In Mathematics, the Laplace transform is a powerful tool that can be used to solve differential equations. Differential equations are mathematical models that describe complex systems. Read here how to use the TI89 Calculator when solving Differential Equations via LaPlace Transforms.
In this blog post, we’ll take a look at the definition, formula, and solved examples of the Laplace transform.
What is the Laplace transform?
The Laplace transform is a branch of mathematics that allows for the solving of differential equations. A Laplace transform is perhaps the second most useful integral transform in solving physical problems after the Fourier transform.
Specifically, the Laplace transform is useful for solving linear ordinary differential equations, such as those associated with electronic circuit analysis.
The Laplace transform can be used to solve problems in physics and engineering. For example, the Laplace transform can be used to solve problems related to motion, heat transfer, and sound propagation. Additionally, the Laplace transform is also used in statistics and economics.
Formula of Laplace transform
The general formula used to calculate the Laplace transform of the function is:
F(s) = L{f(t)} = ∫0∞ e-st f(t) dt
Where,
- f(t) is the given function.
- ∫ Integral notation
- L is the notation of Laplace.
- t is the independent variable of the function.
- F(s) is the result of Laplace transform
All the calculations of the Laplace transform can also be done with the help of the table of the Laplace transform. Here is the basic table of Laplace transform in which all types of inputs are available, you just have to place the value of an independent variable.
Table of Laplace Transform
Here is the table:
| Functions | Laplace transform |
| Laplace of 1 | 1/s |
| Laplace of eat | 1/ (s – a) |
| Laplace of tn, where n = 1, 2, 3, …, n | n / sn+1 |
| Laplace of tp, p > -1 | [Γ(p+1)] / sp+1 |
| Laplace of √t | √π / 2s3/2 |
| Laplace of tn – 1/2 | [1⋅3⋅5⋯(2n−1) √π] / [2n sn + 1/2] |
| Laplace of sin(at) | a / (s2 + a2) |
| Laplace of cos(at) | S / (s2 + a2) |
| Laplace of tsin(at) | 2as / / (s2 + a2)2 |
| Laplace of tcos(at) | (s2 – a2) / (s2 + a2)2 |
| Laplace of sin(at) – at cos(at) | 2a3 / (s2 + a2)2 |
| Laplace of sin(at) + at cos(at) | 2as2 / (s2 + a2)2 |
| Laplace of cos(at) – at sin(at) | [s(s2 – a2)] / (s2 + a2)2 |
| Laplace of cos(at) + at sin(at) | [s(s2 + 3a2)] / (s2 + a2)2 |
| Laplace of sin(at + b) | [a*sin(b)+a*cos(b)] / (s2 + a2) |
| Laplace of cos(at + b) | [a*cos(b) − a*sin(b)] / (s2 + a2) |
| Laplace of sinh(at) | a / (s2 – a2) |
| Laplace of cosh(at) | s / (s2 – a2) |
| Laplace of eat sin(bt) | b / [(s – a)2 + b2] |
| Laplace of eat cos(bt) | (s – a) / [(s – a)2 + b2] |
| Laplace of eat sinh(bt) | b/[(s – a)2 – b2] |
| Laplace of eat cosh(bt) | (s – a) / [(s – a)2 – b2] |
| Laplace of eat tn | n! / (s – a)n+1 |
| Laplace of f(ct) | 1/s F(s/c) |
| Laplace of uc(t) = u(c – t) | e-cs/s |
| Laplace of ect f(t) | F(s – c) |
| Laplace of tn f(t) | (-1)n Fn(s) |
| Laplace of 1/t f(t) | ∫s∞F(u) du |
| Laplace of f’(t) | sF(s) – f(0) |
| Laplace of f’’(t) | s2 F(s) – sf(0) – f’(0) |
Any kind of input can be transformed into Laplace according to the above table easily. You can also try a Laplace transform calculator to get the result within seconds according to the table of Laplace.
How to calculate Laplace Transforms?
You can easily calculate the problems of the Laplace transform with the help of its formula or Laplace table. Let us take a few examples to calculate the Laplace transform.
By using the Laplace formula
Example
Transform the given function f(t) = 2t
Solution
Step 1: First of all, write the given function and apply the formula of Laplace.
Function = 2t
∫0∞ e-st f(t) dt = ∫0∞ e-st (2t) dt
Step 2: Now calculate the integral of the above expression.
∫0∞ e-st (2t) dt = 2∫0∞ te-st dt
Now integrate by parts ∫f * g’ = f*g – ∫f’ g … (1)
f = t and g’ = e-st
f’ = 1 and g = – e-st/s
Step 3: Now place the values in (1).
2∫0∞ (t)e-st dt = 2[-te-st/s – ∫- e-st/s dt] …. (2)
Step 4: Now integrate an integral part of the above expression.
∫- e-st/s dt
Let u = -st
Then du/dt = -s
dt = -1/s du
∫- e-st/s dt = ∫- eu/s (-du/s)
∫- e-st/s dt = ∫ eu/s2 du
∫- e-st/s dt = 1/s2 ∫ eu du
∫- e-st/s dt = 1/s2 (eu) = eu/s2
Substitute u = -st
∫- e-st/s dt = 1/s2 (eu) = e-st/s2
Step 5: Now put the result of the integral in (2).
2∫0∞ (t)e-st dt = 2[-te-st/s – e-st/s2]
2∫0∞ (t)e-st dt = -2te-st/s – 2e-st/s2
2∫0∞ (t)e-st dt = [2ste-st – 2e-st]/s2
2∫0∞ (t)e-st dt = 2e-st[st – 1]/s2
Hence, the Laplace of the given function is 2e-st[st – 1]/s2
An integral calculator to avoid such larger steps to calculate the integral of functions.
By using the Laplace table
Example 1
Transform the given function f(t) = e2t + cos(t)
Solution
Step 1: First of all, separate the function and write them.
Function = e2t + cos(t)
f(t) = e2t
g(t) = cos(t)
Step 2: Now use to the notation of Laplace to the given function.
L [f(t) + g(t)] = L [e2t + cos(t)]
According to the linearity property of Laplace, we can write the given function as:
L [e2t + cos(t)] = L [e2t] + L [cos(t)] … (1)
Step 3: Now use the Laplace table to write the Laplace transform of the given function.
L [e2t] = 1/ (s – 2) : because eat = 1/ s-a
L [cos(t)] = s / (s2 + 12) = s/ (s2 + 1)
Step 4: Substitute the values to the equation (1).
L [e2t + cos(t)] = L [e2t] + L [cos(t)]
L [e2t + cos(t)] = 1/ (s – 2) + s/ (s2 + 1)
Example 2
Transform the given function f(t) = 1 + t3 – sin(t)
Solution
Step 1: First of all, separate the function and write them.
Function = 1 + t3 – sin(t)
f(t) = 1
g(t) = t3
h(t) = sin(t)
Step 2: Now use the notation of Laplace to the given function.
L [f(t) + g(t) – h(t)] = L [1 + t3 – sin(t)]
According to the linearity property of Laplace, we can write the given function as:
L [1 + t3 – sin(t)] = L [1] + L [t3] – L [sin(t)] … (1)
Step 3: Now use the Laplace table to write the Laplace transform of the given function.
L [1] = 1/ s
L [t3] = 3 / (s3+1) : because tn = n/sn+1
L [t3] = 3/s4
L [sin(t)] = 1/ (s2 + 12) = 1/ (s2 + 1)
Step 4: Substitute the values to the equation (1).
L [1 + t3 – sin(t)] = L [1] + L [t3] – L [sin(t)]
L [1 + t3 – sin(t)] = 1/s + 3/s4 – 1 / (s2 + 1)
Example 3
Transform the given function f(t) = t + t4 – e-3t
Solution
Step 1: First of all, separate the function and write them.
Function = t + t4 – e-3t
f(t) = t
g(t) = t4
h(t) = e-3t
Step 2: Now use the notation of Laplace to the given function.
L [f(t) + g(t) – h(t)] = L [t + t4 – e-3t]
According to the linearity property of Laplace, we can write the given function as:
L [t + t4 – e-3t] = L [t] + L [t4] – L [e-3t] … (1)
Step 3: Now use the Laplace table to write the Laplace transform of the given function.
L [t] = 1/ s1+1 = 1/s2
L [t4] = 4 / (s4+1)
L [t4] = 4/s5
L [e-3t] = 1 / (s – (-3))
L [e-3t] = 1 / (s + 3)
Step 4: Substitute the values to the equation (1).
L [t + t4 – e-3t] = L [t] + L [t4] – L [e-3t]
L [t + t4 – e-3t] = 1/s2 + 4/s5 – 1 / (s + 3)
Wrap up
In this post, we have covered almost all the basics of the Laplace transform. We have discussed the basic definition, formula, and Laplace table along with solved examples. Now you can grab all the basics of the Laplace from this post.
