L}\left\{e^{at}\cosh\!\left(kt\right)\rig ht\}=\frac{s-a}{\left(s-a\right)^2-k^2}$

$\mathcal{L}\left\{e^{at}\sinh\!\left(kt\right)\rig ht\}=\frac{k}{\left(s-a\right)^2-k^2}$

$\mathcal{L}\left\{e^{at}\cos\!\left(kt\right)\righ t\}=\frac{s-a}{\left(s-a\right)^2+k^2}$

$\mathcal{L}\left\{e^{at}\sin\!\left(kt\right)\righ t\}=\frac{k}{\left(s-a\right)^2+k^2}$

There is one more interesting Laplace Transform worth considering:

$\mathcal{L}\left\{\int_0^t f\!\left(\tau\right)\,d\tau\right\}=\frac{1}{s}\ma thcal{L}\left\{f\!\left(t\right)\right\}=\frac{F\! \left(s\right)}{s}$

With these fundamental Laplace Transforms, we can now tackle some initial value problems (some of these may require partial fraction techniques).

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Example 36

Use Laplace Transforms to solve the IVP $x^{\prime\prime}+4x^{\prime}+13x=te^{-t};\,x\!\left(0\right)=0,\,x^{\prime}\!\left(0\rig ht)=2$

First, we take the Laplace Transform of both sides:

$\mathcal{L}\left\{x^{\prime\prime}+4x^{\prime}+13x \right\}=\mathcal{L}\left\{te^{-t}\right\}\implies\mathcal{L}\left\{x^{\prime\prim e}\right\}+4\mathcal{L}\left\{x^{\prime}\right\}+1 3\mathcal{L}\left\{x\right\}=\mathcal{L}\left\{te^ {-t}\right\}$

Applying the proper formulas and translations, we have

$s^2X\!\left(s\right)-sx\!\left(0\right)-x^{\prime}\!\left(0\right)+4\left(sX\!\left(s\righ t)-x\!\left(0\right)\right)+13X\!\left(s\right)=\frac {1}{\left(s+1\right)^2}$

Now apply the initial conditions $x\!\left(0\right)=0$ and $x^{\prime}\!\left(0\right)=2$ to get

$\left(s^2+4s+13\right)X\!\left(s\right)-2=\frac{1}{\left(s+1\right)^2}$ $\implies X\!\left(s\right)=\frac{1}{\left(s+1\right)^2\left (s^2+4s+13\right)}+\frac{2}{s^2+4s+13}$

Now here comes the fun part: Take the Inverse Laplace transform of both sides to find the solution $x\!\left(t\right)$.

Lets consider each fraction individually.

First, consider $\mathcal{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2\left(s^2+4s+1 3\right)}\right\}$.

To help us find the Inverse Laplace Transform, we need to apply partial fractions (I will redo this problem in the next post, when I talk about convolution):

$\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)} =\frac{A}{s+1}+\frac{B}{\left(s+1\right)^2}+\frac{ Cs+D}{s^2+4s+13}$.

Our objective now is to find A, B, C, and D.

First, multiply both sides by the common denominator to get

$1=A\left(s+1\right)\left(s^2+4s+13\right)+B\left(s ^2+4s+13\right)+\left(Cs+D\right)\left(s+1\right)^ 2$

If we take $s=-1$, we have

$1=B\left(1-4+13\right)\implies B=\frac{1}{10}$.

If we take $s=0$, we have

$1=13A+\frac{13}{10}+D\implies -\frac{3}{10}-13A=D$

If we take $s=-2$, we have

$1=-9A+\frac{9}{10}-2C+D\implies \frac{1}{10}=-9A-2C+D\implies -\frac{2}{10}-11A=C$

If we take $s=2$, we have

$1=75A+\frac{25}{10}+18C+9D$ $\implies 1=75A+\frac{25}{10}+18\left(-\frac{2}{10}-11A\right)+9\left(-\frac{3}{10}-13A\right)$

This simplifies to $1=75A+\frac{25}{10}-198A-\frac{36}{10}-\frac{27}{10}-117A\implies\frac{48}{10}=-240A\implies A=-\frac{1}{50}$

Thus, $C=\frac{11}{50}-\frac{10}{50}=\frac{1}{50}$ and $D=-\frac{15}{50}+\frac{13}{50}=-\frac{2}{50}$

Thus, $\frac{1}{\left(s+1\right)^2\left(s^2+4s+13\right)} =\frac{-1}{50\left(s+1\right)}+\frac{1}{10\left(s+1\right) ^2}+\frac{s-2}{50\left(s^2+4s+13\right)}$

Therefore,

$\mathcal{L}^{-1}\left\{\frac{-1}{50\left(s+1\right)}+\frac{1}{10\left(s+1\right) ^2}+\frac{s-2}{50\left(s^2+4s+13\right)}\right\}$ $=-\frac{1}{50}\mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\}+\frac{1}{10}\mathca l{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2}\right\}+\fra c{1}{50}\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+4s+13}\right\}$

Note that $\mathcal{L}^{-1}\left\{\frac{s-2}{s^2+4s+13}\right\}=\mathcal{L}^{-1}\left\{\frac{s+2}{\left(s+2\right)^2+9}-\frac{3}{(s+2)^2+9}-\frac{3}{3\left[\left(s+2\right)^2+9\right]}\right\}$ $=e^{-2t}\cos\!\left(3t\right)-\frac{4}{3}e^{-2t}\sin\!\left(3t\right)$

Therefore, we finally have

$\mathcal{L}^{-1}\left\{\frac{1}{\left(s+1\right)^2\left(s^2+4s+1 3\right)}\right\}=-\frac{1}{50}e^{-t}+\frac{1}{10}te^{-t}+\frac{1}{50}\cos\!\left(3t\right)-\frac{4}{150}\sin\!\left(3t\right)$

Now, we need the second half of the solution! (We have only part of it!) XD

We now consider the other Inverse Laplace Transform:

$\mathcal{L}^{-1}\left\{\frac{2}{s^2+4s+13}\right\}$

We see that $\mathcal{L}^{-1}\left\{\frac{2}{s^2+4s+13}\right\}=\mathcal{L}^{-1}\left\{\frac{3}{\left(s+2\right)^2+9}\right\}-\frac{1}{3}\mathcal{L}^{-1}\left\{\frac{3}{\left(s+2\right)^2+9}\right\}$ $=\frac{2}{3}e^{-2t}\sin\!\left(3t\right)$

Therefore, we now see that

$x\!\left(t\right)=-\frac{1}{50}e^{-t}+\frac{1}{10}te^{-t}+\frac{1}{50}e^{-2t}\cos\!\left(3t\right)-\frac{4}{150}e^{-2t}\sin\!\left(3t\right)+\frac{2}{3}e^{-2t}\sin\!\left(3t\right)$ $=\color{red}\boxed{\frac{1}{50}\left[\left(5t-1\right)e^{-t}+e^{-2t}\left[\cos\!\left(3t\right)+32\sin\!\left(3t\right)\righ t]\right]}$

Example 37

Use Laplace Transforms to solve the IVP $x^{\prime\prime}+3x^{\prime}+2x=t;\,x\!\left(0\rig ht)=0,\,x^{\prime}\!\left(0\right)=2$

First apply the Laplace Transform on both sides to get

$s^2X\!\left(s\right)-sx\!\left(0\right)-x^{\prime}\!\left(0\right)+3sX\!\left(s\right)-3x\!\left(0\right)+2X\!\left(s\right)=\frac{1}{s^2 }$

Applying the initial conditions $x\!\left(0\right)=0$ and $x^{\prime}\!\left(0\right)=2$, we have

$\left(s^2+3s+2\right)X\!\left(s\right)-2=\frac{1}{s^2}\implies X\!\left(s\right)=\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}+\frac{2}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}$

This is where the Laplace Transform of an Integral comes into play nicely (to avoid partial fractions)

In finding $\mathcal{L}^{-1}\left\{\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}$, we see that

$\mathcal{L}^{-1}\left\{\frac{1}{s\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}=\int_0^t\mathcal{L}^{-1}\left\{\frac{1}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}\right\}\,d\tau$ $=2\int_0^te^{-3/2 \tau}\sinh\!\left(\tfrac{1}{2}\tau\right)\,d\tau=\ int_0^t e^{-\tau}-e^{-2\tau}\,d\tau=-e^{-t}+\tfrac{1}{2}e^{-2t}+\tfrac{1}{2}$

Therefore,

$\mathcal{L}^{-1}\left\{\frac{1}{s^2\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}=\int_0^t\mathcal{L}^{-1}\left\{\frac{1}{s\left[\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}\right]}\right\}\,d\tau$ $=\int_0^t -e^{-\tau}+\tfrac{1}{2}e^{-2\tau}+\tfrac{1}{2}\,d\tau=e^{-t}-\tfrac{1}{4}e^{-2t}+\tfrac{1}{2}t-\tfrac{3}{4}$

Now, $\mathcal{L}^{-1}\left\{\frac{2}{\left(s+\tfrac{3}{2}\right)^2-\tfrac{1}{4}}\right\}=4e^{-3/2t}\sinh\!\left(\tfrac{1}{2}t\right)=2e^{-t}-2e^{-2t}$.

Therefore, $\color{red}\boxed{x\!\left(t\right)=\tfrac{1}{4}\l eft[2t-3+12e^{-t}-9e^{-2t}\right]}$

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This will conclude the first post on Laplace Transforms. I'm not sure when I will be able to post again, now that I start classes today. I'll try to find some time in the next several weeks to do so.

2. What does the future hold for the Differential Equations Tutorial?

To the MHF Community,

It's been over a year since I have updated this differential equations tutorial of mine.

Some of you have PMed me about errors/typos found in this tutorial, why certain topics aren't covered, etc. But I just haven't had the time lately to fix them or add missing topics since most of my time is devoted to my coursework (since I'm a full time Ph.D. student now). However, this summer (starting in June), I will be redoing this entire differential equations tutorial and re-release it as a mini-book (it may not be mini when I'm through with it... ) of some kind; meaning, it will have the theory explained in a more, precise detail. It will also have examples illustrating various techniques outlined in each section as well as a few exercises for you to attempt. At first I will not include solutions to these exercises, but in the months after releasing this to the public on MHF I will include solutions to selected exercises outlined in each section/chapter (the format is still yet to be decided).

The chapter outline will be something like:

Chapter 0 - Calculus Review

Part 1 - Ordinary Differential Equations (to be released end of summer 2011)

Chapter 1 - First Order Differential Equations
Chapter 2 - Second and Higher Order Differential Equations
Chapter 3 - Solving Differential Equations by Numerical Methods
Chapter 4 - Matrix Methods and Systems of Differential Equations
Chapter 5 - Laplace Transforms
Chapter 6 - Power Series Solutions to Differential Equations

Part 2 - Partial Differential Equations (to be released sometime in 2012)

Chapter 7 - Introduction to Partial Differential Equations
Chapter 8 - Fourier Series
Chapter 9 - The Wave, Heat, and Laplace Equations
Chapter 10 - Solving Partial Differential Equations by Various Methods
Chapter 11 - Green's Functions
Chapter 12 - Solving Partial Differential Equations by Numerical Methods

If you would like to contribute to this project, send me a PM and we can discuss it in more detail. If I use any of your work in my book, you will be acknowledged.

I will close this thread in the meantime, and I will post updates when I'm able.