DE-BOOK Common Transforms

Section Common Transforms

Now that we’ve defined the Laplace transform, the next step is to build a basic toolkit of common transforms. These are formulas for frequently encountered functions like constants, exponentials, powers of \(t\text{,}\) and trigonometric functions. Learning these transforms will allow you to compute Laplace transforms quickly, without evaluating integrals from scratch each time.

🔗

We’ll derive these formulas using straightforward examples and then generalize the results. You’ll see how the Laplace transform simplifies when applied to each type of function and how certain patterns repeat across different cases.

🔗

Checkpoint 188. Exponential Functions Pre-Reading Questions.

(a) Exponential Integration Rules.

Match each exponential integrand with its antiderivative. Assume \(K\) is a constant.

🔗

\(\ds \int e^{5t} dt\)
\(\ds \frac{1}{5} e^{5t} + C\)
\(\ds \int e^{-3t} dt\)
\(\ds \frac{-1}{3} e^{-3t} + C\)
\(\ds \int e^{Kt} dt\)
\(\ds \frac{1}{K} e^{Kt} + C\)
\(\ds \int e^{(7-s)t} dt\)
\(\ds \frac{1}{7 - s} e^{(7-s)t} + C\)

🔗

(b) How is \(s\) Treated During Integration?

In the Laplace transform integral, the variable \(s\) is treated as a during the integration process.

🔗

constant
Correct! When evaluating the integral, \(s\) is treated as a constant, since the integration is with respect to \(t\text{.}\)
variable
No, even though \(s\) is a variable overall, it is treated as a constant during the integration.
coefficient
No, while \(s\) acts like a coefficient in \(e^{-st}\text{,}\) it’s conceptually treated as a constant in the integration.
limit
No, \(s\) is not a limit of integration, it appears inside the integrand.

🔗

(c) Choosing \(u\) and \(dv\) in \(\int t^2 e^{-st} dt\).

Which functions should you choose as \(u\) and \(dv\) when applying integration by parts to \(\int t^2 e^{-st} dt\text{?}\)

🔗

\(u = t^2,\quad dv = e^{-st} dt\)
Correct! This choice simplifies with each integration by parts.
\(u = e^{-st},\quad dv = t^2 dt\)
No, this makes integration more difficult. We want to differentiate \(t^2\) and integrate the exponential.
\(u = st,\quad dv = t dt\)
These choices are not aligned with the integrand \(t^2 e^{-st}\text{.}\)

🔗

Subsection Laplace Transform of a Constant

We begin with the simplest case: the constant function \(f(t) = 1\text{.}\) By the definition of the Laplace transform,

\begin{equation*} \lap{1} = \int_0^\infty e^{-st}\, dt. \end{equation*}

🔗

This integral converges when \(s \lt 0\) and leads to our first Laplace transform rule:

🔗

📜 Laplace Transform of \(1\).

\begin{gather*} \text{L}_1:\quad\lap{1} = \frac{1}{s}, \quad s \gt 0 \end{gather*}

💡 Derivation of L\(_1\).

By definition, we have

\begin{equation*} \lap{1} = \int_0^{\infty} e^{-st} \cdot 1\ dt = \lim_{b \to \infty} \ub{\int_0^b e^{-st}dt}_{I}\text{.} \end{equation*}

🔗

Assuming \(s\) and \(b\) are constant, we integrate \(I\) with respect to \(t\text{:}\)

\begin{equation*} I = \int_0^b e^{-st}dt = -\frac{1}{s}e^{-st}\Big|_{t=0}^{t=b} = -\frac{1}{s}\left[ e^{-sb} - 1 \right]\text{.} \end{equation*}

🔗

Next, we compute the limit of \(I\) as \(b \to \infty\) (with \(s\) constant):

\begin{align*} \lim_{b \to \infty} I \amp = \lim_{b \to \infty} -\frac{1}{s}\left[ e^{-sb} - 1 \right]\\ \amp = -\frac{1}{s} \lim_{b \to \infty} \Big[ e^{-sb} - 1 \Big] = -\frac{1}{s} \Big[ \ub{\lim_{b \to \infty} e^{-sb}}_{L} - 1 \Big] \end{align*}

🔗

Taking \(b\to\infty\text{,}\) \(L\) must go to \(0\) if \(s\) is positive, leaving us with the transform of \(1\text{:}\)

\begin{equation*} \lap{1} = -\frac{1}{s} [0 - 1] = \frac{1}{s} \quad \text{for } s \gt 0\text{.} \end{equation*}

🔗

Although this transform is specific to the contant function \(1\text{,}\) we will see later that it will serve as the transform formula for all constant functions.

🔗

Subsection Laplace Transform of Exponentials

To get the transform formula for exponentials of the form \(f(t) = e^{at}\text{,}\) where \(a\) is a constant, we will first find it for \(e^{7t}\text{:}\)

🔗

\begin{equation*} \lap{e^{7t}} = \frac{1}{s - 7}, \quad s \gt 7 \end{equation*}

Solution.

Using the definition of the Laplace transform we get

\begin{equation*} \lap{ e^{7t} } = \int_0^{\infty} e^{-st} \cdot e^{7t}\ dt = \lim_{b \to \infty}\int_0^b e^{(7-s)t}\ dt\text{.} \end{equation*}

🔗

🔍 Focusing just on the integral, we let \(u=7-s\) and apply \(u\)-substitution:.

👉 Using \(u\)-substitution with

\begin{align*} u \amp = \ob{(7-s)}^{\text{constant}}t\\ du \amp = (7-s)dt \quad \Rightarrow \quad dt = \frac{du}{7-s} \end{align*}

the integral is compute as

\begin{align*} \int_0^b e^{(7-s)t}\ dt = \int_{t=0}^{t=b} e^{u}\frac{du}{7-s} \amp = \frac{1}{7-s}\int_{t=0}^{t=b} e^{u}\ du\\ \amp = \frac{1}{7-s} \left[ e^{u} - 1 \right]\Bigg|_{t=0}^{t=b}\\ \amp = \frac{1}{7-s} \left[ e^{7-s} - 1 \right]\Bigg|_{0}^{b}\text{.} \end{align*}

🔗

\begin{align*} \int_0^b e^{(7-s)t}\, dt = \frac{1}{7-s} e^{(7-s)t} \Bigg|_0^b \amp = \frac{1}{7-s} \left( e^{(7-s)b} - e^{0} \right)\\ \amp = \frac{1}{7-s} \left( e^{(7-s)b} - 1 \right) \end{align*}

🔗

Now, we compute the limit of this expression as \(b \to \infty\text{:}\)

\begin{align*} \lim_{b \to \infty} \frac{1}{7-s} \left( e^{(7-s)b} - 1 \right) \amp = \frac{1}{7-s} \Bigg[ \ub{\lim_{b \to \infty} e^{(7-s)b}}_{L} - \lim_{b \to \infty} 1 \Bigg]\text{.} \end{align*}

🔗

As before, \(L\) goes to \(0\) if the power of the exponential function goes to \(-\infty\text{.}\) This only happens if \(7-s\) is negative. That is, \(\ 7-s \lt 0\ \) or \(\ 7 \lt s\text{.}\)

🔗

So, as long as \(s \gt 7\text{,}\) then \(L \to 0\) and the transform becomes

\begin{equation*} \lap{e^{7t}} = \frac{1}{s - 7}, \quad s \gt 7\text{.} \end{equation*}

🔗

Replacing \(7\) with an \(a\) in this solution leads to the general formula:

🔗

For any constant \(a\text{,}\) the Laplace transform of \(e^{at}\) is given by:

🔗

📜 Laplace Transform of \(e^{at}\).

\begin{gather*} \text{L}_2:\quad\lap{e^{at}} = \frac{1}{s - a}, \quad s \gt a \end{gather*}

💡Derivation of L\(_2\).

By definition,

\begin{equation*} \lap{ e^{at} } = \int_0^{\infty} e^{-st} \cdot e^{at}\ dt = \lim_{b \to \infty}\int_0^b e^{(a-s)t}\ dt\text{.} \end{equation*}

🔗

For the integral, let \(u=a-s\) and apply \(u\)-substitution:

\begin{align*} \int_0^b e^{(a-s)t}\, dt \knowl{./knowl/xref/lt-example-2-details.html}{\text{\(\os{\large ❔}{=}\)}} \frac{1}{a-s} e^{(a-s)t} \Bigg|_0^b \amp = \frac{1}{a-s} \left( e^{(a-s)b} - e^{0} \right)\\ \amp = \frac{1}{a-s} \left( e^{(a-s)b} - 1 \right) \end{align*}

🔗

Now, we compute the limit of this expression as \(b \to \infty\text{:}\)

\begin{align*} \lim_{b \to \infty} \frac{1}{a-s} \left( e^{(a-s)b} - 1 \right) \amp = \frac{1}{a-s} \Bigg[ \ub{\lim_{b \to \infty} e^{(a-s)b}}_{L} - \lim_{b \to \infty} 1 \Bigg]\text{.} \end{align*}

🔗

As before, \(L\) goes to \(0\) if \(a-s\) is negative. That is, \(\ a-s \lt 0\ \) or \(\ a \lt s\text{.}\)

🔗

So, as long as \(s \gt a\text{,}\) then \(L \to 0\) and the transform becomes

\begin{equation*} \lap{e^{at}} = \frac{1}{s - a}, \quad s \gt a\text{.} \end{equation*}

🔗

Checkpoint 189. 📖❓ Give the Transform.

\begin{equation*} \lap{e^{-3t}} = \fillinmath{XX} \end{equation*}

\(\ds\frac{1}{s+3}\)
Correct! The Laplace transform of \(e^{-3t}\) is \(\ds\frac{1}{s+3}\text{.}\)
\(\ds\frac{1}{s-3}\)
No, this would be the transform of \(e^{3t}\text{.}\) Double-check the sign of the exponent.
\(\ds\frac{1}{s-3t}\)
No, \(t\) should not appear in the final expression, only \(s\text{.}\)
\(\ds\frac{3}{s+3}\)
No, the numerator should be 1, not 3. This is not a scaled exponential.

🔗

Subsection Laplace Transform of Powers of \(t\)

Now let’s consider power functions, \(f(t) = t^n\) where \(n\) is a non-negative integer.

\begin{equation*} \lap{t^n} = \frac{n!}{s^{n+1}}, \quad s > 0. \end{equation*}

🔗

We already found the transforms for \(n = 0\) and \(n = 1\) since

\(\displaystyle n=0:\qquad\lap{t^0} = \lap{1} = \dfrac{1}{s}\)

🔗
\(\displaystyle n=1:\qquad\lap{t^1} = \lap{t} = \dfrac{1}{s^2}\)

🔗

We could find \(\lap{t^2}\) next, but it is more instructive to show following relationship:

🔗

\begin{equation*} \lap{t^2} = \frac{2}{s}\cdot \lap{t}, \quad s \gt 0 \end{equation*}

Solution.

Applying the definition and writing the integral as a limit, we have

\begin{equation*} \lap{ t^2 } = \lim_{b \to \infty} \ub{\int_0^b e^{-st} \cdot t^2\ dt}_{I}\text{.} \end{equation*}

🔗

Applying integration by parts to \(I\) shows:.

\begin{align*} u = t^2, \quad\amp dv = e^{-st}dt, \\ du = 2t\ dt, \quad\amp v = -\frac{1}{s}e^{-st} \end{align*}

integration by parts gives

\begin{align*} \int_0^b e^{-st} \cdot t^2\ dt \amp = t^2 \cdot \left( -\frac{1}{s}e^{-st} \right)\Bigg|_0^b - \int_0^b \left( -\frac{1}{s}e^{-st} \right) 2t\ dt\\ \amp = -\frac{b^2}{s}e^{-sb} + \frac{2}{s} \int_0^b e^{-st} \cdot t\ dt \end{align*}

🔗

\begin{align*} \lap{ t^2 } \amp = \lim_{b \to \infty} \left[-\frac{b^2}{s}e^{-sb} + \frac{2}{s}\int_0^b e^{-st}\cdot t\ dt\right]\\ \amp = -\frac{1}{s} \lim_{b \to \infty} \left[-\frac{b^2}{e^{sb}}\right] + \frac{2}{s}\lim_{b \to \infty}\int_0^b e^{-st}\cdot t\ dt\\ \amp = -\frac{1}{s}\ub{\lim_{b \to \infty}\frac{b^2}{e^{sb}}}_{L} + \frac{2}{s}\ub{\int_0^\infty e^{-st}\cdot t\ dt}_{\large\lap{t}} \end{align*}

🔗

Noticing that the improper integral is just \(\lap{t}\text{,}\) we have

\begin{equation*} \lap{ t^2 } = -\frac{1}{s}\cdot L + \frac{2}{s}\cdot \lap{t}\text{,} \end{equation*}

but \(L=0\) if \(s \gt 0\text{,}\) so we have the relationship.

Again, if \(s \lt 0\text{,}\) the limit goes to \(+\infty\text{.}\) So we must have \(s \gt 0\text{.}\) Applying L’Hôpital’s rule twice, we get

\begin{align*} L = \lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b^2}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}\ \amp\us{LH}{=}\ \lim_{b \to \infty}\frac{2b}{se^{sb}}\ = \frac{2}{s}\lim_{b \to \infty}\frac{\os{\infty}{\os{\uparrow}{\boxed{b}}}}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}}\\ \amp\us{LH}{=}\ \frac{2}{s}\lim_{b \to \infty}\frac{1}{se^{sb}} = \frac{2}{s^2}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0 \end{align*}

🔗

\begin{equation*} \lap{t^2} = \frac{2}{s}\cdot \lap{t}, \quad s \gt 0\text{.} \end{equation*}

🔗

You can use this work as a template for a general recursive relationship:

🔗

\begin{equation*} \lap{t^n} = \frac{n}{s}\cdot \lap{t^{n-1}}, \quad s \gt 0 \end{equation*}

Solution.

Replacing \(2\) with \(n\) in the previous solution, we have

\begin{equation*} \lap{ t^n } = \lim_{b \to \infty} \ub{\int_0^b e^{-st} \cdot t^n\ dt}_{I}\text{.} \end{equation*}

🔗

🔍 Apply integration by parts to \(I\).

\begin{align*} u = t^n, \quad\amp dv = e^{-st}dt, \\ du = nt^{n-1}\ dt, \quad\amp v = -\frac{1}{s}e^{-st} \end{align*}

integration by parts gives

\begin{align*} \lap{ t^{n} } \amp = \lim_{b \to \infty} \left[-\frac{t^{n}}{s}e^{-sb}\Bigg|_0^b - \int_0^b \left(-\frac{1}{s}e^{-st}\right)\cdot nt^{n-1}\ dt\right]\\ \amp = \lim_{b \to \infty} \left[\left(-\frac{b^n}{s}e^{-sb} - 0\right) + \frac{n}{s}\int_0^b e^{-st}\cdot t^{n-1}\ dt\right]\\ \amp = \lim_{b \to \infty} \left[-\frac{b^n}{e^{sb}} + \frac{n}{s}\int_0^b e^{-st}\cdot t^{n-1}\ dt\right] \end{align*}

🔗

\begin{align*} \lap{ t^{n} } \amp = \lim_{b \to \infty} \left[-\frac{b^n}{e^{sb}} + \frac{n}{s}\int_0^b e^{-st}\cdot t^{n-1}\ dt\right]\\ \amp = -\frac{1}{s} \lim_{b \to \infty} \left[-\frac{b^n}{e^{sb}}\right] + \frac{n}{s}\lim_{b \to \infty}\int_0^b e^{-st}\cdot t^{n-1}\ dt\\ \amp = -\frac{1}{s}\ub{\lim_{b \to \infty}\frac{b^n}{e^{sb}}}_{L} + \frac{n}{s}\ub{\int_0^\infty e^{-st}\cdot t^{n-1}\ dt}_{\large\lap{t^{n-1}}} \end{align*}

🔗

Noticing that the improper integral is just \(\lap{t^{n-1}}\text{,}\) we have

\begin{equation*} \lap{ t^{n} } = -\frac{1}{s}\cdot L + \frac{n}{s}\cdot \lap{t^{n-1}}\text{.} \end{equation*}

🔗

🔍 Since \(L=0\) if \(s \gt 0\text{,}\) we have.

The reasoning is the same as in previous solution except we apply L’Hôpital’s rule n times. So we have

\begin{align*} \lim_{b \to \infty}\frac{b^n}{e^{sb}}\ \amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}} \lim_{b \to \infty}\frac{nb^{n-1}}{se^{sb}} = \frac{n}{s}\lim_{b \to \infty}\frac{b^{n-1}}{e^{sb}}\\ \amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}} \frac{n}{s}\lim_{b \to \infty}\frac{(n-1)b^{n-2}}{se^{sb}} = \frac{n(n-1)}{s^2}\lim_{b \to \infty}\frac{b^{n-2}}{e^{sb}}\\ \amp \qquad\qquad \vdots \quad LH\ \ n-2 \text{ more times}\\ \amp\os{\large\left(\frac{\infty}{\infty}\right)}{\us{LH}{=}} \frac{n!}{s^n}\lim_{b \to \infty}\frac{1}{se^{sb}} = \frac{n!}{s^{n+1}}\lim_{b \to \infty}\frac{1}{\us{\infty}{\us{\downarrow}{\boxed{e^{sb}}}}} = 0 \end{align*}

🔗

\begin{equation*} \lap{t^{n}} = \frac{n}{s}\cdot \lap{t^{n-1}}, \quad s \gt 0\text{.} \end{equation*}

🔗

To get the general formula we apply this relationship repeatedly for increasing \(n\text{:}\)

\begin{equation*} \begin{array}{crcl} n=1: \amp \DLO\lap{t} \amp = \amp \dfrac{1}{s}\cdot\lap{1} = \dfrac{1}{s}\cdot\dfrac{1}{s} = {\DLO\dfrac{1}{s^2}}\\ n=2: \amp \DLBa\lap{t^2} \amp = \amp \dfrac{2}{s}\cdot{\DLO\lap{t}} = \dfrac{2}{s}\cdot{\DLO\dfrac{1}{s^2}} = {\DLBa\dfrac{2\cdot 1}{s^3}}\\ n=3: \amp \DLGb\lap{t^3} \amp = \amp \dfrac{3}{s}\cdot{\DLBa\lap{t^2}} = \dfrac{3}{s}\cdot{\DLBa\dfrac{2\cdot 1}{s^3}} = {\DLGb\dfrac{3\cdot 2\cdot 1}{s^4}}\\ n=4: \amp \DLRa\lap{t^4} \amp = \amp \dfrac{4}{s}\cdot{\DLGb\lap{t^3}} = \dfrac{4}{s}\cdot{\DLGb\dfrac{3\cdot 2\cdot 1}{s^4}} = {\DLRa\dfrac{4\cdot 3\cdot 2\cdot 1}{s^5}}\\ \vdots \amp \vdots\quad\amp \amp \qquad\qquad\vdots\\ \end{array} \end{equation*}

🔗

This pattern suggests that, in general, we have

🔗

\begin{equation*} \lap{t^n} =\dfrac{n}{s}\cdot\lap{t^{n-1}} = \dfrac{n}{s}\cdot\dfrac{(n-1)\dotsm 4\cdot 3\cdot 2\cdot 1}{s^{n}}\text{,} \end{equation*}

which can be summarized as follows

🔗

📜 Laplace Transform of \(t^n\).

\begin{gather*} \text{L}_3:\quad\lap{t^n} = \frac{n!}{s^{n+1}}, \quad s \gt 0, \quad n = 0, 1, 2, \ldots \end{gather*}

💡Derivation of L\(_3\).

Most of the ideas were described above, but the formal derivation follows by induction on \(n\text{,}\) which we will not cover here.

🔗

Subsection Laplace Transforms of Sine and Cosine

Finally, we derive the Laplace transforms of \(\sin(bt)\) and \(\cos(bt)\text{.}\)

🔗

To make the derivation easier, we will find the transform of \(\cos(3t)\) first.

🔗

\begin{equation*} \lap{\cos(3t)} = \frac{s}{s^2 + 3^2}, \quad s \gt 0 \end{equation*}

Solution.

Applying the definition and writing the integral as a limit, we have

\begin{equation*} \lap{ \cos(3t)} = \int_0^{\infty} e^{-st} \cdot \cos(3t)\ dt = \lim_{b \to \infty} \ub{\int_0^{b} e^{-st} \cos(3t)\ dt}_{I}\text{.} \end{equation*}

🔗

Treating \(s\) constant, we apply integration by parts to \(I\) as follows:

🔗

\begin{equation*} I = -\frac{e^{-st}}{s} \cos(3t) \Bigg|_0^b - \frac{3}{s}\int_0^{b} e^{-st} \sin(3t)\ dt\text{,} \end{equation*}

📝: Integration by Parts.

but this requires a second integration by parts:

\begin{align*} I \amp = -\frac{e^{-st}}{s} \cos(3t) \Bigg|_0^b - \frac{3}{s} \left[ -\frac{e^{-st}}{s} \sin(3t)\Bigg|_0^b + \frac{3}{s} \int_0^{b} e^{-st} \cos(3t)\ dt \right]\\ \amp = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b - \frac{9}{s^2} \ub{\int_0^{b} e^{-st} \cos(3t)\ dt}_{I}\text{.} \end{align*}

🔗

Since the integral \(I\) reappeared, we actually have

\begin{equation*} I = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b - \frac{9}{s^2} I\text{.} \end{equation*}

Evaluating the limits of integration and solving for \(I\text{,}\) we find

\begin{align*} I + \frac{9}{s^2} I \amp = \left(-\frac{e^{-st}}{s} \cos(3t) + \frac{3e^{-st}}{s^2} \sin(3t)\right)\Bigg|_0^b\\ I\left(\frac{s^2+9}{s^2}\right) \amp = \left(-\frac{e^{-sb}}{s} \cos(3b) + \frac{3e^{-sb}}{s^2} \sin(3b) + \frac{1}{s} \right)\\ I \amp = \frac{s^2}{s^2+9}\left(-\frac{e^{-sb}}{s} \cos(3b) + \frac{3e^{-sb}}{s^2} \sin(3b) + \frac{1}{s} \right) \end{align*}

🔗

Finally, the limit of \(I\) as \(b \to \infty\) is the sum of the limits

\begin{align*} \lim_{b \to \infty} I = \amp\\ \frac{s^2}{s^2+9} \amp \Bigg( - \frac{1}{s}\ub{\lim_{b \to \infty} e^{-sb} \cos(3b)}_{L_1} + \frac{3}{s^2}\ub{\lim_{b \to \infty} e^{-sb} \sin(3b)}_{L_2} + \frac{1}{s} \Bigg) \end{align*}

🔗

🔍 Since \(L_1 \to 0\) & \(L_2 \to 0\) if \(s \gt 0\text{,}\) we can conclude that.

👉 If \(s > 0\text{,}\) then we know that \(e^{-sb} \to 0\) as \(b \to \infty\) and both \(\cos(3b)\) and \(\sin(3b)\) oscillate between \(-1\) and \(1\text{,}\) Thus, their product must approach zero. That is,

\begin{equation*} \lim_{b \to \infty} e^{-sb} \cos(3b) = 0 \quad \text{and} \quad \lim_{b \to \infty} e^{-sb} \sin(3b) = 0 \end{equation*}

🔗

\begin{equation*} \lap{\cos(3t)} = \frac{s}{s^2 + 9}, \quad s \gt 0\text{.} \end{equation*}

🔗

leads to the following general formula:

🔗

📜 Laplace Transform of \(t^n\).

\begin{gather*} \text{L}_4:\quad\lap{\cos(bt)} = \frac{s}{s^2 + b^2}, \quad s \gt 0 \end{gather*}

\begin{gather*} \text{L}_5:\quad\lap{\sin(bt)} = \frac{b}{s^2 + b^2}, \quad s \gt 0 \end{gather*}

💡Derivation of L\(_4\) & L\(_5\).

The derivation of L\(_4\) mirrors the solution to \(\lap{\cos(3t)}\text{.}\) Just swap \(3\) with \(b\) and \(9\) with \(b^2\text{.}\)

🔗

The derivation of L\(_5\) follows a similar approach to L\(_4\) and is left as exercise for you to try on your own.

🔗

Subsection 📤 Wrap-Up

🗝️ \(\textbf{Key Takeaways...}\)

The definition is used to derive the Laplace transforms of common functions encountered in differential equations.

🔗
\(\ds\lap{1} = \frac{1}{s}, \quad s \gt 0 \text{,}\)

🔗
\(\ds\lap{e^{at}} = \frac{1}{s - a}, \quad s \gt a \text{,}\)

🔗
\(\ds\lap{t^n} = \frac{n!}{s^{n+1}}, \quad s \gt 0, \quad n = 0, 1, 2, \ldots \text{,}\)

🔗
\(\ds\lap{\cos(bt)} = \frac{s}{s^2 + b^2}, \quad s \gt 0 \text{,}\)

🔗
\(\ds\lap{\sin(bt)} = \frac{b}{s^2 + b^2}, \quad s \gt 0 \text{.}\)

🔗

🔗

You have attempted of activities on this page.

🔗

Prev Top Next